EES Procedures and Functions for Radiation heat transfer calculations

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Absorptivity, Reflectivity and Transmissivity:

where, α = absorptivity = fraction of incident radiation absorbed

ρ = reflectivity = fraction of incident radiation reflected

τ = fraction of incident radiation transmitted

Emissive power (E): The ‘total (or hemispherical) emissive power’ is the total thermal energy radiated by a surface per unit time and per unit area, over all the wave lengths and in all directions.

Solid angle: “Solid angle’ is defined as a region of a sphere, which is enclosed by a conical surface with the vertex of the cone at the centre of the sphere. See Fig. below:

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If a plane area A intercepts the line of propagation of radiation such that the normal to the surface makes an angle θ with the line of propagation, then we project the incident area normal to the line of propagation, such that, the solid angle is now defined as:

ω A cos θ( ).

r2sr.....(13.3)

Note that, A.cos(θ) = An is the projected area of the incident surface, normal to the line of propagation.

Intensity of radiation (Ib): Intensity of radiation for a black body, Ib is defined as the energy radiated per unit time per unit solid angle per unit area of the emitting surface projected normal to the line of view of the receiver from the radiating surface.

Mathematically, this is expressed as:

I bdQ b

dA cos θ( ).( ) dω.W/(m2.sr).....(13.4)

i.e. I bdE b

cos θ( ) dω.W/(m2.sr).....(13.5)

Note that Emissive power Eb of a black body refers to unit surface area whereas Intensity Ib of a black surface refers to unit projected area.

Ibλ is the intensity of blackbody radiation for radiation of a given wave length λ. And, Ib is the summation over all the wave lengths, i.e.

I b0

∞λI bλ d W/(m2.sr).....(13.6)

Lambert’s cosine law: “A diffuse surface radiates energy such that the rate of energy radiated in a direction θ from the normal to the surface is proportional to the cosine of the angle θ”. i.e.

Q θ Q n cos θ( ).

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13.3 Laws of black body radiation:�

Planck’s Law for spectral distribution:

Radiation energy emitted by a black surface depends on the wave length, temperature of the surface and the surface characteristics.

Planck’s distribution Law relates to the spectral black body emissive power, Ebλλλλ defined as ‘the amount of radiation energy emitted by a black body at an absolute temperature T per unit time, per unit surface area, per unit wave length about the wave length λ’. Units of Ebλ are: W/(m2.µm). The first subscript ‘b’ indicates black body and the second subscript ‘λ’ stands for given wavelength, or monochromatic. Planck derived his equation for Ebλ in 1901 in conjunction with his ‘quantum theory’.

Planck’s distribution law is expressed as:

E bλ λ( )C 1 λ 5.

expC 2

λ T.1

W/(m2.µm).....(13.7)

where, C 1 3.742108. W.µm4/m2

and, C 2 1.4387104. µm.K

Plots of Ebλ vs. λ for a few different temperatures are shown in Fig. below:

Wein’s displacement law:

It is clear from the Fig. that the spectral distribution of emissive power of a black body at a given absolute temperature goes through a maximum. To find out the value of λmax, the wave length at which this maximum occurs, differentiate Planck’s equation w.r.t. λ and equate to zero. Finally, we get:

Wein’s displacement law is stated as: “product of absolute temperature and wave length at which emissive power of a black body is a maximum, is constant”.

Value of maximum monochromatic emissive power of a black body at a given temperature is obtained by substituting this value of λmaxT (= 2898 µm.K) in Planck’s equation. And, we get:.

Stefan-Boltzmann Law:

σ is known as ‘Stefan-Boltzmann constant’.

Net radiant energy exchange between two black bodies at temperatures T1 and T2 is given by:

Q net σ T 1

4 T 24. W/m2....(13.14)

Radiation from a wave band:

Often, it is required to know the amount of radiation emitted in a given wave band, i.e. in a wave length

interval between λ1 and λ2. This is expressed as a fraction of the total emissive power and is written as

Fλ1-λ2. Then, we can write:

F λ 1_λ 2λ 1

λ 2λE bλ d

0

∞λE bλ d

1

σ T4. λ 1

λ 2λE bλ d.

i.e. F λ 1_λ 21

σ T4. 0

λ 2λE bλ d

0

λ 1λE bλ d.

i.e. F λ 1_λ 2 F 0_λ 2 F 0_λ 1 ......(13.15)

Above formula is not very convenient to use, since Ebλ depends on absolute temperature T, and it is not practicable to tabulate F0-λ for each T. This difficulty is overcome by expressing F0-λ as follows:

F 0_λ0

λλE bλ d

σ T4.

0

λ T.λ T.( )E bλ d

σ T5.

i.e. F 0_λ f λ T.( ) ....(13.16)

i.e. Now, F0-λ is expressed as a function of the product of wave length and absolute temperature ( = λ.T) only.

Values of F0-λ vs. λ.T are tabulated in Table 13.2 below:

��Table 13.2.

Note in the above Table that the units of product λλλλ.T is (micron.Kelvin).

Therefore,

F λ 1_λ 2 F λ 1.T_λ 2.T1

σ0

λ 2 T.

λ T.( )E bλ

T5d

0

λ 1 T.

λ T.( )E bλ

T5d.

i.e. F λ 1_λ 2 F 0_λ 2.T F 0_λ 1.T ......(13.17)

Relation between Radiation intensity and Emissive power:

i.e. Total emissive power of a black (diffuse) surface is equal to � times the intensity of radiation.

This is an important relation, which will be used while calculating the view factors required to determine net energy exchange between surfaces.

Total hemispherical emissivity (εεεε) of a surface is the average emissivity over all directions and all wave lengths and is expressed as:

ε T( )E T( )

E b T( )

E T( )

σ T4......(13.22)

where E(T) is the emissive power of the real surface.

Similarly, spectral emissivity is defined as:

ε λ T( )E λ T( )

E bλ T( ).....(13.23)

Emissivity values for a few surfaces at room temperature are given in handbooks.

A surface is said to be gray if its properties are independent of wave length, and a surface is diffuse if its properties are independent of direction.

To calculate the average emissivity:

Let the variation of spectral emissivity with wave length be as follows:

ε1 = constant, 0 ≤ λ ≤ λ1

ε2 = constant, λ1 ≤ λ ≤ λ2

ε3 = constant, λ2 ≤ λ ≤ ∞

Then, the average emissivity is calculated using equation (13.24) as follows:

ε T( )

ε 10

λ 1λE bλ T( )d.

σ T4.

ε 2λ 1

λ 2λE bλ T( ) d.

σ T4.

ε 3λ 2

∞λE bλ T( )d.

σ T4....(13.25)

i.e. ε(T) = ε1.F0-λ1(T) + ε2.Fλ1-λ2(T) + ε3.Fλ2-∞(T) ….(13.26)

Factors F0-λ1(T) etc. can easily be determined using Tables in handbooks.

13.3.7 Kirchoff’s Law:

Kirchoff’s Law states that the total hemispherical emissivity, ε of a gray surface at a temperature T is equal to its absorptivity, α for black body radiation from a source at the same temperature T.

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$UnitSystem SI Pa K J PROCEDURE Blackbody_EmissivePower_etc(T:E_b,I_bn,lambda_m) "Finds Black body Emissive Power, E_b (W/m^2), Intensity of normal radiation, I_bn (W/m^2), Wave lengty of max. monochromatic emissive power, lambda_m (microns)" "Input: T (K) Output: E_b (W/m^2), I_bn (W/m^2) and lambda_m (microns)" sigma := 5.67E-08 [W/m^2-K^4]"...Stefan Boltzmann constant"

E_b := sigma * T^4 "[W/m^2]...Total Emissive power" I_bn := E_b / pi "[W/m^2]...Intensity of normal radiation" lambda_m := 2898/T "...by Wein’s displacement Law.. finds lambda_max, microns" END "================================================================" FUNCTION Plancks_Law_E_b_lambda(T,lambda) "Finds monochr. emissive power at given Temp and lambda " "Input: T (K), lambda (microns) Output: E_b_lambda (W/m^2. micron)" C1 := 3.742E08 [W-micron^4/m^2]"...const. in Planck's law eqn" C2 := 1.4388E04 [micron-K]"...const. in Planck's law eqn" "Calculations:" Plancks_Law_E_b_lambda := C1 * lambda^(-5) / (exp(C2 / (lambda * T)) -1) "[W/m^2-micron]...monochr. emissive power at given T and lambda " END "================================================================" "Prob.A.3.3 Write an EES Function to determine the fraction of total emission from a blackbody that is in a certain wave length interval, or band. Also, tabulate and plot the fraction of total blackbody emission in the spectral band from 0 to lambda as a function of (lambda.T)." "First, write a Function F(x) as shown below, and then use it in the next Function as shown:" FUNCTION F (x) "This function is used to calculate the fraction F_0_lambda, Ref: Incropera" C_1 := 374200000 "W.micron^4/m2" C_2 := 14388 "micron.K" sigma := 0.0000000567 "W/m2.K^4" IF (x < 5) THEN x:= 5 ENDIF F := C_1 / (sigma * x ^ 5 * (exp(C_2 / x) - 1))

END "============================================" FUNCTION F_zero_lambdaT_trapezoidal(LambdaT1, LambdaT2) "Gives the fraction F_zero_Lambda by integration using Trapezoidal rule, with N = 500 Trapezoids" "Input: LambdaT1, LambdaT2 ... in micron.K, N is the no. of equal intervals" "If LambdaT1 is less than 5, F(x) goes to extremely low value and gives error" N := 500 "...No. of divisions, i.e. no. of Trapezoids" IF LambdaT1 < 5 THEN LambdaT1 := 5 ENDIF IF LambdaT2 > 100000 THEN LambdaT2 = 100000 ENDIF dx := (LambdaT2 - LambdaT1) / N sum[0] :=(F(LambdaT1) + F(LambdaT2)) / 2 Duplicate i = 1,N-1 y[i] :=F(LambdaT1 + (dx * i)) sum[i] := sum[i-1] + y[i] END sum := sum[N-1] * dx F_zero_lambdaT_trapezoidal := sum END "================================================"

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"Prob.A3.1 For a body, of area 0.15 m^2, at an effective temp of 900 K, calculate the following: (i) Total rate of energy emission (ii) Intensity of normal radiation, and (iii) Wave length of max. monochromatic emissive power" "EES Solution:" "Data:" A = 0.15 [m^2] T = 900 [K] sigma = 5.67E-08 [W/m^2-K^4]"...Stefan Boltzmann constant" "Calculations:" CALL Blackbody_EmissivePower_etc(T:E_b,I_bn,lambda_m) E = E_b * A "[W]...Total rate of energy emission" "-----------------------------------------------------------------" ��"�

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Thus:

Total rate of energy emission = E = 5580 W … Ans.

Intensity of normal radiation = I_bn = 11841 W/m^2 …. Ans.

Wavelength of max. monochromaric emissive power = �_m = 3.22 microns … Ans.

Plot, �_m , E and I_bn against T as T varies from 500 K to 2000 K:

First, compute the Parametric Table:

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"Prob.A.3.2 Write an EES Function for E_b_lambda (Planck's Law). Find E_b at T = 100 K and lambda = 10 micron. Then, plot the E_b_lambda against lambda for T = 100, 300, 500, 1000, 2000, 4000, 5800 K (Solar)." EES Function for Planck’s Law for spectral distribution (i.e. spectral black body emissive power, E_b�) is written above. Now, solve the problem: "EES Solution:"

T = 100 [K] lambda = 10 [micron] E_b_lambda = Plancks_Law_E_b_lambda(T,lambda) "W/m^2-micron" "-----------------------------------------------------------------"

Results:

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Now, plot the graph of E_b� vs � for different T:


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Now, plot the results:

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"Prob.A.3.3 Write an EES Function to determine the fraction of total emission from a blackbody that is in a certain wave length interval, or band. Also, tablulate and plot the fraction of total blackbody emission in the spectral band from 0 to � as a function of �.T." �

EES Solution:

The EES Function is already written above.

Now, to produce the Table of F0-� against �.T:

We use the Function: F_zero_lambdaT_trapezoidal(LambdaT1, LambdaT2)

We take LambdaT1 = 0, and LambdaT2 is the required LambdaT:

So, we have:

LambdaT1 = 0 LambdaT2 = 1000[micron-K]

Using the EES Function, we get: F_0_lambda = F_zero_lambdaT_trapezoidal(LambdaT1, LambdaT2)�

Results:

Now, get the Parametric Table:

Now, plot the graph:

"Prob.A.3.4. Window glass transmits radiant energy in the wave length range 0.4�m to 2.5 �m. Determine the fraction of total radiant energy which is transmitted, when the source temperature is: (a) 5800 K (i.e. Sun's surface temp.), and (b) 300 K (i.e. room temp.) [5]”

EES Solution: "Data:" lambda1 = 0.4[micron] lambda2 = 2.5[micron] T1 = 5800 [K] "...in the first case" T2 = 300 [K] "..in the second case" "Calculations:" "We get the fraction of blackbody radiation in the given wave band directly, by using the EES Function written above:" "Case 1:" Fraction1 = F_zero_lambdaT_trapezoidal(lambda1 *T1, lambda2*T1) "...fraction for case 1" "Case 2:" Fraction2 = F_zero_lambdaT_trapezoidal(lambda1 *T2, lambda2*T2) "...fraction for case 2" Results:

Thus:

In case1: Fraction of energy coming from Sun (i.e. T1 = 5800 K), that is transmitted through the window glass = Fraction1 = 0.8421 = 84.21% …. Ans.

In case2: Fraction of energy coming from the source at 300 K, that is transmitted through the window glass = Fraction2 = 5.948 x 10-6 = almost zero …. Ans.

In other words, glass is 'opaque' to radiation at 300 K in the wave length range 0.4 �m to 2.5 �m. This is the principle of 'Greenhouse effect', wherein radiation from a high temperature source (i.e. Sun) is allowed to pass through the glass into the enclosure of the green house, while radiation at a relatively low temperature from within the enclosure, is not allowed to escape out. This, in effect, causes an increase in the temperature of the space within the enclosure.

======================================================================

"Prob.A.3.5. Consider a large isothermal enclosure that is maintains at a uniform temp of 2000 K. Calculate the emissive power of the radiation that emerges from a small aperture on the enclosure surface. What is the wave length �1below which 10% of the emission is concentrated? What is the wave length �2 above which 10% of the of the emission is concentrated? Determine the max. spectral emissive power and the wavelength at which this emission occurs. What is the irradiation incident on a small object placed inside the enclosure? [4]”

EES Solution:

"Data:" T = 2000 [K] sigma = 5.67E-08 "[W/m^2 - K^4] ... Stefan-Boltzmann const." "Calculations:" "1. Emissive power, E:" "Emission from aperture can be considered as blackbody radiation. So, we have:" E = sigma * T^4 "W/m^2 ...Emissive power" "2. Find lambda1 for which F_0_lambda1 = 0.1:" F_zero_lambdaT_trapezoidal(0, lambda1T) = 0.1"... finds lambd1T, microns-K" "Therefore:" lambda1 = lambda1T / T ".... microns" "Let lambda2 be the lower limit of the wave band, upper limit being infinity, from which the radiation is 10%.

i.e. F -0_lambda2 = 0.9. Using the EES Function again:" F_zero_lambdaT_trapezoidal(0, lambda2T) = 0.9"... finds lambd2T, microns-K" "Therefore:" lambda2 = lambda2T / T ".... microns" "3. From Wein's Law:" lambda_max = 2898 / T "[micron] ... " "Spectral power associated with this wave length (i.e. at lambda_max) is:" E_lambda_b = Plancks_Law_E_b_lambda(T,lambda_max) "W/m^2-micron" “4. Irradiation of a small object inside the enclosure is approximated as equal to emission from a blackbody at the enclosure surface temp. i.e.” G = E Results:

Thus:

Wave length below which 10% of the emission is concentrated = �1 = 1.098 microns … Ans.

Wave length above which 10% of the of the emission is concentrated= �2 = 4.687 microns … Ans.

Max. spectral emissive power and the wavelength at which this emission occurs:

�_max = 1.449 microns, E�_b = 411734 W … Ans.

Irradiation incident on a small object placed inside the enclosure = G = E = 907200 W … Ans.

"Prob.A.3.6. In Prob.A.3.3, we wrote an EES Function to determine the fraction of total emission from a blackbody that is in a certain wave length interval, or band. Also, we tabulated and plotted the fraction of total blackbody emission in the spectral band from 0 to � as a function of (�.T) This Table and plot are given in most of the standard textbooks on Heat Transfer. While solving numerical problems by hand, we generally refer to these Tables and interpolate the value of F_0_� for given (�.T). Now, write an EES program to do this interpolation from the Table.

First, save the Radiation Function Table as a lookup Table, by name:

Blackbody Radn function .lkt""

Following is the EES Function:

FUNCTION Radiation_function(lambdaT) {Gives the fraction of radiation between T = 0 and T, (F_0-lambda) as function of (lambda.T). (lambda.T) in micron-Kelvin. Input: lambda.T, in micron-K Output: Fraction F_0_lambda} "Between (lamda.T1) and (lambda.T2), the fraction is (F_0_lambda.T2 - F_0_lambda.T1)" IF (lambdaT < 200) THEN Radiation_function := 0 RETURN ENDIF IF (lambdaT >100000) THEN Radiation_function := 1 RETURN ENDIF Radiation_function := INTERPOLATE1('Blackbody Radn function', 'lambdaT','f0lambdaT','lambdaT'= lambdaT)"linear interpolation from the Table" END "========================================================"

"Ex: See Prob.A.3.4. Window glass transmits radiant energy in the wave length range 0.4micro-m to 2.5 micro-m. Determine the fraction of total radiant energy which is transmitted, when the source temperature is: (a) 5800 K (i.e. Sun's surface temp.), and (b) 300 K (i.e. room temp.). Use the above Function to solve it:"

"Data:" lambda1 = 0.4[micron] lambda2 = 2.5[micron] T1 = 5800 [K] "...in the first case" T2 = 300 [K] "..in the second case" "Calculations:" "We get the fraction of blackbody radiation in the given wave band directly, by using the EES Function written above:" Fraction1 = Radiation_function(lambda2 * T1) - Radiation_function(lambda1*T1) "... for case 1" Fraction2 = Radiation_function(lambda2*T2) - Radiation_function(lambda1*T2) "..for case 2"

Results:

Thus:

In case1: Fraction of energy coming from Sun (i.e. T1 = 5800 K), that is transmitted through the window glass = Fraction1 = 0.8421 = 84.21% …. Ans.

In case2: Fraction of energy coming from the source at 300 K, that is transmitted through the window glass = Fraction2 = 12.35 x 10-6 = almost zero …. Ans.

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The View factor:

View factor is also known by other names such as: ‘configuration factor’, ‘shape factor’, ‘angle factor’ etc.

View factor is defined as the fraction of radiant energy leaving one surface which strikes a second surface directly.

Here, ‘directly’ means that reflection or re-radiated energy is not considered. View factor is denoted by F12, where the first subscript, 1 stands for the emitting surface, and the second subscript, 2 stands for the receiving surface.

View factor between Infinitesimal areas:

(Such a situation is encountered even when finite areas are involved, when the distance between these two areas ‘r’, is very large.)�

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Note that the View factor involves geometrical quantities only.�

View factor between Infinitesimal and finite area:

)32.13()cos()cos(

22

22121 ��=−

AAdA r

dAF

πφφ

Practical situation of calculating view factors between infinitesimal to finite areas are encountered the case of a small thermocouple bead located inside a pipe or a small, spherical point source radiator located by the side of a wall etc.

View factor from Finite to finite area:

In general, we write eqn. (13.33) compactly as:

Reciprocity theorem:

This is a very useful and important relation. It helps one to find out one of the view factors when the other one is known. In practice, one of the view factors which is easier to calculate is obtained first, and the other view factor is found out next, by using the reciprocity theorem.

Note: It is easier to remember the view factor relation given in eqn. (13.34) as:��

�� =1

2212

211211 )37.13(

)cos()cos(A

A

dAdAr

FA ��π

φφ�

Radiation energy exchange between black bodies:

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Properties of View factor and view factor algebra:

Properties of view factor:

(i) The view factor depends only on the geometrics of bodies involved and not on their temperatures or surface properties.

(ii) Between two surfaces that exchange energy by radiation, the mutual shape factors are governed by the ‘reciprocity relation’, viz. A1.F12 = A2.F21.

(iii) When a convex surface 1 is completely enclosed by another surface 2, it is clear from Fig. 13.14(a) that all of the radiant energy emitted by surface 1 is intercepted by the enclosing surface 2. Therefore, view factor of surface 1 w.r.t. surface 2 is equal to unity. i.e. F12 = 1. And, the view factor of surface 2 w.r.t. surface 1 is then easily calculated by applying the reciprocity relation, i.e. A1.1 = A2.F21, or, F21 = A1/A2.

(iv) Radiation emitted from a flat surface never falls directly on that surface (see Fig. 13. 14 (b)), i.e. View factor of a flat surface w.r.t. itself is equal to zero, i.e. F11 = 0. This is valid for a convex surface too, as shown in Fig. 13.14(c).

(v) For a concave surface, it is clear from Fig. 13.14(d) that F11 is not equal to zero since some fraction of radiation emitted by a concave surface does fall on that surface directly.

(vi) If two, plane surfaces A1 and A2 are parallel to each other and separated by a short distance between them, practically all the radiation issuing from surface 1 falls directly on surface 2, and vice –versa. Therefore, F12 = F21 = 1

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(vii) When the radiating surface 1 is divided into, say, two sub-areas A3 and A4, as shown in Fig. 13.15(a), we have:

A 1 F 12

. A 3 F 32. A 4 F 42

. .....(13.41)

Obviously, F12 ≠ F32 + F42.

(viii) Instead, if the receiving surface A2 is sub-divided into parts A3 and A4 as shown in Fig. 13.15 (b), we have:

A 1 F 12. A 1 F 13

. A 1 F 14. .....(13.42)

i.e. F 12 F 13 F 14

i.e. view factor from the emitting surface 1 to a sub-divided receiving surface is simply equal to the sum of the individual shape factors from the surface 1 to the respective parts of the receiving surface. This is known as ‘Superposition rule’.

(ix) Symmetry rule: If two (or more) surfaces are symmetrically located w.r.t. the radiating surface 1, then the view factors from surface 1 to these symmetrically located surfaces are identical. A close inspection of the geometry will reveal if there is any symmetry in a given problem.

(x) Summation rule: Since radiation energy is emitted from a surface in all directions, invariably, we consider the emitting surface to be part of an enclosure. Even if there is an opening, we consider the opening as a surface with the radiative properties of that opening.

Then, the conservation of energy principle requires that sum of all the view factors from the surface 1 to all other surfaces forming the enclosure, must be equal to 1. See Fig. 13.16, where the interior surface of a completely enclosed space is sub-divided into n parts, each of area A1, A2, A3…An.

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Then,�

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F 11 F 12 ..... F 1n 1

F 21 F 22 ..... F 2n 1

..............................................

F n1 F n2 ..... F nn 1/ / / / / / �01*�2*3�

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��1

n

j

F ij=

1 i = 1, 2, 3,.....n......(13.44)

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(xi) In an enclosure of ‘n’ black surfaces, maintained at temperatures T1, T2,….Tn, net radiation from any surface, say surface 1, is given by summing up the net radiation heat transfers from surface 1 to each of the other surfaces of the enclosure:

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Q 1netA 1 F 12

. σ. T 14 T 2

4. A 1 F 13. σ. T 1

4 T 34. A 1 F 14

. σ. T 14 T 4

4.

A 1 F 1n. σ. T 1

4 T n4.+

...

....(13.45)��

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Note: Often, while solving radiation problems, determination of the view factor is the most difficult part. It will be useful to keep in mind the definition of view factor, summation rule, reciprocity relation, superposition rule and symmetry rule while attempting to find out the view factors.

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View factor relations for a few geometries are given below: (Ref: Cengel)

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View factor for two aligned parallel, rectangles of equal size:

View factor between two coaxial parallel discs:

View factor between two perpendicular rectangles with a common edge:

View factors for two concentric cylinders of finite length: (a) outer cylinder to inner cylinder (b) outer cylinder to itself:

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Let us write EES Functions to find out View Factors for three practically important geometries:

(a) Aligned, parallel rectangles,

(b) Coaxial, parallel disks, and

(c) Perpendicular rectangles with a common edge

---------------------------------------------------------------

(a) Aligned, parallel rectangles:

$UnitSystem SI Pa J Radians FUNCTION F12_parallel_rectangles(X, Y, L) { Gives View factor for coaxial, parallel rectangles separated by a distance L Note: Trigonometric functions are used with Radians; so, set the angles to Radians in Unit Systems} XX := X / L YY := Y / L A := 2 / (pi * XX * YY) B := ln(sqrt(( 1 + XX^2) * (1 + YY^2) / (1 + XX^2 + YY^2))) C := XX * (1 + YY^2)^ (1/2) * arctan(XX / (1 + YY^2)^(1/2)) D := YY * (1 + XX^2)^(1/2) * arctan(YY / (1 + XX^2)^(1/2)) E := XX * arctan(XX) F := YY * arctan(YY) F12_parallel_rectangles := A * (B + C + D - E - F) END "---------------------------------------------------------------------------"

(b) Coaxial, parallel disks:

FUNCTION F12_parallel_disks(r_1, r_2, L) { Gives View factor for coaxial, parallel disks separated by a distance L} RR_1 := r_1 / L RR_2 := r_2 / L S := 1 + (1 + RR_2^2) / RR_1^2 F12_parallel_disks := (1/2) * ( S - (S^2 - 4 * (r_2 / r_1)^2)^(1/2)) END "---------------------------------------------------------------------------"

(c) Perpendicular rectangles with a common edge:

$UnitSystem SI Pa J Radians

FUNCTION F12_perpendicular_rectangles(X, Y, Z) { Gives View factor for coaxial, perpendicular rectangles with a common edge Note: Trigonometric functions are used with Radians; so, set the angles to Radians in Unit Systems} H := Z / X W := Y / X A := 1 / (pi * W) B := W * arctan(1 / W)

C := H * arctan(1 / H) D := (H^2 + W^2)^(1/2) * arctan(1 / (H^2 + W^2)^(1/2)) E :=(1 + W^2) * (1 + H^2) / (1 + W^2 + H^2) F :=( W^2 * (1 + W^2 + H^2) / ((1 + W^2) * (W^2 + H^2)))^(W^2) G := ( H^2 * (1 + H^2 + W^2) / ((1 + H^2) * (H^2 + W^2)))^(H^2) F12_perpendicular_rectangles := A * (B + C - D + (1/4)*ln(E * F * G)) END "---------------------------------------------------------------------------"

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Let us write EES Functions for the view factors for:

(a) From the outer cylinder to itself, (b) From the outer cylinder to inner cylinder, and (c) From the inner cylinder to itself

We use the view factor equations provided by Modest. (Ref: ‘Radiative Heat Transfer’ by Modest)

Note: In the EES Functions we write below, height of the cylinders (i.e. h in the above figs.) is taken as L.

--------------------------------------------------------------------------------

FUNCTION F22_CoaxialCyl_outer_to_itself (R1, R2, L) "Finds the View factor F_22 between two coaxial parallel cylinders--outer surface to itself" " Ref: Radiative Heat Transfer by Modest" R := R2/R1 H := L / R1 A = (sqrt(H ^ 2 + 4 * R ^ 2) - H) / (4 * R) B = (2 * sqrt(R ^ 2 - 1)) / H C = sqrt(4 * R ^ 2 + H ^ 2) / H D = H ^ 2 + 4 * (R ^ 2 - 1) - 2 * H ^ 2 / R ^ 2 E = H ^ 2 + 4 * (R ^ 2 - 1) F = (R ^ 2 - 2) / R ^ 2 G = (2 / R) * arctan(B) F22_CoaxialCyl_outer_to_itself = 1 - 1 / R - A + (1 / pi) * (G - (H / (2 * R)) * (C *arcsin(D / E) - arcsin(F))) END

"---------------------------------------------------------------------------------------------" FUNCTION F21_coaxialCyl_outer_to_inner(R1, R2, L) "Finds the View factor F_21 between two coaxial parallel cylinders--outer surface to inner surface Ref: Radiative Heat Transfer by Modest" R := R2/R1 H := L / R1 A := (H ^ 2 + R ^ 2 - 1) / (4 * H) B := (H ^ 2 - R ^ 2 + 1) / (H ^ 2 + R ^ 2 - 1) C := sqrt((H ^ 2 + R ^ 2 + 1) ^ 2 - 4 * R ^ 2) / (2 * H) D := (H ^ 2 - R ^ 2 + 1) / (R * (H ^ 2 + R ^ 2 - 1))

E := (H ^ 2 - R ^ 2 + 1) / (2 * H) F21_coaxialCyl_outer_to_inner = (1 / R) * (1 - A - (1 / pi) * (arccos(B) - C * arccos(D) - E * arcsin(1 / R))) END "---------------------------------------------------------------------------------------------" FUNCTION F12_Coaxial_cyl_inner_to_outer (R1, R2, L) { Finds the View factor F12 between two coaxial parallel cylinders--inner surface to outer surface. Ref: F_21 From Modest, see above. Use: A2 * F_21 = A1 * F_12; i.e. F_12 = F_21*(A2/A1); And, A2/A1 = R2/R1 } F12_Coaxial_cyl_inner_to_outer = (R2/R1) * F21_coaxialCyl_outer_to_inner(R1, R2, L) END "--------------------------------------------------------------------------------------------------------"

20�� "

Prob.B3.1. Plot the shape factors for two concentric parallel disks.

Consider a truncated cone with top and bottom diameters of 10 and 20 cm and a height of 10 cm. Calculate:

(a) the shape factor from the top surface to the sides and also from the side surface to itself. (b) If surface 1 is at 1000 K and surface 2 is at 500 K, find the net exchange of heat between 1 and 2

(Q_12), if all the surfaces are black. (c) Plot F_12 and Q_12 for L varying from 0.1 to 1 m

EES Solution:

Let the top surface be designated as 1, bottom surface as 2, and the side surfaces as 3. See the fig. above.

Then:

First, write the EES Function as follows, to easily plot the View factor graphs:

FUNCTION F12_parallel_disks_2(RR_1, RR_2) { Gives View factor for coaxial, parallel disks separated by a distance L} { RR_1 = r_1 / L RR_2 = r_2 / L } S := 1 + (1 + RR_2^2) / RR_1^2 F12_parallel_disks_2 := (1/2) * ( S - (S^2 - 4 * (RR_2 /RR_1)^2)^(1/2)) END "---------------------------------------------------------------------------"

Now, write the program:

RR_2 = 8 {LbyR1 = 1} F12 = F12_parallel_disks_2(1/LbyR1, RR_2) "-----------------------------------"

And, the Parametric Table:

And, plot the graphs:

Now, solve the problem:

"Data:" r_1 = 0.05 [m] r_2 = 0.1 [m] L = 0.1 [m] "We have to find out F13 and F33:" "Calculations:" A1 = pi * r_1^2 "[m^2].... area of top surface" A2 = pi * r_2^2 "[m^2].... area of bottom surface" A3 = pi * (r_1 + r_2) * sqrt((r_2 - r_1)^2 + L^2) "[m^2]....area of the side surface" "Now: F11 + F12 + F13 = 0 ...... by Summation rule

But, F11 = 0... since it is a flat surface. We find F12 by the EES Function for parallel disks, written above. So, we have:" F11 = 0 F12 = F12_parallel_disks(r_1, r_2, L) “using the EES Function for parallel disks” F11 + F12 + F13 = 1"....finds the view factor from top surface to side surface, F13" "Now, by reciprocal relation between surfaces 1 and 3:" A1 * F13 = A3 * F31"...finds view factor F31" "Now, by reciprocal relation between surfaces 1 and 2:" A1 * F12 = A2 * F21 "...finds view factor from surface 2 to 1, F21" "And, by reciprocal relation between surfaces 2 and 3:" F22 = 0 "...since surface 2 is flat" "For surface 2, by Summation rule:" F22 + F21 + F23 = 1"...determines F23" A2 * F23 = A3 * F32 "...finds view factor from surface 3 to 2, F32" "Now, applying Summation rule to surface 3:" F33 + F31 + F32 = 1"....determines F33" Results:

Thus:

Shape factor from the top surface to the sides = F13 = 0.5311 …. Ans.

Shape factor from the side surface to itself = F33 = 0.3944 …. Ans.

-----------------------------------------------------------------------------------------------------

In addition: (b) If surface 1 is at 1000 K and surface 2 is at 500 K, find the net exchange of heat between 1 and 2, if all the surfaces are black.

Add the following code at the end of the above program and run it again:

--------------------------------------------------------

"If surface 1 is at 1000 K and surface 2 is at 500 K, find the net exchange of heat between 1 and 2, if all the surfaces are black." T1 = 1000 [K] T2 = 500 [K] sigma = 5.67E-08 [W/m^2-K^4] "...Stefan Boltzmann const." Q_12 = A1 * F12 * sigma * (T1^4 - T2^4) "[W] .... net radiant exchange between surfaces 1 and 2" ----------------------------------------------------------------------

We get:

i.e. Q_12 = 195.7 W … net heat exchange between surfaces 1 and 2 … Ans.

----------------------------------------------------------------------------------------------

Also, plot F_12 and Q_12 against L, with L varying from 0.1 m to 1 m:

First, compute the Parametric Table. (Here, we have included F_33 also)


======================================================================

Prob.B3.2. Consider two coaxial, parallel rectangles, each of size: 1 m x 1m, with a distance of 2 m from each other. If surface 1 is at 800 K and the surface 2 at 400 K, find the net heat interchange between the two surfaces. Consider the surfaces as black.

EES Solution:

"Data:" X = 1 [m] Y = 1 [m] L =2 [m]

T1 = 800 [K] T2 = 400 [K] sigma = 5.67E-08[W/m^2-K^4] "... Stefan Boltzmann constant" "We have to find out F 12 :" "Calculations:" A1 = X * Y "[m^2].... area of top surface" A2 =A1 "[m^2].... area of bottom surface" F_12 = F12_parallel_rectangles(X, Y, L) "Net exchange of heat between A1 and A2:" Q_12 = A1 * F_12 * sigma * (T1^4 - T2^4) "[W] ... net exchange of heat between surfaces1 and 2" Results:

Thus:

Net exchange of heat between A1 and A2 = Q_12 = 1493 W … Ans. >��"�

Plot the variation of F_12 and Q_12 as the distance between the plates (L) varies from 0.5 m to 2.5 m: First, compute the Parametric Table:

Prob.B3.3. Two concentric cylinders have diameters of 10 and 20 cm. Calculate the View factor between the two open ends.

Now, let us use the Functions for concentric cylinders written above, in solving this problem:

EES Solution:

Let: Surface 1: outer surface of inner cylinder

Surface 2 : inner surface of outer cylinder

Surfaces 3 and 4 : surfaces at the two ends

Then, we have:

"Data:" R1 = 0.05 [m] R2 = 0.1 [m] L = 0.2 [m] "Calculations:" A1 = 2 * pi * R1 * L "[m^2] ... area of inner cylinder surface" A2 = 2 * pi * R2 * L "[m^2] ... area of outer cylinder surface" A3 = pi * (R2^2 - R1^2) "[m^2] ... area of open, end surface"

A4 = A3 "[m^2] ... area of open, end surface" F_21 = F21_CoaxialCyl_outer_to_inner(R1, R2, L)"...finds View factor from outer to inner surface" F_22 = F22_CoaxialCyl_outer_to_itself (R1, R2, L)"...finds View factor from outer surface to itself" "Now, by reciprocity:" A1 * F_12 = A2 * F_21 "..finds F_12" "For surface 2, Summation rule gives:" F_21 + F_22 + F_23 + F_24 = 1 "But, by symmetry:" F_23 = F_24 "...therefore, F_23 and F_24 are calculated." "Again, applying reciprocity between surfaces 2 and 3:" A2 * F_23 = A3 * F_32 "...finds F_32" "Now," F_33 = 0 "...since surface is flat" F_44 = 0 "...since surface is flat" F_11 = 0"...since surface is convex" "Applying Summation rule for surface 3:" F_31 + F_32 + F_33 + F_34 = 1"....eq. (A)" "In the above, F_31 and F_34 are unknown." "To find F_31:" F_11 + F_12 + F_13 + F_14 = 1 "...by Summation rule on surface 1" "But," F_13 = F_14 "...by Symmetry" "Therefore, F_13 and F_14 are determined, since F_11 = 0 and F_12 is already determined" "Then, by reciprocity:" A1 * F_13 = A3 * F_31 "...finds F_31" "And, from eqn.(A), F_34 is immediately calculated." Results:

Thus: View factor between the two open ends = F_34 = 0.07694 … Ans.

Also, plot F_34 for different values of L, i.e. 0.1 < L < 0.5 m:

Parametric Table:


=============================================================

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Definitions:

1. Irradiation, (G): is the total radiation incident upon a surface per unit time, per unit area (W/m2). 2. Radiosity, (J): is the total radiation leaving a surface, with no regard for its origin (i.e., reflected

plus emitted from the surface) per unit time, per unit area (W/m2).

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3. Surface resistance:

By analogy with Ohm’s law:

4. Space resistance:

By analogy with Ohm’s law:

Rij is known as ‘space resistance’ and it represents the resistance to radiative heat flow between the radiosity potentials of the two surfaces, due to their relative orientation and spacing.

For a N surface enclosure, net heat transfer from surface i should be equal to the sum of net heat transfers from that surface to the remaining surfaces. i.e.

�

��

Q i1

N

j

Q ij= 1

N

j

A i F ij. J i J j

.

= 1

N

j

J i J j

R ij=

...W....(13.54)

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�

��

i.e.E bi J i

R i 1

N

j

J i J j

R ij=

W...(13.55)

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where Ri is the surface resistance and Rij is the space resistance.�

This situation is shown in Fig. (13.27).�

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-----------------------------------------------------------------------------------------------------------------

Radiation heat exchange in two-zone enclosures: (Ref: 1, 3)

----------------------------------------------------------------------

A few special cases of two-surface enclosure:

Here, basic, governing eqn. is eqn.(13.56), which is modified depending upon the case considered.

Few special cases of two-surface enclosures: (Ref: Cengel)�

Case (i): Radiant heat exchange between two black surfaces:

For a black body, ε = 1, and J = Eb, as explained earlier. i.e. surface resistance

[= (1 - ε)/(A.ε)] of a black body is zero.

Next, we shall consider four cases of practical interest where the view factor between the inner surface 1 and the outer surface 2 (i.e. F12) is equal to 1, and also the net radiation from a gray cavity:

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Case (ii): Radiant heat exchange for a small object in a large cavity: A practical example of a small object in a large cavity is the case of a steam pipe passing through a large plant room.�

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Case (iii): Radiant heat exchange between infinitely large parallel plates: �

In this case, A1 = A2 = A, say, and F12 = 1

Then, eqn. (13.56) becomes:

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Case (iv): Radiant heat exchange between infinitely long concentric cylinders: �

In this case:

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��F 12 1

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Then, eqn. (13.56) becomes:

��

Q 12A 1 σ. T 1

4 T 24.

1

ε 1

A 1

A 2

1

ε 21.

....for infinitely long concentric cylinders.....(13.60)

where

A 1

A 2

r 1

r 2 �

Remember that A1 refers to the inner (or enclosed) surface.�

Case (v): Radiant heat exchange between concentric spheres:�In this case:

��F 12 1

�

Then, eqn. (13.56) becomes�

��

Q 12A 1 σ. T 1

4 T 24.

1

ε 1

A 1

A 2

1

ε 21.

....for concentric spheres.....(13.61)

where

A 1

A 2

r 1

r 2

2

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Remember, again, that A1 refers to the inner (or enclosed) surface.

Case (vi): Energy radiated from a gray cavity: �

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Eqn. (13.62) is an important result, which gives net radiation from a gray cavity to surrounding space. If it is desired, for example, to calculate the net radiation from a blind hole drilled in a flange, then the relation to use is the eqn. (13.62).�

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Temperature of each surface is known; i.e. emissive power Eb for each surface is known:

Then, the problem reduces to determining the radiosities J1, J2 and J3.

This is done by applying Kirchoff’s law of d.c. circuits to each node: i.e. sum of the currents (or, rate of heat transfers) entering into each node is zero:

��

Node J1:E b1 J 1

R 1

J 2 J 1

R 12

J 3 J 1

R 130 .....(13.63,a)

Node J2:E b2 J 2

R 2

J 1 J 2

R 12

J 3 J 2

R 230 .....(13.63,b)

Node J3:E b3 J 3

R 3

J 1 J 3

R 13

J 2 J 3

R 230 .....(13.63,c)

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Solving these three eqns. simultaneously, we get J1, J2 and J3.

Once the radiosities are known, expressions for net heat flows between the surfaces are:

��

Q 12J 1 J 2

R 12

J 1 J 2

1

A 1 F 12.

....(13.64,a)

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Q 13J 1 J 3

R 13

J 1 J 3

1

A 1 F 13.

....(13.64,b)

Q 23J 2 J 3

R 23

J 1 J 2

1

A 2 F 23.

....(13.64,c)

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And, net heat flow from each surface is:

��

Q 1E b1 J 1

R 1

E b1 J 1

1 ε 1

A 1 ε 1.

.....(13.65,a)

Q 2E b2 J 2

R 2

E b2 J 2

1 ε 2

A 2 ε 2.

.....(13.65,b)

Q 3E b3 J 3

R 3

E b3 J 3

1 ε 3

A 3 ε 3.

.....(13.65,c)

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To the general eqns (13.64), apply the constraints such as:

If the surface is black or re-radiating: Ji = Ebi = σ.Ti4.

And, Qi = 0 for a re-radiating surface.

If Qi at any surface is specified instead of the temperature (i,e. Ebi), then, (Ebi – Ji)/Ri is replaced by Qi.

Few special cases of three-zone enclosures:

1. Two black surfaces connected to a third refractory surface:

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Fig. (13.33,a) shows the radiation network for this case. The radiation network is drawn very easily by remembering the usual principles: for a black surface, the surface resistance is zero, i.e. Eb = J. For a re-radiating surface too, Eb = J, as already explained; further, for a re-radiating surface, Q = 0. Between two given surfaces, the radiosity potentials are connected by the respective space resistances, as shown. It may be observed that the system reduces to a series-parallel circuit of resistances as shown in Fig. (13.33,b).

We get:

Here, Q12 is the net radiant heat transferred between surfaces 1 and 2. Similar expressions can be written for heat transfer between surfaces 2 and 3 (= Q23) and the heat transfer between surfaces 1 and 3 (= Q13).

--------------------------------------------------------------------------------------------------------------

2. Two gray surfaces surrounded by a third re-radiating surface:

Here, we have:

Q 3 0

i.e.E b3 J 3

1 ε 3

A 3 ε 3.

0

i.e. E b3 J 3 �

i.e. once the radiosity of the re-radiating surface is known, its temperature can easily be calculated, since Eb3 = σ.T3

4. Further, note that T3 is independent of the emissivity of surface 3.

Expression for heat flow rate is:

Q 1 Q 2E b1 E b2

R tot

where, Rtot is the total resistance, given by:�

��

R tot R 11

1

R 12

1

R 13 R 23

R 2

i.e. R tot1 ε 1

A 1 ε 1.

1

A 1 F 12. 1

1

A 1 F 13.

1

A 2 F 23.

1 ε 2

A 2 ε 2.

....(13.67)

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Radiation heat exchange in four zone enclosures:

(a) When all the four surfaces are black: Remembering the principles already explained, if the radiation network for an enclosure comprising of four black surfaces is drawn, it will look as shown in Fig. (13.35), below:�

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Expression for net radiant heat flow rate from surface 1 is:�

��

Q 1J 1 J 2

R 12

J 1 J 3

R 13

J 1 J 4

R 14

i.e. Q 1E b1 E b2

R 12

E b1 E b3

R 13

E b1 E b4

R 14...since E b1 = J1 ec.

i.e. Q 1 A 1 F 12. E b1 E b2

. A 1 F 13. E b1 E b3

. A 1 F 14. E b1 E b4

. ....(13.68)�

Similar expressions can be written for the net heat flow from other three surfaces.

(b) When all the four surfaces are gray: Now, for each surface, a surface resistance has also to be included, and the radiation network for this system will be as shown in Fig. (13.36):

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Expression for net radiant heat flow rate from surface 1 is:�

Q 1E b1 J 1

R 1

J 1 J 2

R 12

J 1 J 3

R 13

J 1 J 4

R 14

i.e. Q 1E b1 J 1

1 ε 1

A 1 ε 1.

A 1 F 12. J 1 J 2

. A 1 F 13. J 1 J 3

. A 1 F 14. J 1 J 4

. ....(13.69)

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Similar expressions can be written for the net heat flow from other three surfaces.

-----------------------------------------------------------------------------------------------------

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Write EES Functions for a general two-surface enclosures, and also for few special cases of two-surface enclosures:

Following are the EES Functions. Later, we shall use them in solving some problems.

---------------------------------------------------------------------------------------------------------------------------------- $UnitSystem SI Pa J K FUNCTION Q12_TwoSurfaceEnclosure(A_1, A_2, F_12, epsilon_1, epsilon_2, T_1, T_2) { Returns Q12 from surface 1 to 2 in a general, two-surface enclosure; 1 is the internal surface. Inputs: A1, A2 (m^2), T1, T2 (K)} sigma := 5.67E-08 "[W/m^2-K^4]....Stefan Boltzmann constant" R_12 := 1/(A_1 * F_12) "[1/m^2]...space resistance" R_1 := (1 - epsilon_1) / (A_1 * epsilon_1) "[1/m^2]..surface resistance of surface 1" R_2 := (1 - epsilon_2) / (A_2 * epsilon_2) "[1/m^2]..surface resistance of surface 2" R_tot := R_1 + R_2 + R_12 "...total resistance" E_b1 := sigma * T_1^4 "[W/m^2]" E_b2 := sigma * T_2^4 "[W/m^2]" Q12_TwoSurfaceEnclosure := (E_b1 - E_b2) / R_tot "[W]" END "-----------------------------------------------------------------------------------------------------" FUNCTION Q12_SmallObject(A_1, epsilon_1, T_1, T_2) { Returns Q12 from small object 1 to a large surrounding surface 2; 1 is the internal surface. Inputs: A1 (m^2), T1, T2 (K)} sigma := 5.67E-08 "[W/m^2-K^4]....Stefan Boltzmann constant" Q12_SmallObject := sigma * A_1 * epsilon_1 * (T_1^4 - T_2^4) "[W]" END "--------------------------------------------------------------------------------------------------"

FUNCTION Q12_parallel_plates(A, epsilon_1, epsilon_2,T_1, T_2) { Returns net Q12 from plate 1 to a parallel plate 2; Inputs: A (m^2), T1, T2 (K)} sigma := 5.67E-08 "[W/m^2-K^4]....Stefan Boltzmann constant" Q12_parallel_plates := sigma * A * (T_1^4 - T_2^4) / (1/epsilon_1 + 1/epsilon_2 - 1)"[W]" END "--------------------------------------------------------------------------------------------------" FUNCTION Q12_long_concentric_cylinders(L, R_1, R_2, epsilon_1, epsilon_2,T_1, T_2) { Returns net Q12 from surface 1 to surface 2; 1 is the inner cylindrical surface. Inputs: L, R_1, R_2 (m), T1, T2 (K)} sigma := 5.67E-08 "[W/m^2-K^4]....Stefan Boltzmann constant" Q12_long_concentric_cylinders := sigma * (2 * pi * R_1 * L) * (T_1^4 - T_2^4) / (1/epsilon_1 + ( R_1 / R_2) * (1/epsilon_2 - 1))"[W]" END "--------------------------------------------------------------------------------------------------" FUNCTION Q12_concentric_spheres( R_1, R_2, epsilon_1, epsilon_2,T_1, T_2) { Returns net Q12 from surface 1 to surface 2; 1 is the inner spherical surface. Inputs: R_1, R_2 (m), T1, T2 (K)} sigma := 5.67E-08 "[W/m^2-K^4]....Stefan Boltzmann constant" Q12_concentric_spheres := sigma * (4 * pi * R_1^2) * (T_1^4 - T_2^4) / (1/epsilon_1 + ( R_1 / R_2)^2 * (1/epsilon_2 - 1))"[W]" END "--------------------------------------------------------------------------------------------------" FUNCTION Q12_from_gray_cavity( A_1,F_11, epsilon_1, T_1) { Returns net Q12 from surface 1 of cavity to a closing surface 2; Inputs: A_1 (m^2), T1 (K)} sigma := 5.67E-08 "[W/m^2-K^4]....Stefan Boltzmann constant" Q12_from_gray_cavity := sigma * A_1 * epsilon_1 * T_1^4 * ((1 - F_11) / (1 - (1 - epsilon_1) * F_11))"[W]" END "--------------------------------------------------------------------------------------------------"

For three zone enclosure: The EES Procedure is given below. Read the comments in the program:

$UnitSystem SI Pa J K PROCEDURE HTrans_Three_surface_enclosure(A_1,A_2,A_3,T_1,T_2,T_3,eps_1,eps_2,eps_3,F_12,F_13,F_23:Eb_1,Eb_2,Eb_3,R_1,R_2,R_3,R_12,R_13,R_23) { Inputs: A1, A2 ,...etc Outputs: Emissive powers: Eb_1,Eb_2,Eb_3 (W/m^2), and Surface and Space resistances: R_1, R_2, R_3, R_12, R_13, R_23 (1/m^2) } "Radiation Heat transfer in Three-Zone enclosure:" "Note: For a insulated/re-radiating surface: Q_i = 0, and J_i = Eb_i ( = sigma * T_i^4)" "For a black surface: R_i = 0; so, J_i = Eb_i " sigma := 5.67e-08 “[W/m^2-K^4]….Stefan Boltzmann const.” Eb_1 := sigma * T_1^4 Eb_2 := sigma * T_2^4 Eb_3 := sigma * T_3^4 R_1 := (1 - eps_1)/(A_1 * eps_1) "surface resist of surface 1" R_2 := (1 - eps_2)/(A_2 * eps_2) "surface resist of surface 2" R_3 := (1 - eps_3)/(A_3 * eps_3) "surface resist of surface 2" R_12 := 1/(A_1 * F_12) "space resist bet surfaces 1 and 2" R_13 := 1/(A_1 * F_13) "space resist bet surfaces 1 and 3" R_23 := 1/(A_2 * F_23) "space resist bet surfaces 2 and 3" END "==========================================================================="

40�� "� Prob. C3.1. Two large parallel plates are at 1000 K and 800 K. Determine the heat exchange per unit area, when:

(i) The surfaces are black (ii) The hot surface has an emissivity of 0.9 and the cold surface has emissivity = 0.6 [VTU –

Dec. 2006/Jan. 2007]

In addition, plot the variation of Q12 with e2, all other conditions remaining the same.

EES Solution: "Data:" A = 1 [m^2] epsilon_1 = 0.9 epsilon_2 = 0.6 T_1 = 1000 [K] T_2 = 800 [K] "Case 1: when both surfaces are black:" Q12_both_black = 5.67E-08[W/m^2-K^4] * A * (T_1^4 - T_2^4) "Case 2: When epsilon_1 and epsilon_2 are 0.9 and 0.6 respectively:" "Using the EES Function for parallel plates, written above:" Q12_both_gray = Q12_parallel_plates(A, epsilon_1, epsilon_2,T_1, T_2) Results:

Thus: Q12 when both surfaces are black: 33476 W …. Ans. Q12 when �1 and �2 are 0.9 and 0.6, respectively = 18830 W … Ans. ------------------------------------------------------------------------------------------------------------ In addition, plot the variation of Q12 with e2, all other conditions remaining the same:

First, compute the parametric Table:


==================================================================

Prob. C3.2. A solid copper sphere of 10 cm dia at 1000 C is suspended in a large enclosure. If the walls of the enclosure are at 30 C, determine: (i) the initial rate of cooling of the sphere (ii) time taken by the sphere to cool to 900 C. Use the following properties for copper: density, � = 8954 kg/m^3, specific heat, cp = 381 J/kg.K, thermal conductivity, k = 386 W/m.K, emissivity, � = 0.78.[VTU – Aug. 2001] (b) Additionally, plot the variation of these quantities as emissivity of surface of sphere varies from 0.1 to 0.9. EES Solution:

This is the case of a small object in a large surrounding.

So, to get Q12, use the EES Function written above.

"Data:" R = 0.05 [m]"....radius of sphere" epsilon_1 = 0.78"...emissivity of sphere surface" T_1 = 1273 [K]"...sphere temp." T_2 = 303 [K]"....surroundings temp." rho = 8954[kg/m^3]"...density of copper" cp = 381 [J/kg-K]"...sp. heat of copper" k = 386 [W/m-K]"...thermal cond. of copper" "Calculations:"

A_1 = 4 * pi * R^2 " [m^2]... surface area of sphere" "To find Q12, use the EES Function already written for a small object in a large enclosure:" Q12 = Q12_SmallObject(A_1, epsilon_1, T_1, T_2)"[W]...finds rate of initial heat loss from the sphere" "But, Q12 is also equal to: Q12 = mass * cp * dTbydtau, where dTbydtau is the initial rate of cooling" "Therefore:" mass = rho * (4/3) * pi * R^3 "...[kg]... mass of sphere" Q12 = mass * cp * dTbydtau "...[C/s]....finds initial rate of cooling" "Time taken to cool to 900 C:" "Find the rate of heat loss when the sphere temp is 900 C. Then, take the average rate of heat loss from 1000 C to 900 C, and the find the approx. time taken to cool from 1000 C to 900 C, knowing the total amount of heat removed in the process:" Q12_2 = Q12_SmallObject(A_1, epsilon_1, 1173, T_2)"[W]...finds rate of heat loss from the sphere when it is at 900 C" "Therefore:" Q12_avg = (Q12 + Q12_2 ) / 2 "And:" Q_lost = mass * cp * (1000 - 900)"[J]" "Therefore, approx. time taken to cool to 900 C:" time = Q_lost / Q12_avg "[s]......time taken to cool to 900 C" Results:

Thus:

Initial rate of heat loss, Q12 = 3637 W….Ans.

Initial rate of cooling, dT/d� = 2.036 C/s … Ans.

Approx. time taken to cool to 900 C = time = 57.11 s …. Ans.

----------------------------------------------------------------------------------------------------

Additionally, plot the variation of these quantities as emissivity of surface of sphere varies from 0.1 to 0.9:



==========================================================================

Prob. C3.3. Refer to Fig. below. Three thin walled, long, circular cylinders 1, 2 and 3, of diameters 15 cm, 25 cm and 35 cm respectively, are arranged concentrically as shown. Temperature of cylinder 1 is 80 K and that of cylinder 3 is 300 K. Emissivities of cylinders 1, 2 and 3 are 0.05, 0.1 and 0.2 respectively. Assuming that there is vacuum inside the annular spaces, determine the steady state temperature attained by cylinder 2.

(b) In addition, plot Q12 and T2 as �2 varies from 0.1 to 0.9.

EES Solution:

This is the case of long, concentric cylinders.

To find T2, use the fact that: in steady state, net radiant heat transfer between cylinders 1 and 2 must be equal to the net radiant heat transfer between cylinders 2 and 3.

To find Q12 between long, concentric cylinders, use the EES Function already written.

"Data:" R1 = 0.075 [m]"....radius of inner cyl" R2 = 0.125 [m]"....radius of middle cyl" R3 = 0.175 [m]"....radius of outer cyl" L = 1[m] "...assumed" epsilon_1 = 0.05"...emissivity of surface 1" epsilon_2 = 0.1"...emissivity of surface 2" epsilon_3 = 0.2"...emissivity of surface 3" T_1 = 80 [K]"...inner cylinder temp." T_3 = 300 [K]"....outer cylinder temp." "Calculations:" "Net heat transfer between surfaces 1 and 2: Using the EES Function already written:" Q12 = Q12_long_concentric_cylinders(L, R1, R2, epsilon_1, epsilon_2,T_1, T_2) "Similarly, Net heat transfer between surfaces 2 and 3:" Q23 = Q12_long_concentric_cylinders(L, R2, R3, epsilon_2, epsilon_3,T_2, T_3) "Equating Q12 and Q23, we get the unknown temp, T2:" Q12 = Q23 "..equate Q12 to Q23 to get T2" Results:

Thus:

Temp of middle cylinder = T2 = 280.9 K … Ans.

Net heat transfers Q12 = Q23 = -6.503 W…. Ans. (-ve sign indicates heat flow from outer cylinder to inner cylinder)

-----------------------------------------------------------------------------------------------------------------------

In addition, plot Q12 and T2 as �2 varies from 0.1 to 0.9:



===================================================================

Prob.C3.4. For a hemispherical furnace, the flat floor is at 700 K and has an emissivity of 0.5. The hemispherical roof is at 1000 K and has an emissivity of 0.25. Find the net radiant heat transfer from the roof to floor. [V.T.U--.July-Aug. 2003] (b) Also, plot Q12 as the emissivity of base, �2 varies from 0.1 to 0.9, all other conditions remaining the same.

EES Solution:

Assume the radius of hemisphere as 1 m.

This is a two-surface enclosure problem.

Use the EES Function written earlier, to find Q12 of a two-surface enclosure:

-----------------------------------------------------------

"Data:" R1 = 1 [m]"....radius of hemisphere" epsilon_1 = 0.25"...emissivity of hemisph. surface 1" epsilon_2 = 0.5"...emissivity of base surface 2" T_1 = 1000 [K]"...hemisph. surface temp." T_2 = 700 [K]"....base temp." "Calculations:" A_1 = 4 * pi * R1^1 / 2 "[m^2]....area of hemisph. surface" A_2 = pi * R1^2 "[m^2]..area of base surface" "View factors:" F_21 = 1 "...since all the energy leaving surface 2 is intercepted by surface 1" "To find F_12, View factor from surface 1 to surface 2:"

A_1 * F_12 = A_2 * F_21 "...by reciprocity... finds F_12" "Now, to get Q12 use the EES Function for a general, two-surface enclosure, already written:" Q12 = Q12_TwoSurfaceEnclosure(A_1, A_2, F_12, epsilon_1, epsilon_2, T_1, T_2) Results:

Thus:

Net radiant heat transfer from roof to floor = Q12 = 38674 W …. Ans.

---------------------------------------------------------------------------------------------------------

(b) Also, plot Q12 as the emissivity of base, �2 varies from 0.1 to 0.9:


And, plot the results:

======================================================================

Three zone enclosures:

Prob. C3.5. Two parallel plates, 0.5 m x 1 m each, are spaced 0.5 m apart. The plates are at temperatures of 1000 C and 500 C and their emissivities are 0.2 and 0.5 respectively. The plates are located in a large room, the walls of which are at 27 C. The surfaces of the plates facing each other only exchange heat by radiation. Determine the rates of heat lost by each plate and heat gain of the walls by radiation. Use radiation network for solution. Assume shape factor between parallel plates: F12 = F21 = 0.285. [M.U. 1996]

(b) Also, plot Q1 and Q3 as epsilon1 varies from 0.1 to 0.9

�

EES Solution:

The schematic fig. and the radiation network are shown in Fig (a) and (b) above.

We shall, use the EES PROCEDURE written above, to calculate the Emissive powers and Resistances involved; then, in the Main EES program we find out the Radiosities by applying the Kirchoff’s Law to the three nodes J1, J2 and J3 and solve the resulting three equations simultaneously to get J1, J2 and J3. Then, the heat transfers are easily calculated by applying the Ohm’s Law between any two given radiosity potentials.

"Data:" A_1 = 0.5 [m^2] A_2 = 0.5 [m^2] A_3 = 1e10 [m^2] "...very large room" T_1 = 1273 [K] T_2 = 773 [K] T_3 = 300 [K] eps_1 = 0.2 "emissivity of plate 1" eps_2 = 0.5 "emissivity of plate 2"

eps_3 = 0.9999 "...emissivity of surroundings = 1, taken as 0.9999 to avoid division by zero" F_12 = 0.285 "...View Factor from surface 1 to 2.... by data" F_21 = 0.285 "...View Factor from surface 2 to 1... by data" F_11 = 0 "...since surface 1 is flat" "F_11 + F_12 + F_13 = 1 ... by Summation rule for surface 1" "Therefore:" F_13 = 1- F_12 "...View Factor from surface 1 to 3" "Similarly, for surface 2, F_22 = 0, and F_21 + F_22 + F_23 = 1 by Summation rule for surface 2, and we get:" F_23 = 1- F_21 "...View Factor from surface 2 to 3" "Now, calculate various Emissive powers and Resistances, using the EES PROCEDURE:" CALL HTrans_Three_surface_enclosure(A_1,A_2,A_3,T_1,T_2,T_3,eps_1,eps_2,eps_3,F_12,F_13,F_23:Eb_1,Eb_2,Eb_3,R_1,R_2,R_3,R_12,R_13,R_23) "Kirchoff's Law for Nodes J_1, J_2 and J_3:" "For J_1:" (Eb_1 - J_1)/R_1 + (J_2 - J_1)/R_12 + (J_3 - J_1)/R_13 = 0 "For J_2:" (Eb_2 - J_2)/R_2 + (J_1 - J_2)/R_12 + (J_3 - J_2)/R_23 = 0 "For J_3:" (Eb_3 - J_3)/R_3 + (J_1 - J_3)/R_13 + (J_2 - J_3)/R_23 = 0 "Net heat transfer from each surface:" Q_1 = (Eb_1 - J_1)/R_1 Q_2 = (Eb_2 - J_2)/R_2 Q_3 = (Eb_3 - J_3)/R_3 "Net heat transfer between surface 1 and 2:" Q_12 = (J_1 - J_2)/R_12 "Net heat transfer between surface 1 and 3:" Q_13 = (J_1 - J_3)/R_13 "Net heat transfer between surface 2 and 3:" Q_23 = (J_2 - J_3)/R_23 "Check : Q_1 + Q_2 + Q_3 = 0 " SumQ1Q2Q3 = Q_1 + Q_2 + Q_3 -----------------------------------------------------------------------

Results:

Thus:

Heat leaving surface 1 = Q1 = 14428 W …. Ans.

Heat leaving surface 2 = Q2 = 2594 W … Ans.

Heat received by surface 3, i.e. ambient = Q3 = -17022 W … Ans. (negative sign indicates that heat is going into the surface).

Also, check: Q1 + Q2 + Q3 = 0 (which is satisfied… see SumQ1Q2Q3 = -0.0001729 = almost zero)

In addition, values of radiosities J1, J2 and J3, and also the various resistances, and heat transfers between surfaces, viz. Q12, Q13 and Q23 are also calculated.

--------------------------------------------------------------------------------------------------------------

Now, plot Q1 and Q3 as epsilon1 varies from 0.1 to 0.9:

First, produce the Parametric Table:

Note in the above Table that:

(i) For eps1 = > 0.5, Q2 is –ve, i.e. surface 2 receives heat (ii) Q3 is the heat received by the ambient (-ve) (iii) Sum of Q1, Q2 and Q3 is equal to almost zero, as it should be

And, Plot the results:

====================================================================

Prob. C3.6. A square room, 3 m x 3 m, has a floor heated to 27 C and has a ceiling at 10 C. Walls are perfectly insulated. Height of room is 2.5 m. Emissivity of all surfaces is 0.8. Determine:

(i) Net heat transfer from the ceiling (ii) Net heat transfer between ceiling and floor, and (iii) Temp of the walls [M.U.]

(b) Also, plot the variation of Q1 and T3 as eps1 varies from 0.1 to 0.9:

EES Solution:

This is a three-surface enclosure problem.

Important: Walls act as re-radiating surfaces, i.e. Eb_3 = J3, and Q_3 = 0.

Radiation network is shown below:

Let ceiling be designated as surface 1, floor as surface 2, and the surrounding walls as surface 3.

We shall use the EES PROCEDURE written above to calculate Emissive powers, various resistance etc.

We shall also use the EES Function written earlier to determine the View Factor F_12 between the ceiling and the floor.

-----------------------------------------------------------

"Data:" A_1 = 9 [m^2]"....area of ceiling" A_2 = 9 [m^2]"...area of floor" A_3 = 30 [m^2] "...area of 4 walls" T_1 = 283 [K]"...temp of ceiling, by data" T_2 = 300 [K]"...temp of floor, by data" {T_3 = ...to be found out} sigma = 5.67E-08 [W/m^2-K^4]"...Stefan - Boltzmann constant" eps_1 = 0.8 "emissivity of ceiling, i.e. surface 1" eps_2 = 0.8 "emissivity of floor, i.e. surface 2" eps_3 = 0.8 "...emissivity of surrounding walls, i.e. surface 3 " "Determine View Factors: F_12, F_13 etc." "Use the EES Function written earlier for parallel rectangles to get F_12 :"

"We have: for the ceiling and floors: X = 3 m, Y = 3 m, L = 2.5 m" X = 3 [m] Y = 3 [m] L = 2.5 [m] F_12 = F12_parallel_rectangles(X, Y, L) "...View Factor from surface 1 to 2.... " F_21 = F_12"...View Factor from surface 2 to 1... by symmetry" F_11 = 0 "...since surface 1 is flat" "F_11 + F_12 + F_13 = 1 ... by Summation rule for surface 1" "Therefore:" F_13 = 1- F_12 "...View Factor from surface 1 to 3"

"Similarly, for surface 2, F_22 = 0 , and F_21 + F_22 + F_23 = 1 by Summation rule for surface 2, and we get:" F_23 = 1- F_21 "...View Factor from surface 2 to 3" "Now, calculate various Emissive powers and Resistances, using the EES PROCEDURE:" CALL HTrans_Three_surface_enclosure(A_1,A_2,A_3,T_1,T_2,T_3,eps_1,eps_2,eps_3,F_12,F_13,F_23:Eb_1,Eb_2,Eb_3,R_1,R_2,R_3,R_12,R_13,R_23) "Kirchoff's Law for Nodes J_1, J_2 and J_3:" "For J_1:" (Eb_1 - J_1)/R_1 + (J_2 - J_1)/R_12 + (J_3 - J_1)/R_13 = 0 "For J_2:" (Eb_2 - J_2)/R_2 + (J_1 - J_2)/R_12 + (J_3 - J_2)/R_23 = 0 "For J_3:" (J_1 - J_3)/R_13 + (J_2 - J_3)/R_23 = 0 Eb_3 = J_3 "....since surface 3 is re-radiating" “Also:” Eb_3 = sigma * T3 ^4 "[W/m^2]...by definition of Eb" "Net heat transfer from each surface:" Q_1 = (Eb_1 - J_1)/R_1"[W]...net heat transfer from surface 1" Q_2 = (Eb_2 - J_2)/R_2"[W]...net heat transfer from surface 2" Q_3 = (Eb_3 - J_3)/R_3"[W]...net heat transfer from surface 3" "Net heat transfer between surface 1 and 2:" Q_12 = (J_1 - J_2)/R_12 "[W]" "Net heat transfer between surface 1 and 3:" Q_13 = (J_1 - J_3)/R_13 "[W]" "Net heat transfer between surface 2 and 3:" Q_23 = (J_2 - J_3)/R_23 "[W]" "Check : Q_1 + Q_2 + Q_3 = 0 " SumQ1Q2Q3 = Q_1 + Q_2 + Q_3 Results:

Thus:

(i) Net heat transfer from surface 1 = Q1 = -409.8 W …. Ans…. negative sign indicates heat coming in to the surface.

(ii) Net heat transfer between ceiling and floor = Q12 = -164.4 W ….Ans. …again, negative sign indicates heat coming in to the surface 1.

(iii) Temp of re-radiating walls = T3 = 291.9 K … Ans.

Note: Sum of Q1, Q2 and Q3 = 0, as it should be.

-----------------------------------------------------------------------------------------------------Plot Plot (b) Plot the variation of Q1 and T3 as eps1 varies from 0.1 to 0.9:



=================================================================

Prob. C3.7. In a cylindrical furnace with R = 0.75 m, L = 2 m as shown. Emissivities and temperatures of surfaces 1, 2 and 3 are: eps1 = 0.65, eps2 = 0.55, eps3 = 0.7, T1 = 800 K, T2 = 600 K and T3 = 1400 K respectively. Assume surface 3 to be re-radiating. Determine the net rate of radiation to / from each surface.

(b) If surface 3 is black, what are the values of Q1, Q2 and Q3?

(c) For the case (a) plot the variation of Q3 as emissivity of surface 3 varies from 0.1 to 1.

EES Solution:

This is a general, three-surface enclosure, for which the electrical network is shown below:

Writing Kirchoff’s Law to Nodes J1, J2 and J3:

While writing the above equations, remember to consider all heat currents as flowing in to the respective node.

First, get the Emissive powers and various resistances using the EES Function written earlier.

Then, solve simultaneously the three equations obtained by applying the Kirchoff’s Law to the three Nodes, written above.

To get the View Factors F12, F13 and F23 use the EES Function for two parallel disks, written earlier.

Following is the EES program:

"Data:" R = 0.75 [m] L = 2 [m] A_1 = pi * R^2 " [m^2]...area of top disk" A_2 = A_1" [m^2]...area of bottom disk" A_3 = 2 * pi * R * L " [m^2] ...area of walls" T_1 = 800 [K]"...temp of top disk" T_2 = 600 [K]"...temp of bottom disk" T_3 =1400 [K]"....temp of walls" sigma = 5.67E-08 [W/m^2-K^4]"...Stefan - Boltzmann constant" eps_1 = 0.65 "emissivity of surface 1" eps_2 = 0.55 "emissivity of surface 2" eps_3 = 0.7 "...emissivity of surface 3 " "Determine View Factors: F_12, F_13 etc." "Use the EES Function written earlier for parallel disks to get F_12 :

i.e. F_12 = F12_parallel_disks(r_1, r_2, L) " "We have: for the top and bottom disks, in the above Function: r_1 = R, r_2 = R = 0.75 m, L = 2 m" F_12 = F12_parallel_disks(R, R, L) "...View factor from top disk to bottom disk" F_21 = F_12 "...by symmetry" "F_11 + F_12 + F_13 = 1 ... by Summation rule for surface 1" "Therefore:" F_13 = 1- F_12 "...View Factor from surface 1 to 3" "Similarly, for surface 2, F_22 = 0 , and F_21 + F_22 + F_23 = 1 by Summation rule for surface 2, and we get:" F_23 = 1- F_21 "...View Factor from surface 2 to 3" "Now, calculate various Emissive powers and Resistances, using the EES PROCEDURE:" CALL HTrans_Three_surface_enclosure(A_1,A_2,A_3,T_1,T_2,T_3,eps_1,eps_2,eps_3,F_12,F_13,F_23:Eb_1,Eb_2,Eb_3,R_1,R_2,R_3,R_12,R_13,R_23) "Kirchoff's Law for Nodes J_1, J_2 and J_3:" "For J_1:" (Eb_1 - J_1)/R_1 + (J_2 - J_1)/R_12 + (J_3 - J_1)/R_13 = 0 "For J_2:" (Eb_2 - J_2)/R_2 + (J_1 - J_2)/R_12 + (J_3 - J_2)/R_23 = 0 "For J_3:" (Eb_3 - J_3)/R_3 + (J_1 - J_3)/R_13 + (J_2 - J_3)/R_23 = 0 "Net heat transfer from each surface:" Q_1 = (Eb_1 - J_1)/R_1 “[W]” Q_2 = (Eb_2 - J_2)/R_2 “[W]” Q_3 = (Eb_3 - J_3)/R_3 “[W]” "Net heat transfer between surface 1 and 2:" Q_12 = (J_1 - J_2)/R_12 "Net heat transfer between surface 1 and 3:" Q_13 = (J_1 - J_3)/R_13 "Net heat transfer between surface 2 and 3:" Q_23 = (J_2 - J_3)/R_23 "Check : Q_1 + Q_2 + Q_3 = 0 "

SumQ1Q2Q3 = Q_1 + Q_2 + Q_3 Results:

Thus: Net heat transfer from surface 1 = Q1 = -190056 W … negative sign indicates heat flowing in to the surface 1. .. Ans. Net heat transfer from surface 2 = Q2 = -175443 W … negative sign indicates heat flowing in to the surface 2. .. Ans. Net heat transfer from surface 3 = Q3 = 365499 W … heat flowing from the surface 3. .. Ans. Also, SumQ1Q2Q3 = 0.0122, i.e. almost equal to zero, as it should be. -----------------------------------------------------------------------------------------------------

(b) If surface 3 is black, what are the values of Q1, Q2 and Q3?

Now, eps3 = 1. In EES data, enter: eps3 = 0.9999 to avoid division by zero. Press F2 to calculate, and we get:

i.e. Net heat transfer from surface 1 = Q1 = -207904 W … negative sign indicates heat flowing in to the surface 1. .. Ans. Net heat transfer from surface 2 = Q2 = -190385 W … negative sign indicates heat flowing in to the surface 2. .. Ans. Net heat transfer from surface 3 = Q3 = 398289 W … heat flowing from the surface 3. .. Ans. Also, SumQ1Q2Q3 = 0.0002028, i.e. almost equal to zero, as it should be. ------------------------------------------------------------------------------------------------------------------- (c) For the case (a), plot the variation of Q3 as emissivity of surface 3 varies from 0.1 to 1:

First, prepare the Parametric Table:

In the above Table, in addition to Q3, we have also obtained Q1, Q2, Q13 and SumQ1Q2Q3. Note that SumQ1Q2Q3 is almost equal to zero, in all cases.

As said earlier, negative sign indicates that heat is flowing in to the surface.

Plot the result:

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Prob. C3.8. A long duct of equilateral triangular section, of side w = 0.75 m, shown in Fig., has its surface 1 at 700 K, surface 2 at 1000 K, and surface 3 is insulated. Further, surface 1 has an emissivity of 0.8 and surface 2 is black. Determine the rate at which energy must be supplied to surface 2 to maintain these operating conditions.

(b) Plot the variation of Q1 and T3 as epsilon1 varies from 0.1 to 0.9.

EES Solution:

Since the duct is very long, end effects can be neglected, and this is a three-surface enclosure problem.

Surface 1 is gray, surface 2 is black and surface 3 is insulated.

Radiation network is shown in Fig.(b) below:

We use the EES Function written earlier to calculate Emissive powers and resistances.

T3 is unknown, but is calculated from the relations: Eb3 = J3 = �.T3^4, where � is the Stefan-Boltzmann constant.

EES code is given below:

---------------------------------------------------------------

"Data:" L = 1 [m]"...assumed length of duct" W = 0.75 [m]"...side of equilateral triangle" A_1 = W * L " [m^2]....area of side 1" A_2 = A_1"[m^2]...area of side 2" A_3 = A_1 " [m^2] ...area of side 3" T_1 = 700 [K]"...temp of ceiling, by data" T_2 = 1000 [K]"...temp of floor, by data" {T_3 = ...to be found out} sigma = 5.67E-08 [W/m^2-K^4]"...Stefan - Boltzmann constant" eps_1 = 0.8 "emissivity of surface 1" eps_2 = 0.9999 "emissivity of surface 2 .... black" eps_3 = 0.9999 "...emissivity of surface 3 .... " " View Factors: F_12, F_13 etc." "F_11 + F_12 + F_13 = 1 ... by Summation rule for surface 1" "But, " F_11 = 0 "...since surface 1 is flat; Also, F_12 = F_13, by symmetry. "

"Therefore:" F_12 = 0.5 "...View Factor from surface 1 to 2" F_13 = 0.5 "...View Factor from surface 1 to 3" "Similarly, for surface 2, F_22 = 0 , and F_21 + F_22 + F_23 = 1 by Summation rule for surface 2, and we get:" F_23 = 0.5 "...View Factor from surface 2 to 3" "Now, calculate various Emissive powers and Resistances, using the EES PROCEDURE:" CALL HTrans_Three_surface_enclosure(A_1,A_2,A_3,T_1,T_2,T_3,eps_1,eps_2,eps_3,F_12,F_13,F_23:Eb_1,Eb_2,Eb_3,R_1,R_2,R_3,R_12,R_13,R_23) "Kirchoff's Law for Nodes J_1:" "For J_1:" (Eb_1 - J_1)/R_1 + (J_2 - J_1)/R_12 + (J_3 - J_1)/R_13 = 0 Eb_2 = J_2 "....since surface 2 is black" Eb_3 = J_3 "....since surface 3 is insulated" {Eb_3 = sigma * T_3 ^4 "[W/m^2]...by definition of Eb"} "Net heat transfer from each surface:" Q_1 = (Eb_1 - J_1)/R_1"[W]...net heat transfer from surface 1" Q_2 = -Q_1"[W]...net heat transfer from surface 2" Q_3 = 0 "[W]...net heat transfer from surface 3" "Net heat transfer between surface 1 and 2:" Q_12 = (J_1 - J_2)/R_12 "[W]" "Net heat transfer between surface 1 and 3:" Q_13 = (J_1 - J_3)/R_13 "[W]" "Net heat transfer between surface 2 and 3:" Q_23 = (J_2 - J_3)/R_23 "[W]" "Check : Q_1 + Q_2 + Q_3 = 0 " SumQ1Q2Q3 = Q_1 + Q_2 + Q_3 "======================================================================"

Results:

Thus:

Net heat transfer from surface 1 = Q1 = -20409 W … negative sign indicates heat coming into the surface … Ans.

Net heat transfer from surface 2 = Q2 = 20409 W … positive sign indicates heat leaving the surface … Ans.

Temp of insulated surface 3 = T3 = 908.1 K … ans.

Note that as a check: SumQ1Q2Q3 = 0…is satisfied.

-----------------------------------------------------------------------------------------------------------------

(b) Plot the variation of Q1 and T3 as epsilon1 varies from 0.1 to 0.9: First, compute the Parametric Table:

Note: Negative sign for Q1 denotes heat flowing into the surface 1.


Note: The four problems solved above with EES cover the important variations of the three-surface enclosure problem.

� � ��"�

�'�� E'+�%F"�

Radiation shielding:

For two infinite, parallel plates 1 and 2, radiation shield 3, be placed between the plates :

��

�

When there is no shield, the radiation heat transfer between plates 1 and 2 is already shown to be:�

Q 12A σ. T 1

4 T 24.

1

ε 1

1

ε 21

....for infinitely large parallel plates.....(13.59)

With one shield placed between plates 1 and 2, the radiation network will be as shown in Fig. (b) above. Note that now all the relevant resistances are in series. Net heat transfer between plates 1 and 2 is given as:

�

Q12-one shield = (Eb1-Eb2)/Rtot where Rtot is the total resistance.

��

��

Q 12_one_shieldE b1 E b2

1 ε 1

A 1 ε 1.

1

A 1 F 13.

1 ε 3_1

A 3 ε 3_1.

1 ε 3_2

A 3 ε 3_2.

1

A 3 F 32.

1 ε 2

A 2 ε 2.

....for two gray surfaces with one radiation shield placed in between.....(13.70) �

Now, for two large parallel plates, we note:

F 13 F 32 1 and, A 1 A 2 A 3 A

Then, eqn. (13.70) simplifies to:

Q 12_one_shieldA σ. T 1

4 T 24.

1

ε 1

1

ε 21

1

ε 3_1

1

ε 3_21

....(13.71)

Note that as compared to eqn. (13.59) for the case of no-shield, we have, with one shield, an additional term appearing in the denominator of eqn. (13.71). Therefore, if there are N radiation shields, we have, for net radiation heat transfer:

��

Q 12N_shields

A σ. T 14 T 2

4.

1

ε 1

1

ε 21

1

ε 3_1

1

ε 3_21 .....

1

ε N_1

1

ε N_21

...(13.72)

�

�

If emissivities of all surfaces are equal, eqn. (13.72) becomes:

��

��

Q 12N_shields

A σ. T 14 T 2

4.

N 1( )1

ε1

ε1.

1

N 1( )Q 12no_shield

. ....(13.73)

�

Note this important result, which implies that, when all emissivities are equal, presence of one radiation shield reduces the radiation heat transfer between the two surfaces to one-half, two radiation shields reduce the heat transfer to one-third, 9 radiation shields reduce the heat transfer to one-tenth etc.

For a more practical case of the two surfaces having emissivities of ε1 and ε2, and all shields having the same emissivity of εs, eqn. (13.72) becomes:

��

Q 12N_shields

A σ. T 14 T 2

4.

1

ε 1

1

ε 21 N

2

ε s1.

....(13.74)

�

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++�

To determine the equilibrium temperature of the radiation shield:

Once Q12 is determined from eqn. (13.71), the temperature of the shield is easily found out by applying the condition that in steady state:

Q 12 Q 13 Q 32 ...(13.75)

We can use either of the conditions: Q12 = Q13 or Q12 = Q32.

Q13 or Q32 is determined by applying eqn. (13.59); i.e. we get:

�

��

Q 12 Q 13A σ. T 1

4 T 34.

1

ε 1

1

ε 31

...(13.76,a)

or,

Q 12 Q 32A σ. T 3

4 T 24.

1

ε 3

1

ε 21

....(13.76,b)

�

�

In both the above eqns. T3 is the only unknown, which can easily be determined.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++�

For a cylindrical radiation shield placed in between two, long concentric cylinders:

The radiation heat transfer between two long, concentric cylinders is:

�

�

Now, let a cylindrical radiation shield, 3, be placed in between the inner cylinder (1) and the outer cylinder (2), as shown in Fig.(13.38).

The radiation network for this system is shown in Fig. (13.38,b) and it is exactly the same as shown in Fig. (13.37,b).

And, the radiation heat transfer between cylinders 1 and 2, when the shield is present, is given by:

�

��

Q 12_one_shieldE b1 E b2

1 ε 1

A 1 ε 1.

1

A 1 F 13.

1 ε 3_1

A 3 ε 3_1.

1 ε 3_2

A 3 ε 3_2.

1

A 3 F 32.

1 ε 2

A 2 ε 2.

....for two gray surfaces with one radiation shield placed in between.....(13.70) �

�

Now, for the cylindrical system, we have:

�

��

F 13 F 32 1

A 1 2 π. r 1. L.

A 2 2 π. r 2. L.

and, A 3 2 π. r 3. L.

�

�

Then, eqn. (13.70) reduces to:

�

��

Q 12one_shield

A 1 σ. T 14 T 2

4.

1

ε 1

A 1

A 2

1

ε 21.

A 1

A 3

1

ε 3_1

1

ε 3_21.

...for concentric cylinders with one radiation shield...(13.77) �

In eqn. (13.77), we have: (A1/A2) = (r1/r2), and (A1/A3) = (r1/r3).

Note that as compared to the relation for two concentric cylinders with no shield (i.e. eqn. 13.60), an additional term appears in the denominator of eqn. (13.77) (i.e. the third term) when one radiation shield is introduced; if there is a second radiation shield, say (4), then one more similar term will have to be added in the denominator to take care of the resistance of that shield.

In this case too, the equilibrium temperature of the shield is determined by applying the principle that, in steady state,

��Q 12 Q 13 Q 32�

For a spherical radiation shield placed in between two concentric spheres:�

This case is also represented by Fig. (13.38,a), where inner sphere 1 is enclosed by an outer sphere 2, and a radiation shield 3, is placed in between. The radiation network for this system is shown in Fig. (13.38,b).

When there is no radiation shield, radiation heat transfer between surfaces 1 and 2 is given by eqn. (13.61), viz.

�

�

Again, when the radiation shield is present, the general relation for radiation heat transfer between surfaces 1 and 2 is eqn. (13.70). Remembering that for concentric spheres,

�

��

F 13 F 32 1

A 1 4 π. r 12.

A 2 4 π. r 22.

and, A 3 4 π. r 32.�

Relation for radiant heat transfer between surfaces 1 and 2, is exactly as eqn. (13.77), i.e.

�

��

Q 12one_shield

A 1 σ. T 14 T 2

4.

1

ε 1

A 1

A 2

1

ε 21.

A 1

A 3

1

ε 3_1

1

ε 3_21.

...for concentric spheres with one radiation shield...(13.78)�

�

In eqn. (13.78), we have:

�

��

A 1

A 2

r 1

r 2

2

and,A 1

A 3

r 1

r 3

2

�

�

In this case also, equilibrium temperature of the shield is determined by applying the principle that, in steady state,�

��Q 12 Q 13 Q 32�

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++�

�

�

�%�� "�

Prob.5.D.12. Write EES Functions for radiation heat transfer without and with radiation shield for the cases of: (i) infinite, parallel plates (ii) infinite, concentric cylinders, and (iii) concentric spheres. Also write Functions to determine the equilibrium temp of the radiation shield.

Equations for these cases are already given.

Now, the EES Functions:

In the following Functions:

Inputs:

In case of cylinders and spheres:

1…. denotes the inner cylinder / sphere, and

2… denotes outer cylinder / sphere

R1 …. Radius of inner cylinder / sphere, (m)

R2 …. Radius of outer cylinder / sphere, (m)

R_shield …. Radius of radation shield placed between 1 and 2, (m)

A … area of plates, (m^2)

eps1, eps2 …emissivities of two plates or cylinders or spheres

eps_s1, eps_s2 .. emissivities of radiation shield, on the faces looking towards surfaces 1 and 2 respectively

T1, T2 … temps of two plates or cylinders or spheres, (K)

Output:

Q12 … net heat transfer between surfaces 1 and 2, (W)

T_shield … equilibrium temp of shield, (K)

$UnitSystem SI Pa J C FUNCTION Q12_parallel_plates(A,eps_1,eps_2,T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_parallel_plates:=(A * sigma * (T1^4 - T2^4)) / (1/eps_1 + 1/eps_2 -1) END "------------------------------------------------------------------------------"

FUNCTION Q12_parallel_plates_one_shield(A, eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_parallel_plates_one_shield:=(A * sigma * (T1^4 - T2^4)) / ((1/eps_1 + 1/eps_2 -1) + (1 / eps_s1 + 1/eps_s2 -1)) END "------------------------------------------------------------------------------" FUNCTION Temp_Shield_parallel_plates(A, eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Temp_Shield_parallel_plates:=(T1^4 - Q12_parallel_plates_one_shield(A, eps_1, eps_2, eps_s1, eps_s2, T1,T2) * (1 / eps_1 + 1 / eps_s1 - 1) / (A * sigma))^0.25 END "------------------------------------------------------------------------------" FUNCTION Q12_parallel_plates_N_shields(A, N, eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_parallel_plates_N_shields:=(A * sigma * (T1^4 - T2^4)) / ((1/eps_1 + 1/eps_2 -1) + N * (1 / eps_s1 + 1/eps_s2 -1)) END "------------------------------------------------------------------------------" FUNCTION Q12_concentric_cyl(R1, R2, L , eps_1, eps_2,T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_concentric_cyl:=(2 * pi * R1 * L * sigma * (T1^4 - T2^4)) / (1/eps_1 + (R1 / R2) * (1/eps_2 -1)) END "------------------------------------------------------------------------------" FUNCTION Q12_concentric_cyl_one_shield(R1, R2,R_shield, L , eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_concentric_cyl_one_shield:=(2 * pi * R1 * L * sigma * (T1^4 - T2^4)) / (1/eps_1 + (R1 / R2) * (1/eps_2 -1)+ (R1 / R_shield) * (1/eps_s1 + 1/eps_s2 -1)) END "------------------------------------------------------------------------------"

FUNCTION Temp_Shield_concentric_cyl(R1, R2,R_shield, L , eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Temp_Shield_concentric_cyl:=(T1^4 - Q12_concentric_cyl_one_shield(R1, R2,R_shield, L , eps_1, eps_2, eps_s1, eps_s2, T1,T2) *(1/eps_1 + (R1 / R_shield) * (1/eps_s1 -1)) / (2 * pi * R1 * L * sigma))^0.25 END "------------------------------------------------------------------------------" FUNCTION Q12_concentric_spheres(R1, R2, eps_1, eps_2,T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_concentric_spheres :=(4 * pi * R1^2 * sigma * (T1^4 - T2^4)) / (1/eps_1 + (R1 / R2)^2 * (1/eps_2 -1)) END "------------------------------------------------------------------------------" FUNCTION Q12_concentric_sph_one_shield(R1, R2,R_shield, eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Q12_concentric_sph_one_shield:=(4 * pi * R1^2 * sigma * (T1^4 - T2^4)) / (1/eps_1 + (R1 / R2)^2 * (1/eps_2 -1)+ (R1 / R_shield)^2 * (1/eps_s1 + 1/eps_s2 -1)) END "------------------------------------------------------------------------------" FUNCTION Temp_Shield_concentric_spheres(R1, R2,R_shield, eps_1, eps_2, eps_s1, eps_s2, T1,T2) sigma:=5.67E-08 "[W/m^2-K^4]" Temp_Shield_concentric_spheres:=(T1^4 - Q12_concentric_sph_one_shield(R1, R2,R_shield, eps_1, eps_2, eps_s1, eps_s2, T1,T2) *(1/eps_1 + (R1 / R_shield)^2 * (1/eps_s1 -1)) / (4 * pi * R1^2 * sigma))^0.25 END "------------------------------------------------------------------------------"

�

�0�� "�

Now, let us use the above EES Functions to solve some problems:

Prob.D3.1. Two very large parallel planes, with emissivities 0.3 and 0.8 exchange heat. Find the percentage reduction in heat transfer when a polished Aluminium shield (eps = 0.04) is placed between them. [VTU – May/June 2010]

EES Solution:

We have:

Therefore, taking the ratio of Q12 with shield to that without shield:

In EES, data and this eqn for Ratio are entered:

"Data:" eps_1 = 0.3 eps_2 = 0.8 eps_s1 = 0.04 eps_s2 = 0.04 Ratio = (1/eps_1 + 1/eps_2 - 1) / ((1/eps_1 + 1/eps_2 - 1) + (1/eps_s1 + 1/eps_s2 - 1) ) And, Results:

i.e. Q12 with shield is 6.815 % of that without shield…. Ans.

=======================================================================

Prob.D3.2. Two large parallel planes with emissivity 0.6 are at 900 K and 300 K. A radiation shield with one side polished and having emissivity of 0.05 and the other side unpolished with emissivity 0.4 is proposed to be used between them. Which side of the shield should face the hotter plane if the temp of the shield is to be kept minimum? Justify your answer. [VTU – June 2012]

(b) Also, for the case 1, plot Q12_with_shield and T_shield as eps_s1 (i.e. emissivity of shield surface facing the 900 K surface) varies from 0.05 to 0.4:

EES Solution:

We shall use the EES Functions written above:

"Data:" A = 1"[m^2]" eps_1 = 0.6 eps_2 = 0.6 eps_s1 = 0.05 "..facing surface 1" eps_s2 = 0.4 ".... facing surface 2" T1 = 900"[K]" T2 = 300 "[K]" "Let: Case 1: polished surface of shield is facing the hot surface 1 Case 2: unpolished surface of shield is facing the hot surface 1" Q12 = Q12_parallel_plates(A,eps_1,eps_2,T1,T2)"....when there is no shield" Q12_shield_case1 = Q12_parallel_plates_one_shield(A, eps_1, eps_2, eps_s1, eps_s2, T1,T2) "..polished surface facing hotter plate" Q12_shield_case2 = Q12_parallel_plates_one_shield(A, eps_1, eps_2, eps_s2, eps_s1, T1,T2) "..unpolished surface facing hotter plate" T_shield_case1 = Temp_Shield_parallel_plates(A, eps_1, eps_2, eps_s1, eps_s2, T1,T2) "...for case 1" T_shield_case2 = Temp_Shield_parallel_plates(A, eps_1, eps_2, eps_s2, eps_s1, T1,T2) "...for case 2" Results:

Thus:

For case 1: Temp of shield = 554 K … Ans.

For case 2: Temp of shield = 868.9 K …. Ans.

Therefore, use option 1 for minimum temp of shield.

--------------------------------------------------------------------------------------------------------------

(b) Also, for case 1, plot Q12_with_shield and T_shield as eps_s1 (i.e. emissivity of shield surface facing the 900 K surface) varies from 0.05 to 0.4:


Next, plot the results:

==========================================================================

"Prob.D3.3. A cryogenic fluid flows in a pipe of 10 mm OD at a temp of 100 K. It is surrounded by another coaxial pipe of 13 mm OD. Space between the pipes is evacuated. The outer pipe is at 280 K. Emissivities of both surfaces is 0.3. (b) If a shield of emissivity of 0.05 on either face and dia of 11.5 mm is placed between the pipes, determine the percentage reduction in heat flow. Determine the radiant heat flow for a 3 m length. Also, calculate the equilibrium temp of the shield. (c) Plot the Percentage reduction and the equilibrium temp of the shield as emissivity of shield varies from 0.05 to 0.3"

EES Solution: "Data:" R1 = 0.005 "[m]" R2 = 0.0065"[m]" R_shield = 0.00575 "[m]" L = 3 "[m]" eps_1 = 0.3 eps_2 = 0.3 eps_s1 = 0.05 eps_s2 = 0.05 T1 = 100 "[K]" T2 = 280 "[K]" “Calculations:” Q12 = Q12_concentric_cyl(R1, R2, L , eps_1, eps_2,T1,T2) Q12_shield = Q12_concentric_cyl_one_shield(R1, R2,R_shield, L , eps_1, eps_2, eps_s1, eps_s2, T1,T2) T_shield = Temp_Shield_concentric_cyl(R1, R2,R_shield, L , eps_1, eps_2, eps_s1, eps_s2, T1,T2) PercentReduction = (Q12 - Q12_shield) * 100 / Q12 Results:

Thus: Q12 without shield = -6.301 W…. negative sign indicates flow towards the inner pipe which is at a lower temp …. Ans. Q12 with radiation shield = - 0.8276 W … negative sign indicates flow towards the inner pipe which is at a lower temp …. Ans. Equilibrium temp of shield = T_shield = 237.4 K … Ans. -------------------------------------------------------------------------------------

To plot the Percentage reduction and the equilibrium temp of the shield as emissivity of shield varies from 0.05 to 0.3:

First, compute the Parametric Table, keeping both eps_s1 = eps_s2 = 0.05, 0.1, 0.15 ….. 0.4.


=========================================================================

"Prob.D3.4. Two coaxial cylinders of diameters D1 = 0.1 m and D2 = 0.3 m and emissivities eps1 = 0.7 and eps2 = 0.4 are maintained at temps T1 = 800 K and T2 = 500 K respectively. A coaxial shield of diameter D3 = 0.2 m and emissivity eps_s on both its surfaces is placed between the cylinders. Determine eps_s if the radiation heat transfer between the cylinders is to be reduced to 15 % of that without the radiation shield. What is the equilibrium temp of the shield at that time?"

EES Solution: "Data:" R1 = 0.05 "[m]" R2 = 0.15"[m]" R_shield = 0.1 "[m]" L = 1 "[m]" eps_1 = 0.7 eps_2 = 0.4 "Let: eps_s1 = eps_s2 = eps_s ... emissivity of shield" T1 = 800 "[K]" T2 = 500 "[K]" “Calculations:” Q12 = Q12_concentric_cyl(R1, R2, L , eps_1, eps_2,T1,T2) " ..... heat transfer without radiation shield" Q12_shield = Q12_concentric_cyl_one_shield(R1, R2,R_shield, L , eps_1, eps_2, eps_s, eps_s, T1,T2) "....heat transfer with shield" Q12_shield = 0.15 * Q12 "....given condition that heat transfer with shield is 15 % of that without shield" T_shield = Temp_Shield_concentric_cyl(R1, R2,R_shield, L , eps_1, eps_2, eps_s, eps_s, T1,T2) Results:

Thus: Q12 without shield = 3206 W …. Ans. Q12 with shield = 480.9 W …. Ans. Their ratio is = (480.9 * 100) / 3206 = 15 % …. Verified Emissivity of shield to achieve this = eps_s = 0.0875 …. Ans. Temp of shield ( eps_s = 0.0875) is: 692.7 K … Ans. ========================================================================= "Prob.D3.5. Liquid nitrogen (LN2) is stored in a dewar made of two concentric spheres, with the space between them evacuated. The inner sphere has an outer dia D1 = 1 m and for outer sphere the inner dia D2 = 1.2 m and emissivities eps1 = 0.2 and eps2 = 0.2. Temperatures are maintained at temps T1 = 78 K and T2 = 300 K respectively. A coaxial shield of diameter D3 = 1.1 m and emissivity eps_s = 0.05 on both its surfaces is placed between the two spheres. If the latent heat of vaporization of LN2 is 2 x 10^5 J/kg, determine: (a) the boil-off rate of liquid nitrogen when there is no shield, (b) the boil-off rate of liquid nitrogen when the shield is present. What is the equilibrium temp of the shield at that time? (c) Plot the variation of boil-off rate and the shield temp as emissivity of shield (on its either side) varies from 0.05 to 0.4."

EES Solution:

"Data:" R1 = 0.5 "[m]" R2 = 0.6"[m]" R_shield = 0.55 "[m]" eps_1 = 0.2 eps_2 = 0.2 eps_s1 = 0.05 eps_s2 = 0.05 T1 = 78 "[K]" T2 = 300 "[K]" h_fg = 2E05 [J/kg] “Calculations:” Q12 = Q12_concentric_spheres(R1, R2, eps_1, eps_2,T1,T2)"[W]...Q12 with no shield" Q12_shield = Q12_concentric_sph_one_shield(R1, R2,R_shield, eps_1, eps_2, eps_s1, eps_s2, T1,T2) "[W]....Q12 with shield" T_shield = Temp_Shield_concentric_spheres(R1, R2,R_shield, eps_1, eps_2, eps_s1, eps_s2, T1,T2)"[K]....Temp of shield" BoilOff_with_shield = Abs(Q12_shield) / h_fg * 3600"[kg/h]....boil off rate of liq. nitrogen, when there is shield" BoilOff_no_shield = Abs(Q12) / h_fg * 3600"[kg/h]....boil off rate of liq. nitrogen, when there is no shield" Results:

Thus:

LN2 Boil off rate without shield = 3.324 kg/h …. Ans. LN2 Boil off with shield = 0.6462 kg/h …. Ans. Equilibrium temp of shield = 254.7 K … Ans. -----------------------------------------------------------------------------------------------------

To plot the variation of boil-off rate and the shield temp as emissivity of shield (on its either side) varies from 0.05 to 0.4: First, compute the Parametric Table, keeping both eps_s1 = eps_s2 = 0.05, 0.1, 0.15 ….. 0.4.


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�'�� E'+�%F"�

�

�

Let the reading shown by the thermometer be Tc. This reading, however, does not represent the true temperature of the fluid Tf, since the thermometer bulb will lose heat by radiation to the walls of the channel which are at a lower temperature Tw (which is usually the case).

The temperature Tc shown by the thermometer will be some value in between Tf and Tw .

We wish to find out the true temperature of the fluid Tf , by knowing the thermometer reading Tc.

Making an energy balance on the thermometer bulb, in steady state, we have:�

Without radiation shield:�

qconv to the bulb = qrad from the bulb�

�

��

i.e. h A c. T f T c

. ε c A c. σ. T c

4 T w4.

i.e. T f T cε c σ. T c

4 T w4.

h.....(13.79)

�

�

where Ac = surface area of thermometer bulb,

εc = emissivity of thermometer bulb surface�

Eqn. (13.79) gives the true temperature of the fluid Tf.

Second term on the RHS of eqn. (13.79) represents the error in temperature measurement due to radiation effect.��

Energy balance on the thermometer bulb:

�

Heat transferred to the bulb from the fluid by convection =

Heat transferred from the bulb to the shield by radiation

i.e. h A c. T f T c

.σ T c

4 T s4.

1 ε c

A c ε c.

1

A c F cs.

1 ε s

A s ε s.

....(13.80)

In eqn. (13.80), Fcs = view factor of thermometer bulb w.r.t the shield and is, generally equal to 1.

Making an energy balance on the shield:�

heat transferred to shield from the fluid by convection +

heat transferred to shield from bulb by radiation =

heat transferred from shield to walls by radiation

�

i.e. 2 A s. h. T f T s

.σ T c

4 T s4.

1 ε c

A c ε c.

1

A c F cs.

1 ε s

A s ε s.

ε s A s. σ. T s

4 T w4. ....(13.81)

��

�

where, A s = area of shield on one side

ε s = emissivitty of shield surface

A c = area of bulb surface

ε c = emissivity of bulb surface

F cs = view factor of bulb w.r.t. shield�

In the first term of the above eqn. factor 2 appears since convective heat transfer to the shield occurs on both surfaces of the shield. Also, in writing the RHS, the inherent assumption is that:

F sw 1 ...view factor between the shield and the walls

and,

A s

A w0 ...i.e. surface area of shield is negligible compared to the area of the channel

walls

Solving eqns. (13.80) and (13.81) simultaneously, we obtain the shield temperature Ts and the thermometer reading Tc, (if Tf is known), or Tf (if Tc is known).

�

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"Prob.E2.1. Temp of hot exhaust gases flowing in to a room whose walls are maintained at a temperature Tw = 20 C is measured with a thermocouple(TC) whose emissivity is 0.6. The thermocouple shows a reading of 500 C. If the convective heat transfer coefficient between the flowing gases and the thermocouple is h = 200 W/(m2.C), find out the true temperature of the flowing stream, and the radiation error. (b) Now, if a radiation shield (�s = 0.3) is placed between the thermocouple and the walls, what will be new value of Tc read by the thermocouple? And how much is the temperature error? (c) Plot the radiation error and shield temp as emissivity of shield varies from 0.05 to 0.5." EES Solution: "Data:" T_w = 293 [K] "...temp of walls" T_c = 773 [K] "..TC reading" epsilon_c = 0.6 "...emissivity of TC" h = 200 [W/m^2-K] "...conv. heat transfer coeff." epsilon_s = 0.3 "...emissivity of shield"

sigma = 5.67E-08 [W/m^2-K^4] "..Stefan-Boltzmann const." "Calculations:" "When there is no radiation shield:" T_f = T_c + epsilon_c * sigma * (T_c^4 - T_w^4) / h "...finds true temp of fluid" "When radiation shield is present:" "By heat balance on the Thermocouple bead:" h * (T_f - T_c_with_shield) = sigma * epsilon_c * (T_c_with_shield^4 - T_s^4) ".... eqn, (A)." "By heat balance on the Shield:" 2 * h * (T_f - T_s) = epsilon_s * sigma * (T_s^4 - T_w^4) "...eqn. (B)." "Solving eqns. (A) and (B) simultaneously, we get thermocouple reading (T_c) and the Shield temp. ( T_s)" "Radiation error:" Error_no_shield = T_f - T_c "[deg.C].... error when there is no shield" Error_with_shield = T_f - T_c_with_shield "[deg.C].... error when the shield is present" Results:

Thus: True temp of fluid = T_f = 832.5 K …. Ans. TC reading when the shield is present = 827.5 deg.C … Ans. Shield temp = T_s = 814.1 K … Ans. Radiation error when there is no shield = 59.48 deg.C … Ans. Radiation error when shield is present = 5.021 deg.C … Ans. --------------------------------------------------------------------------------------------------- (c) Plot the ‘Radiation error’ and shield temp as emissivity of shield varies from 0.05 to 0.5: First, compute the Parametric Table:

EES Procedures and Functions for Radiation heat transfer calculations

Engineering

Transcript of EES Procedures and Functions for Radiation heat transfer calculations