Lect - 4 Steady conduction One Dimensional(1).pptx

Steady Conduction One DimensionalDr. Senthilmurugan S. Department of Chemical Engineering IIT Guwahati - Part 4

Slabs, cylinders and spheres; Critical thickness of insulation

April 28, 2023 | Slide 2

Objectives

• Understand the concept of thermal resistance and its limitations, and develop thermal resistance networks for practical heat conduction problems

• Solve steady conduction problems that involve multilayer rectangular, cylindrical, or spherical geometries

• Develop an intuitive understanding of thermal contact resistance, and circumstances under which it may be significant

• Identify applications in which insulation may actually increase heat transfer

• Analyze finned surfaces, and assess how efficiently and effectively fins enhance heat transfer

• Solve multidimensional practical heat conduction problems using conduction shape factors


Contents

Steady state conduction heat transfer one dimension Steady Heat Conduction in Plane Walls Generalized Thermal Resistance Networks Thermal Contact Resistance Heat conduction in cylinders and spheres The Overall Heat-Transfer Coefficient Critical Thickness of Insulation Steady Heat Conduction with Heat-Source Systems Cylinder with Heat Sources Conduction-Convection Systems

Fins Thermal Contact Resistance


Steady State Heat Conduction in Plane Walls

Rate of heat transfer into the wall - Rate of heat transfer out of the wall = Rate of change of the energy of the wall

Therefore, the rate of heat transfer into the wall must be equal to the rate of heat transfer out of it. In other words, the rate of heat transfer through the wall must be constant

q𝑞𝑖𝑛+𝑞𝑜𝑢𝑡=

𝜕𝐸𝜕𝑡

0


Steady State Heat Conduction in Plane Walls

If the system is in a steady state, i.e., if the temperature does not change with time, then the problem is a simple one, and we need only integrate Fourier’s law of heat conduction equation and substitute the appropriate values to solve for the desired quantity.

When k is constant

One Dimension & Constant Area

𝑞𝐴=−𝑘 𝑑𝑇

𝑑𝑥

𝑞∫0

𝑥

𝑑𝑥=−𝑘∫𝑇 1

𝑇 2

𝑑𝑇

𝑞=𝑘 𝐴𝑥 (𝑇1−𝑇 2)


Thermal Resistance Concept

The rate of heat conduction through a plane wall is proportional to the average thermal conductivity, the wall area, and the temperature difference, but is inversely proportional to the wall thickness.

heat conduction through a plane wall can be rearranged as

The thermal resistance of the wall against heat conduction or simply the conduction resistance of the wall.

The thermal resistance of a medium depends on the geometry and the thermal properties of the medium

This equation for heat transfer is analogous to the relation for electric current flow I, expressed as (Ohm’s law in electric-circuit theory)

Similarity with Electric resistance

𝑞=𝑘 𝐴𝑥 (𝑇1−𝑇 2)

𝑞=𝑇1−𝑇 2

𝑅𝑤𝑎𝑙𝑙

𝐼=𝑉 1−𝑉 2

𝑅𝑒

𝑅𝑊𝑎𝑙𝑙=𝑥𝑘𝐴=

𝑇 1−𝑇 2

𝑞

𝑅𝑒=𝐿𝜎𝑒

=𝑉 1−𝑉 2

𝐼electrical conductivity

Electric resistance


Analogy between thermal and electrical resistance concepts.

𝑅𝑒=𝐿𝜎𝑒

=𝑉 1−𝑉 2

𝐼

𝑅𝑊𝑎𝑙𝑙=𝑥𝑘𝐴=

𝑇 1−𝑇 2

𝑞

q


One-dimensional heat transfer through a composite wall and electrical analog

T1

T2

T3

T4

Assumption: Resistance for wall contact is negligible


Generalized thermal resistance networks

The thermal resistance concept or the electrical analogy can also be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements.

q

q = q1+q2

Thermal resistance network for two parallel layers.


Generalized thermal resistance networks

The total rate of heat transfer through this composite system can again be expressed as

The result obtained is somewhat approximate, since the surfaces of the third layer are probably not isothermal, and heat transfer between the first two layers is likely to occur.

q


Generalized Thermal Resistance Networks

𝑞=𝑇 1−𝑇5

𝑅𝐴+𝑅𝐵𝐶𝐷+𝑅𝐸+𝑅𝐹𝐺

𝑅𝐵𝐶𝐷=1𝑅𝐵

+1𝑅𝐶

+1𝑅𝐷

𝑅𝐹𝐺=1𝑅𝐹

+1𝑅𝐺

𝑅𝑇=𝑅𝐴+𝑅𝐵𝐶𝐷+𝑅𝐸+𝑅𝐹𝐺


Generalized Thermal Resistance Networks Method

Two assumptions in solving complex multidimensional heat transfer problems by treating them as one-dimensional using the thermal resistance network are

(1) any plane wall normal to the x-axis is isothermal (i.e., to assume the temperature to vary in the x-direction only)

(2) any plane parallel to the x-axis is adiabatic (i.e., to assume heat transfer to occur in the x-direction only)

These two assumptions result in different networks (different results).

The actual result lies between these two results

AnalysisAny plane wall normal to the

x-axis is isothermal

Any plane wall parallel to the x-axis is adiabatic

x

x

y

y

T1

T2

T1

T1

q

q



Newton’s law of cooling for convection heat transfer rate can be rearranged as

Convection resistance

With convection heat transfer

𝑞=h𝐴𝑠 (𝑇 𝑠−𝑇 ∞ )

𝑞=𝑇 𝑠−𝑇 ∞

𝑅𝐶𝑜𝑛𝑣

𝑅𝑐𝑜𝑛𝑣=1h𝐴𝑠=

𝑇 𝑠−𝑇 ∞

𝑞 H∞ , , then

At surfaces where boiling and condensation occur



When the wall is surrounded by a gas, the radiation effects. The rate of radiation heat transfer between a surface of emissivity and area As at temperature Ts and the surrounding surfaces at some average temperature Tsurr can be expressed as

Radiation resistance

Radiation heat transfer coefficient

Temperature must be in kelvin in the evaluation of hrad.

With radiation heat transfer

h𝑟𝑎𝑑=ϵ σ 𝐴𝑠 (𝑇𝑠+𝑇 𝑠𝑢𝑟𝑟 ) (𝑇 𝑠2+𝑇 𝑠𝑢𝑟𝑟

2 )

𝑅𝑟𝑎𝑑= 1h𝑟𝑎𝑑 𝐴𝑠=

𝑇 𝑠−𝑇 𝑠𝑢𝑟𝑟

𝑞


Thermal contact resistance

The temperature drop at plane 2, the contact plane between the two materials, is said to be the result of a thermal contact resistance

Thermal-contact-resistance effect physical situation

Temperature profile.

where the quantity 1/hcA is called the thermal contact resistance and hc is called the contactcoefficient.



No real surface is perfectly smooth, and the actual surface roughness is believed to play a central role in determining the contact resistance.

There are two principal contributions to the heat transfer at the joint:

The solid-to-solid conduction at the spots of contact

The conduction through entrapped gases in the void spaces created by the contact

The second factor is believed to represent the major resistance to heat flow, because the thermal conductivity of the gas is quite small in comparison to that of the solids.

TheoryActual (imperfect) thermal contact

𝑞=𝑞𝑐𝑜𝑛𝑡𝑎𝑐𝑡+𝑞𝑔𝑎𝑝



Designating the contact area by Ac and the void area by Av, we may write for the heat flow across the joint

where Lg is the thickness of the void space and kf is the thermal conductivity of the fluid which fills the void space. The total cross-sectional area of the bars is A. Solving for hc, the contact coefficient, we obtain

Theory


Thermal Contact Resistance

In most instances, air is the fluid filling the void space and kf is small compared with kA and kB. If the contact area is small, the major thermal resistance results from the void space.

The main problem with this simple theory is that it is extremely difficult to determine effective values of Ac, Av, and Lg for surfaces in contact.

Thermal contact resistance can be reduced markedly, perhaps as much as 75 percent, by the use of a “thermal grease” like “Dow 340”

From the physical model, we may tentatively conclude:

The contact resistance should increase with a decrease in the ambient gas pressure when the pressure is decreased below the value where the mean free path of the molecules is large compared with a characteristic dimension of the void space, since the effective thermal conductance of the entrapped gas will be decreased for this condition.

The contact resistance should be decreased for an increase in the joint pressure since this results in a deformation of the high spots of the contact surfaces, thereby creating a greater contact area between the solids.

Theory


Thermal Contact ResistanceMeasurement Technique

𝑅𝑐=∆𝑇 𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒

𝑞𝐴

The thermal contact resistance can be determined from equation given below by measuring the temperature drop at the interface and dividing it by the heat fluxunder steady conditions


Thermal contact conductance of some metal surfaces in air (from various sources)


Heat conduction in cylinders

Consider a long cylinder of inside radius ri, outside radius ro, and length L,

We expose this cylinder to a temperature differential Ti −To

For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction, so that the only space coordinate needed to specify the system is r.

Again, Fourier’s law is used by inserting the proper area relation. The area for heat flow in the cylindrical system is


Heat conduction in cylinders

Fourier’s law

With the boundary conditions

The solution to Fourier’s law Equation is

Thermal resistance

𝑞𝑟∫𝑟 𝑖

𝑟 𝑜 𝑑𝑟𝑟 =−2 𝜋 𝐿𝑘∫

𝑇 𝑖

𝑇 𝑜

𝑑𝑇


One-dimensional heat flow through multiple cylindrical sections and electrical analog.

The thermal-resistance concept may be used for multiple-layer cylindrical walls just as it was used for plane walls.


Fourier’s law

With the boundary conditions

The solution to Fourier’s law Equation is

Heat conduction in Spheres

𝑞𝑟∫𝑟 𝑖

𝑟 𝑜 𝑑𝑟𝑟 2 =−4 𝜋𝑘∫

𝑇 𝑖

𝑇𝑜

𝑑𝑇

ri

ro

𝐴=4 𝜋𝑟 2

𝑞𝑟=−4 𝜋𝑟2𝑘 𝑑𝑇𝑑𝑟

𝑅=

1𝑟 𝑖− 1𝑟𝑜

4 𝜋𝑘


System Flux Resistance Driving force

Electrical I R V

Cartesian Conduction q T

Cylindrical Conduction q T

Conduction through sphere q T

Convection q T

Radiation q T


Heat Conduction in Composite Cylinder

Net Heat transfer rate

Net Heat transfer rate among the individual layer is equal

One-dimensional heat flow through multiple cylindrical sections

Electrical analogy


Over all Heat Transfer Coefficient

Consider the plane wall shown in Figure, exposed to a hot fluid A on one side and a cooler fluid B on the other side. The heat transfer is expressed by


Plane wall

Electrical analogy

Overall heat transfer through a plane wall



The overall heat transfer by combined conduction and convection is frequently expressed in terms of an overall heat-transfer coefficient U,

where A is some suitable area for the heat flow


Hollow cylinder with convection boundaries

Electrical analogy

Hollow cylinder with convection boundaries.



Note that the area for convection is not the same for both fluids in this case, these areas depending on the inside tube diameter and wall thickness. The overall heat transfer would be expressed by

Variable heat transfer area system


Critical Insulation Thickness

We know that adding more insulation to a wall always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate. This is expected, since the heat transfer area A is constant, and adding insulation always increases the thermal resistance of the wall without increasing the convection resistance.

Adding insulation to a cylindrical pipe or a spherical shell, however, is a different matter. The additional insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface because of the increase in the outer surface area for convection. The heat transfer from the pipe may increase or decrease, depending on which effect dominates.

Thermal heat transfer

An insulated cylindrical pipe

exposed to convection from the outer surface


Critical Insulation Thickness

The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as

Cylinder

Rconv

Rins

Rtotal

q

q

q

q

q

Rtotal=Rconv+Rins

𝑑𝑞𝑑𝑟 2

=0=

2𝜋 𝐿 (𝑇 1−𝑇 ∞ )( 1𝑘𝑟2

− 1h𝑟2

2 )[ 𝑙𝑛 (𝑟 2/𝑟 1 )

𝑘 + 1𝑟 2h ]

2

𝑟𝑐𝑟 ,𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟=𝑟 2=𝑘h 𝑟𝑐𝑟 , h𝑠𝑝 𝑒𝑟𝑒=𝑟2=

2𝑘h


Importance of Critical Insulation Thickness

Should we always check and make sure that the outer radius of insulation sufficiently exceeds the critical radius before we install any insulation? Probably not, as explained here

The value of the critical radius rcr is the largest when k is large and h is small.

Lowest value of h encountered in practice is ~ 5 W/m2·K for the case of natural convection of gases, and k ~ 0.05 W/m·K for common insulator

The largest value of the critical radius we are likely to encounter is

Do we need to be concerned about the critical radius of insulation when insulating hot-water pipes or even hot-water tanks.

This value would be even smaller when the radiation effects are considered. The critical radius would be much less in forced convection, often less than 1 mm, because of much larger h values associated with forced convection. 1 mm in very small considering practice

Therefore, we can insulate hot-water or steam pipes freely without worrying about the possibility of increasing the heat transfer by insulating the pipes.

But, the radius of electric wires may be smaller than the critical radius. Therefore, the plastic electrical insulation may actually enhance the heat transfer from electric wires and thus keep their steady operating temperatures at lower and thus safer levels.

𝑟𝑐𝑟=𝑘h=

0.055 =0.01𝑚=1𝑐𝑚


Steady Heat Conduction with Heat-Source Systems

Consider the plane wall with uniformly distributed heat sources shown in Figure. The thickness of the wall in the x direction is 2L.

Assumptions: It is assumed that the dimensions in the other directions are sufficiently large that the heat flow may be considered as one dimensional.

The heat generated per unit volume is , and we assume that the thermal conductivity does not vary with temperature.

Practical Applications: Passing a current through an electrically conducting material. Nuclear fuel rod, nuclear weapons

One dimensional – Plane Wall



The differential equation that governs the heat flow is

BC1: Interface temperature

BC2 : Thermal Symmetry

Solution

Substituting BC2 C1=0 Substituting BC1

Substitute Equation (2) in (1)

One dimensional – Plane Wall

�̇�𝑘 +

𝑑2𝑇𝑑𝑥2 =0

𝑥=∓𝐿→𝑇=𝑇𝑤

𝑥=0→ 𝑑𝑇𝑑𝑥 =0∧𝑇=𝑇0

𝑑𝑇𝑑𝑥 =− �̇�

𝑘 𝑥+𝐶1

𝑇=− �̇�2𝑘 𝑥2+𝑥𝐶1+𝐶2

𝐶2=𝑇𝑤+�̇�

2𝑘 𝐿2

𝑇 −𝑇𝑤=�̇�

2𝑘 (𝐿2− 𝑥2 )

𝑇𝑤−𝑇 0=−�̇�

2𝑘 𝐿2

𝑇 −𝑇0=−�̇�

2𝑘 𝑥2

(2)(3)→

𝑇 −𝑇0

𝑇𝑤−𝑇 0= 𝑥2

𝐿2

𝑥=0→𝑇=𝑇 0

(1)

(2)

(3)

(4)


Steady Heat Conduction with Heat-Source SystemsOne dimensional – Cylinder



The differential equation that governs the heat flow for cylindrical coordinates is

BC1: Interface temperature

BC2 : Thermal Symmetry

Solution

One dimensional – Cylinder

𝑟 𝑑2𝑇𝑑𝑟2 +

𝑑𝑇𝑑𝑟 =

𝑑𝑑𝑟 (𝑟 𝑑𝑇

𝑑𝑟 )=− �̇�𝑘 𝑟

�̇�𝑘 +

𝜕2𝑇𝜕𝑟2 +

1𝑟𝜕𝑇𝜕𝑟 +

1𝑟 2

𝜕2𝑇𝜕𝜙2 +

𝜕2𝑇𝜕 𝑧 2 =

1𝛼

𝜕𝑇𝜕𝑡

𝑟=𝑅→𝑇=𝑇𝑤

𝑟=0→ 𝑑𝑇𝑑𝑟 =0∧𝑇=𝑇0

𝑇=− �̇�4𝑘 𝑟 2+𝐶1ln 𝑟 +𝐶2

𝑟 𝑑𝑇𝑑𝑟 =− �̇�

𝑘 𝑟2+𝐶1

Substituting BC2 C1=0Substituting BC1 𝐶2=𝑇𝑤+

�̇�4 𝑘 𝑅2

𝑇 (𝑟 )−𝑇𝑤=− �̇�4 𝑘 (𝑅2−𝑟 2)

𝑇 0−𝑇𝑤=− �̇�4 𝑘 𝑅2𝑟=0→𝑇=𝑇 0

(2)(3)→

𝑇 −𝑇0

𝑇𝑤−𝑇 0= 𝑟2

𝑅2

(1)

(2)

(3)

(4)

Substitute Equation (2) in (1)𝑇 (𝑟 )−𝑇 0=−�̇�4𝑘 𝑟2

= ¿𝑇𝑤



Derive an expression for the temperature distribution in a hollow cylinder with heat sources that vary according to the linear relation

With the generation rate per unit volume at r =ri. The inside and outside temperatures are T =Ti at r =ri and T =To at r =ro.

Derive an expression for the temperature distribution in a hollow sphere with heat sources that vary according to the linear relation

With the generation rate per unit volume at r =ri. The inside and outside temperatures are T =Ti at r =ri and T =To at r =ro.

Assignment problem - Hollow Cylinder and Sphere


Conduction-Convection Systems

The heat that is conducted through a body must frequently be removed (or delivered) by some convection process

Energy in left face=energy out right face + energy lost by convection

Convection Heat loss (Newton’s Law of Cooling)

FIN

2

2

co

x c

x dx c cx dx

nv

dTq kAdxdT dT d Tq kA kA dxdx dx dx

q hP T T dx

cox x dx nvq q q


Conduction-Convection Systems

The energy balance yields

Let

To solve second order differential equation 2 BC are must.

BC1 : x=0 =

The other boundary condition (BC2) depends on the physical situation. Several cases may be considered:

Case 1: The fin is very long, and the temperature at the end of the fin is essentially that of the surrounding fluid.

Case 2: The end of the fin is insulated so that dT/dx=0 at x=L. adiabatic fin tip

Case 3: The fin is of finite length and loses heat by convection from its end

Case 4: Specified temperature at fin tip

FIN

2

2 0p

c

hAd T T Tdx kA

2

2 0c

d hPdx kA

22

2 0d mdx

2

c

hPmkA


Solution for Case 1: Very Long FIN


BC1 : = at x=0 BC2 : θ =0 at x = ∞ Solution for differential equation

Temperature at the end of the fin is essentially that of the surrounding fluid

22

2 0d mdx

2

c

hPmkA

1 2mx mxC e C e

0 1 21:BC C C 22 : 0BC C

0 0

mxT T eT T

0 0

00 0

L

cx

mc c

dq kA hP dxdx

kA m e hPkA

∫


Case 2: The end of the fin is insulated


BC1 : = at x=0 BC2 : at x = L Solution for differential equation

Solving for the constants C1 and C2, we obtain

Adiabatic Fin tip

22

2 0d mdx

2

c

hPmkA

1 2mx mxC e C e 0 1 21:BC C C

1 22 : 0 mx mxdBC m C e C edx

2 2

0 0

cosh1 1 cosh

mx mx

mL mL

m L xT T e eT T e e mL

ApplicationsTo provide safe temperature

range at end of fins

0 0

0 tanh

L

cx

c

dq kA hP dxdx

hPkA mL

∫

L


Case 3: Convection at End


BC1 : = at x=0 BC2 :at x = L Solution for differential equation

The fin is of finite length and loses heat by convection from its end

22

2 0d mdx

2

c

hPmkA

2 20 0 1 1

cosh sinh

cosh sinh

mx mx

mL mL

T T e eT T e e

hm L x m L xmkhmL mLmk

0 0

0

sinh cosh

cosh sinh

L

cx

c

dq kA hP dxdx

hmL mLmkhPkAhmL mLmk

∫


Case 3: Convection at End

Extension of Case 2’s solution to case 3

The fin is of finite length and loses heat by convection from its end

2 20 0 1 1

coshcosh

c c

mx mx

mL mL

c

c

T T e eT T e e

m L xmL

0 0

0 tanh

cL

cx

c c

dq kA hP dxdx

hPkA mL

∫

cc

AL Lp


Case 4:Specified temperature at fin tip


BC1 : = at x=0 BC2 : =at x = L Solution for differential equation

Heat flow

When L∞ solution will be similar to infinite long FIN

22

2 0d mdx

2

c

hPmkA

0

0

sinh sinh

sinh

L mx m L x

mL

0 0

00

coshsinh

L

cx

Lc

dq kA hP dxdx

mLhPkA

mL

∫

0 0

mxT T eT T

0cq hPkA


Fin Temperature Profile All four case


FIN Efficiency

Efficiency of the fin is defined as ratio between actual heat transfer rate from the fin and Ideal heat transfer rate from the fin if the entire fin were at base temperature

For Case 1: Very Long FIN

For Case 1: Very Long FIN

,max

finfin

fin

qq

0

0

1cfin

hPkAhPL mL

2 phAm

kA

0

0

tanh tanhcfin

hPkA mL mLhPL mL


Fin Effectiveness

Fins are used to enhance heat transfer, and the use of fins on a surface cannot be recommended unless the enhancement in heat transfer justifies the added cost and complexity associated with the fins.

In fact, there is no assurance that adding fins on a surface will enhance heat transfer. The performance of the fins is judged on the basis of the enhancement in heat transfer relative to the no-fin case.

Defined as ratio between Heat transfer rate from the total fin area (Afin) and Heat transfer rate from the surface of area Ab

finfin

no fin

qq


Fin Effectiveness vs Fin Efficiency

Both the fin efficiency and fin effectiveness are related to the performance of the fin, but they are different quantities. However, they are related to each other by

For case 1: Long Fin (single strip)

For case 1: Long Fin with multi strips

0

0

fin fin fin finfin fin

no fin b b

q hA Aq hA A

finfin fin

b c

A kpA hA

0 0

0

fin un fin un fin fin finfin

no fin no fin

q q hA hAq hA

un fin fin finfin

no fin

A AA


0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

mL

qfin

/qlo

ng fi

n

1%

mL=

2.5,

99%

23.8 %

m

L=1,

76

.2 %

Optimal Fin Length

tanhfin

long fin

qmL

q

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