EME3233

THERMODYNAMICS & HEAT TRANSFER ASSIGNMENT 1

Student Name : Shakil Ahmed

Student ID : SCM - 022732

Date of submission : 6/11/2015

Lecturer : Dr. Yau

Programme : B.Eng (Hons) Mechanical Engineering

X= 7

Using 7 in all the answers

A1.

Solution

T1=32℃

A= 27 m2

Thickness = 0.30 m

Sample calculation

Table 1 : Heat loss Q (W) and Outside wall temperature ( ℃) values

Outside wall temperature ( ℃)

Heat loss Q (W) for thermal conductivity =1

Heat loss Q (W) for thermal conductivity =0.75

Heat loss Q (W) for thermal conductivity =1.25

-8 3600 2700 4500

-3 3150 2362.2 3937.5

0 2880 2160 3600

2 2700 2025 3375

12 1800 1350 2250

17 1350 1012.5 1687.5

22 900 675 1125

27 450 337.5 562.5

32 0 0 0

37 -450 -337.5 -562.5

45 -1170 -877.5 -1462.5

-20 -10 0 10 20 30 40 50

-2000

-1000

0

1000

2000

3000

4000

5000

Heat loss Q (W) vs. Outside wall temper-ature T2(℃)

Y-Valuesy2y3

oUTSIDE WALL TEMPERATUE T2( )℃

Heat

loss

Q (W

)

(b) Assumptions :

1. One dimensional conduction in the x direction 2. Steady state conditions3. Constant properties4. Outside wall temperature is that of the ambient air

As looking at the results from graph and table, observing that q x varies linearly with T2 , and the steeper temperature gradient is larger the heat transfer rate. The magnitude of the heat rate increases with increasing thermal conductivity for the same T1 and T2

For steady state conditions with constant thermal conductivity, the temperature in a plane wall will be linear , since from Fourier’s law , dT/dx is a constant

A2

Solution:

(a) Assume the heat flow is positive if it is from the surface to the fluid. From Newton’s law of cooling

h= 457 w /mπ (0.032m)(307℃−47℃)

=17.48W /m2 K

in which P’=q/L is the heat loss per unit length. The above equation may be used to calculate the value of h at each velocity, and the results are shown in the table and the figure below.

Air velocity, V(m/s) 1 2 3 4 5

Power, P’ (W/m) 457 665 990 1514 1970

Convection coefficient, h W/m2K 17.48 25.44 37.86 57.92 75.56

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.50

10

20

30

40

50

60

70

80

h (W/m2.K) vs. V (m/s).

Air velocity V (m/s)

Coeffi

cient

, h (W

/m^2

.K)

Graph . graph of h (W/m2.K) vs. V (m/s).

(b)

plotting h vs. V on log-log coordinates. Selecting C =17.38 W/m2 ⋅K(s/m)n , assuring a match at V = 1.7, and easily find the exponent n from the slope of the h vs. V curve. With different values of n = 0.8, 0.6 and 0.5, choosing n = 0.6 is a valid choice because it over laps original curve . Hence, C = 17.38 and n = 0.6.

1 10 10010

100

1000

Series2Linear (Series2)n=0.5Power (n=0.5)n=0.6Linear (n=0.6)n=0.8Linear (n=0.8)

Air velocity V (m/s)

Coeffi

cient

. h

(W/m

^2.K

)

Graph of h (W/m2.K) vs. V (m/s).

A3

i) The process of heat transfer from the kitchen air to the refrigerated space is transient as the thermal activity in the kitchen change with time. But we would see this situation as a steady heat transfer problem if the conditions are changed such as thermostat settings are lowest and very high temperature of kitchen. The Heat transfer in the refrigerated space is 3D because heat will enter from all six sides of the refrigerator. But the heat transfer takes place in the direction normal to the surface, and thus it can be described as one-dimensional. This problem can also be solved by taking heat transfer from all the six sides, and then lastly can add up all the values and give a total heat transfer.

ii) Heat transfer of an oven through the walls, door, the top and bottom parts is transient in general since the thermal conditions in kitchen and oven, change with time. But we can analyze this problem as a steady heat transfer problem if the conditions are changed such as the highest temperature setting for the oven and lowest temperature of the kitchen. The Heat transfer in the oven is 3D because heat will enter from all six sides of the refrigerator. But the heat transfer takes place in the direction normal to the surface, and thus it can be described as one-dimensional. This problem can also be solved by taking heat transfer from all the six sides, and then lastly can add up all the values and give a total heat transfer.

iii) Assumptions

1 Heat transfer is given to be steady and one-dimensional.

2 Thermal conductivity is given to be constant.

3 There is no heat generation in the medium.

4 The top surface at x = L is subjected to convection and the bottom surface at x = 0 is subjected to uniform heat flux.

Lets start with the general heat conduction equation

We are told that the process is steady and one-dimensional so we can ignore the time, y and z derivative terms. We are also told that there is a heat input of 1257 W from an electric heater of which 92% enters the bottom of the pan

Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as

A4

(a)

Assumptions:

- Steady state heat conduction

- No contact resistance at interfaces

- No radiation heat transfer in the system

Analysis (a) The representative surface area is A = 7.12 x 15 = 106.8 m2. The thermal resistance network and the individual thermal resistances are

KA = KF = 9 KB = 15 KC = 27 KD = 22KE = 42

h = 7.12m L = 15m t = 7.22m

Area for section B and C: A2 = 15 x (7.12/3) = 35.6m2

Area for section D and E: A3 = 15 x (7.12/2) = 53.4m2

RA ¿(L

kA)A ¿(

0.328(9)(106.8)

) = 0.00034 K/W

RB ¿(L

kA)B ¿(

1.641(15)(35.6)

) = 0.003 K/W

RC ¿(L

kA)C ¿(

1.641(27)(35.6)

) = 0.0017 K/W

RD ¿(L

kA)D ¿(

3.282(22)(53.4 )

) = 0.0028 K/W

RE ¿(L

kA)E ¿(

3.282(42)(53.4)

) = 0.0015 K/W

RF ¿(L

kA)A ¿(

1.942(9)(106.8)

) = 0.002 K/W

1R '

= 1RB

+ 1RC

+ 1RC = 1

0.003+ 1

0.0017+ 1

0.0017 = 0.00066 K/W

1R ' '

= 1RD

+ 1ℜ = 1

0.0028+ 1

0.0015 = 0.000977 K/W

Rtotal = RA + R’ + R’’ + RF = 0.00034 + 0.00066 + 0.000977 + 0.002 = 0.00198 K/W

Q '=T 1−T 2R total = 643−443

0.00198 = 1.01 x 105 W

Qtotal=(1.01 x105) 7.12 x 15106.8 = 1.01 x 105 W

(b)

Rtotal = RA + R’ + R’’ + RF = 0.00034 + 0.00066 + 0.000977 + 0.002 = 0.00198 K/W

Q '=T 1−T 2R total = 643−443

0.00198 = 1.01 x 105 W

Qtotal=(1.01 x105) 7.12 x 15106.8 = 1.01 x 105 W

Rtotal = RA + R’ = 0.00034 + 0.00066 = 0.001 K/W

Q '=T 1−TR total => T = T1 – (Q’)(Rtotal) => 643 – (1.01 x 105)(0.001) = 542K

(c)

Q’ = ∆ TR F => ∆ T=¿ (Q’)(RF) => (1.01 x 105)(0.002) = 202K

A5

Properties: The thermal conductivity of fiberglass insulation is given to be k = 0.035 W/m.°C.

(a)

Ao¿ π DL=π (0.1 ) (50 )=15.71 m2

Q’bare = hoA(Ts – Tair) ¿ (20 ) (15.71 ) (150−15 )=42,412W

(b) Loss per year

Q=Q' ∆ t=42,390 (365 ) (24 ) (3600 )=1.336 x109 KJ/Year

Gas consumption at 75% efficiency

Qgas = 1.336 x 109

0.75 ( 1 therm

105,500 KJ ) = 16,885 therms/Year

Annual Cost = 16,885 (0.52) = 8780 MYR /Year

Assumptions

1. Heat transfer is steady since there is no indication of any change with time.

2. Heat transfer is one-dimensional since there is thermal

symmetry about the center line and no variation in the axial

direction.

3. Thermal conductivity is constant.

4. The combined heat transfer coefficient on the outer surface remains constant even

after the pipe is insulated.

In order to save 90% of the heat loss and thus to reduce it to 0.1 x 42,412 = 4241 W, the

thickness of insulation needed is determined from

Q’insulated = T s – T air

R0+Rinsulation

=

T s – T air

1ho Ao

+ln (r 2

r1)

2 πkL

(150−15)

120(2 πr 2)(50)

+ln ( r 2

0.05)

2 π (0.035)(50)

=¿4241 w

r 2=0.0692 m

Then the thickness of insulation becomes tinsulation = r2 – r1 = 6.92 – 5 = 1.9 cm