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2.8 One-dimensional, steady state heat conduction with heat generation

45

The governing equation is

d 2T

+

qv

= 0

 

 

dx2

k

 

 

 

 

 

 

 

d 2T

= − qv

 

 

 

dx2

k

 

 

 

dT

= − qv

x + c1

 

dx

k

 

 

 

 

T = − qv

x2

+ c1x + c2

 

 

2

 

 

 

k

 

Boundary condition 1: At x = 0; dT /dx = 0

 

 

 

 

c1 = 0

 

 

Boundary condition 2: At x = L = 0.025m; T = T1

T1 = 200 2000[0.0252 ]

T1 = 198.7 °C

d 2T

= − qv = 2b

dx2

 

k

 

qv = 200 kW/m3

 

 

= −k

dT

 

 

 

q

 

 

 

 

x =+ L

dx

x =+ L

 

 

 

 

 

 

 

 

 

= −50(2(2000)0.025) = 5000 W/m2

 

= k

dT

 

 

 

q

 

 

 

 

dx

x =− L

 

x =− L

 

 

 

 

 

 

 

= 50(2(2000)(0.025)) = 5000 W/m2

The total heat generation is

q = qv × 2L × A = 200 × 103 × 2 × 25 × 103 q = 10000 W/m2

From Eqs. (2.152) and (2.153)

qleft + qright = 10000 W/m2

Therefore, Heat generated = Heat lost

(2.146)

(2.147)

(2.148)

(2.149)

(2.150)

(2.151)

(2.152)

(2.153)

We still use q instead of Q as we are assuming A=1m2.

46CHAPTER 2  One-dimensional, steady state heat conduction

Example 2.8: Consider steady state conduction in a composite slab with heat generation in the middle portion of the slab. The problem description is shown in Fig. 2.24.

Given T2 = 270 °C, T3 = 220 °C, determine the thermal conductivity of the middle portion of the slab and the volumetric heat generation in the middle portion of the slab, for steady state one dimensional conduction with constant properties of all the materials.

Solution:

 

 

 

 

 

 

 

 

 

 

Under steady sate

 

T2 30

 

 

 

Q =

 

 

 

 

 

 

1

 

+

 

L

a

 

 

 

 

 

 

 

 

 

 

 

 

hA

ka A

 

 

 

qleft =

 

270 30

1

 

+

0.03

 

1000

 

16

 

 

 

 

 

qleft = 83478W/m2

Similarly from the right side, we have

 

 

 

 

 

 

 

 

 

 

qright =

 

 

220 30

 

 

1

 

+

0.02

 

 

1000

50

 

 

 

 

 

 

qright = 135714 W/m2

qtotal = 83478 + 135714 = 219192 W/m2

FIGURE 2.24

Schematic representation of a composite plane wall with heat generation in the middle and convection boundary conditions at the two ends.

 

 

2.8 One-dimensional, steady state heat conduction with heat generation

47

In order to obtain the volumetric heat generation rate, qv, we use energy balance.

Total heat transferred = Total heat generated

qtotal A = qv AL

 

 

qv =

219192

= 7306.4 kW/m3

(2.154)

 

0.03

 

 

Now, for the thermal conductivity calculation, we consider only the middle wall.

The governing equation for the middle wall is

d 2T

+

qv

 

= 0

 

 

 

 

 

 

 

 

 

 

 

dx2

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dT

 

= − qv

 

x + c1

 

 

 

 

 

 

 

 

 

dx

 

kb

 

 

 

 

 

 

 

 

 

 

 

 

 

 

T = − qv

x2

 

+ c1x + c2

 

 

 

 

 

 

 

 

 

 

 

k

b

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

At the left of the middle wall: x = 0; T = T2; qleft = − k

 

dT

 

 

 

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

At the right of middle wall: x = L = 0.03m;T = T3

; qright = − k

dT

dx

At x = 0; T = T2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c2 = T2 = 270 °C

 

 

 

 

 

 

 

At x = L = 0.03m;T = T3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

220 = 7.306 ×106 × 0.032

+ c1 × 0.03+ 270

 

kb

 

 

 

 

 

 

 

 

 

 

 

 

 

At the left of the middle wall: x = 0;T = T2 ; qleft = −k

dT

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k dT

 

 

= 83478 = c k

b

 

dx

 

x=0

 

 

 

 

 

 

1

 

= 83478

 

 

 

 

 

 

 

 

 

 

 

kb

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c1

 

 

 

 

 

 

 

 

 

 

Substituting for kb in Eq. (2.158), we have

 

 

 

 

 

 

 

 

 

 

 

c1 = 1026.7 °C/m

 

 

 

 

 

 

 

kb

= 81.3 W/mK

 

 

 

 

 

 

 

(2.155)

(2.156)

(2.157)

(2.158)

(2.159)

(2.160)

What we have solved above is a typical “inverse” problem. A direct problem is one in which all causes are given and the effects are determined. An inverse problem

48 CHAPTER 2  One-dimensional, steady state heat conduction

is one in which the effects are known (typically temperature measurements in heat transfer problems), and the causes (like thermal conductivity) have to be estimated.

2.9  Fin heat transfer

For maintaining the temperatures of equipment as low as possible, it is necessary to maximize the heat transfer from them, subject to various constraints such as cost, pressure drop, and space. One practical way of increasing heat transfer from surfaces is by providing more surface area, and these extended surfaces are called fins.

The convection heat transfer from a surface is given by

Q = hAs (Tw T)

(2.161)

Here the goal is to increase “Q.”

Methods to increase Q:

1.Increase heat transfer coefficient (h): when velocity increases, h will increase. However, the pressure drop also increases.

2.Increase surface area (A): the passive method.

3.Increase (Tw T): restrictions may be there on the capabilities of the material or legislation or safety in respect of the maximum temperatures allowed.

Maximizing the heat transfer with an increase in surface area is an effective method. We accomplish this by using extended surfaces or fins on the base surface (Kraus et al., 2002). Addition of fins may marginally decrease the heat transfer coefficient. However, (hA) with fins is usually much greater than (hA) without fins, if they are engineered based on sound scientific principles to be discussed here.

Examples: Electrical transformers, condensers, economizers, motor cycle, IC engines, etc.

Desirable properties of fin material:

1.High thermal conductivity to maximize heat transfer.

2.Density of fin material is as low as possible; otherwise, the weight of the system increases significantly, outweighing the benefits.

3.Reasonable cost.

2.10  Analysis of fin heat transfer

Fig. 2.25 shows a schematic representation of a typical variable area fin with a rectangular cross-section.

The following assumptions are in order:

1.One-dimensional, steady state heat transfer, i.e., T = f(x) alone

2.Constant properties

 

 

2.10 Analysis of fin heat transfer

49

FIGURE 2.25

Schematic representation of a rectangular cross-section fin with variable area.

3.No heat generation in the fin

4.(h,T) are constant

5.Radiation is negligible

Consider the energy balance across a control volume of a fin, as shown in Fig. 2.25

 

 

Qx

= Qx+dx + Qconv

 

 

 

Qx

= Qx +

x

d

 

Qx +Qconv

 

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

 

 

 

 

dT

 

 

 

x

 

 

 

kAc

 

 

+ hpdx(T T) = 0

 

 

 

 

 

dx

 

 

 

 

 

dx

 

 

 

 

 

 

 

 

 

d

 

 

 

dT

 

 

 

 

 

 

k

 

 

 

 

Ac

hp(T T) = 0

 

 

 

dx

 

 

 

 

 

 

 

 

dx

 

 

 

Let θ = T T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

dθ

hp

θ = 0

 

 

 

 

 

 

 

 

Ac

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

dx

 

 

(2.162)

(2.163)

(2.164)

(2.165)

(2.166)

Eq. (2.166) is the fin equation and is a second-order differential equation. It supports two boundary conditions.

1.Boundary condition 1: at x = 0; T = Tb or θ = θb

2.Boundary condition 2: Three possibilities exist

50 CHAPTER 2  One-dimensional, steady state heat conduction

a. Insulated condition: at x = L;

dθ

 

= 0 (adiabatic fin tip)

 

dx

 

 

 

 

dθ

 

 

 

 

 

b. Convection boundary: at x = L; -k

 

 

= hθ |x=L

 

 

 

 

dx

 

 

 

 

 

 

x=L

 

c. Long fin: at x = L; (rather x → ∞) θ = 0

 

For a constant area fin, Eq. (2.166) reduces to

 

d 2θ

 

hp

θ

= 0

 

(2.167)

 

 

dx2

 

kA

 

 

 

 

 

 

 

 

 

c

 

 

 

 

d2θ

2

 

 

 

 

 

dx2

m θ = 0

(2.168)

 

 

 

 

 

m2 =

hp

 

(2.169)

 

 

 

 

 

kA

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c

 

In Eq. (2.169), m is the fin parameter and is known for any fin a priori.

For a rectangular fin of constant cross sectional area shown in Fig. 2.26, the perimeter, p and the cross sectional area Ac of the fin are given by,

p = 2(W + t)

 

Ac = W t

 

m = 2h(W + t)

(2.170)

k W t

 

FIGURE 2.26

Schematic representation of a rectangular fin with constant cross section area.

 

 

2.10 Analysis of fin heat transfer

51

If w t, the expression for m for a constant area rectangular fin reduces to

m =

2h

(2.171)

 

kt

 

2.10.1  Case 1: Insulated tip

Assumption: Tb > T

 

θ = θb

 

 

 

 

Boundary condition 1: At x = 0;

 

 

 

 

Boundary condition 2: At x = L;

 

dθ

=

0

 

 

 

 

 

 

 

 

 

dx

 

 

 

 

 

 

 

 

Let

d

= D

 

 

 

 

 

 

 

 

 

 

 

 

 

dx

 

Therefore Eq. (2.168) becomes

 

 

 

 

 

 

 

 

 

 

[D2 m2 ]θ = 0

 

The solution to the above equation is given by

 

 

θ = c1 cosh(mx) + c2 sinh(mx)

 

Boundary condition 1: At x = 0; θ = θb

 

 

 

 

θb = c1 + 0

 

 

 

 

c1 = θb

 

 

 

 

 

 

 

dθ

= c1 (sinh(mx))m + c2 (cosh(mx))m

 

dx

 

 

 

 

 

 

 

 

Boundary condition 2: At x = L;

 

dθ

=

0

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

 

 

0 = m[θb sinh(mL) + c2 cosh(mL)]

 

c2 = −θb tanh(mL)

 

θ = θb cosh(mx) θbtanh(mL) sinh(mx)

 

cosh(mx) cosh(mL) − sinh(mL) sinh(mx)

θ = θb

 

 

 

 

cosh(mL)

 

 

 

 

 

 

 

cosh(m(L x))

 

θ = θb

cosh(mL)

 

 

 

 

 

 

 

θ= cosh(m(L x))

θb cosh(mL)

(2.172)

(2.173)

(2.174)

(2.175)

(2.176)

(2.177)

(2.178)

(2.179)

(2.180)

(2.181)

(2.182)

52 CHAPTER 2  One-dimensional, steady state heat conduction

Eq. (2.182) indicates that the temperature distribution across the fin is exponential.

The heat transfer from the fin can now be calculated as

Q = −kAc

dθ

 

 

 

 

(2.183)

 

dx

 

 

 

 

 

x=0

 

 

 

 

 

1

sinh(m(L x))(m) |x=0

 

Q = −kAcθb

 

 

(2.184)

cosh(mL)

Q = kAcθb

 

 

hp

 

(2.185)

 

 

kA tanh(mL)

 

 

 

 

 

 

 

 

c

 

Q = hpkAcθb tanh(mL)

(2.186)

Fig. 2.27A represents the temperature variation along the length of the fin for the ideal case. In this case, the fin temperature is equal to the base temperature everywhere.

However for a typical fin, the temperature varies along the length, exponentially, and is shown in Fig. 2.27B.

As seen in Fig. 2.27C, the unhatched portion indicates the penalty we pay for the fin not being isothermal. If the fin is not maintained isothermally throughout the length, its efficiency to transfer heat (we will formally introduce the term efficiency in the next sub-section) will get reduced and is shown with the unhatched portion in Fig. 2.27C. Therefore, we try to make the exponential variation as close to the horizontal line as possible in order to get maximum efficiency (close to ideal efficiency).

cosh(m(L x))

θ = θb cosh (2.187)

(mL)

The fin tip temperature can be obtained by evaluating θ at x = L.

 

1

 

 

θ = θb

 

 

(2.188)

 

cosh(mL)

 

FIGURE 2.27

Graphical representation of fin performance. (A) ideal case, (B) typical fin case, and (C) reduction in θ for a fin not being isothermal.

 

 

2.10 Analysis of fin heat transfer

53

The fin tip temperature is a kind of ‘bell-weather’ to capture the non-isothermality of the fin and represents the lowest temperature in the fin.

Fin efficiency

The fin efficiency is defined as the ratio of actual heat transfer to the maximum possible heat transfer (happens when the fin is isothermal) and is mathematically given by

 

 

η = Qactual

 

 

(2.189)

 

 

 

Qmax

 

 

 

For the insulated tip fin considered above

 

 

 

η =

hpkAcθb tanh(mL)

 

 

 

(2.190)

h( pL)θb

 

 

 

 

 

 

 

 

tanh(mL)

 

hp

 

η =

 

mL

, Since m =

 

 

(2.191)

 

 

 

 

 

kAc

 

Effectiveness of the fin

The effectiveness of a fin is defined as the ratio of heat transfer with fins to the heat transfer without fins. The effectiveness is a key engineering metric for making a decision as to whether fins can or need to be used in a particular situaration or not.

ε =

 

Qfin

(2.192)

Q

withoutfin

 

 

 

As a general, it is advisable to use fins, only if ε 2.

For a general case with n fins, each of which has an adiabatic tip and with a fin parameter m, as shown in Fig. 2.28, the overall effectiveness is given by

ε = (Heat transfer from fins + Heat transfer from the unfinned area) Heat transfer from the base area

We obtain an expression for ε as

ε = n hpkAcθb tanh(mL) + h[ A nAc ]θb hAθb

ε = n hpkAcθb tanh mL + h[ A nAc ]θb nhAcθb + h[ A nAc ]θb

(2.193)

(2.194)

(2.195)

For a single fin with an insulated tip and no unfinned area, Eq. (2.195) reduces to

ε =

hpkAcθb tanh(mL)

(2.196)

hAcθb

 

 

54 CHAPTER 2  One-dimensional, steady state heat conduction

FIGURE 2.28

Schematic representation of n number of fins over a base plate.

ε =

pk

tanh mL

(2.197)

 

 

hAc

 

Eq. (2.197) confirms that it is more advantageous to use high thermal conductivity fins in situations where the heat transfer coefficient “h” is low. In so far as the geometry of the fin is concerned, lower the fin cross section and higher the perimeter, higher is the effectiveness. We can now work out the effectiveness relation for a rectangular fin.

Rectangular fin

For a rectangular fin, the effectiveness which turns out to be the key engineering metric, as already discussed is

ε =

2k

(2.198)

ht tanh mL

 

We would like to have an effectiveness as high as possible. Hence, we choose a material with a high thermal conductivity and we try to make a fin thin, as already discussed. A surprising and counterintuitive result is that as h is increasing while Qfin keeps increasing, ε comes down. So, we are confronted with a situation where a fin appears to be good but not effective. In simple English, effficiency tells us if a fin is thermally the best of many competing fin designs, while effectiveness adresses the fundamental question as to whether we need to use a fin, however efficient it may be in an engineering problem.