Transferencia de calor -7th-Kreith-Sol, Notas de estudo de Engenharia Elétrica

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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE

MUCHOS DE ESTOS LIBROS GRATIS EN

DESCARGA DIRECTA

SIGUENOS EN:

VISITANOS PARA DESARGALOS GRATIS.

An Instructor’s Solutions Manual to Accompany

PRINCIPLES OF HEAT TRANSFER, 7

TH

EDITION, SI

FRANK KREITH

RAJ M. MANGLIK

MARK S. BOHN

SI E DITION PREPARED BY : SHALIGRAM TIWARI

INSTRUCTOR'S SOLUTIONS MANUAL TO

ACCOMPANY

PRINCIPLES OF

HEAT TRANSFER

SEVENTH EDITION, SI

FRANK KREITH

RAJ M. MANGLIK

MARK S. BOHN

SI EDITION PREPARED BY :

SHALIGRAM TIWARI

Indian Institute of Technology Madras

T ABLE OF C ONTENTS

CHAPTER PAGE

x Contents

Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

Contents xi

Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1

Chapter 1

PROBLEM 1.

The outer surface of a 0.2m-thick concrete wall is kept at a temperature of –5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2 W/(m K). Determine the heat loss through a wall 10 m long and 3 m high.

GIVEN

10 m long, 3 m high, and 0.2 m thick concrete wall Thermal conductivity of the concrete ( k ) = 1.2 W/(m K) Temperature of the inner surface ( T (^) i ) = 20°C Temperature of the outer surface ( T (^) o ) = –5°C

FIND

The heat loss through the wall ( q (^) k )

ASSUMPTIONS

One dimensional heat flow The system has reached steady state

SKETCH

L = 0.2 m

Ti = 20°C To = – 5°C

q (^) k

L^ = 10 m H = 3 m

SOLUTION

The rate of heat loss through the wall is given by Equation (1.2)

q (^) k = A K L

(Δ T )

q (^) k =

(10 m) (3m) 1.2 W/(m K)( )

0.2 m

(20°C – (–5°C))

q (^) k = 4500 W

COMMENTS

Since the inside surface temperature is higher than the outside temperature heat is transferred from the inside of the wall to the outside of the wall.

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2

PROBLEM 1.

The weight of the insulation in a spacecraft may be more important than the space required. Show analytically that the lightest insulation for a plane wall with a specified thermal resistance is that insulation which has the smallest product of density times thermal conductivity.

GIVEN

Insulating a plane wall, the weight of insulation is most significant

FIND

Show that lightest insulation for a given thermal resistance is that insulation which has the smallest product of density ( ρ) times thermal conductivity ( k )

ASSUMPTIONS

One dimensional heat transfer through the wall Steady state conditions

SOLUTION

The resistance of the wall ( R (^) k ), from Equation (1.13) is

R (^) k =

L A k

where

L = the thickness of the wall A = the area of the wall

The weight of the wall ( w ) is

w = ρ A L

Solving this for L

L =

w ρ A

Substituting this expression for L into the equation for the resistance

R (^) k = (^2)

w ρ k A

∴ w = ρ k R (^) k A 2

Therefore, when the product of ρ k for a given resistance is smallest, the weight is also smallest.

COMMENTS

Since ρ and k are physical properties of the insulation material they cannot be varied individually. Hence in this type of design different materials must be tried to minimize the weight.

PROBLEM 1.

A furnace wall is to be constructed of brick having standard dimensions 22.5 cm × 11 cm × 7.5 cm. Two kinds of material are available. One has a maximum usable temperature of 1040°C and a thermal conductivity of 1.7 W/(m K), and the other has a maximum temperature limit of 870°C and a thermal conductivity of 0.85 W/(m K). The bricks cost the same and can be laid in any manner, but we wish to design the most economical wall for a furnace with a temperature on the hot side of 1040°C and on the cold side of 200°C. If the maximum amount of heat transfer permissible is 950 W/m^2 for each square foot of area, determine the most economical arrangements for the available bricks.

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4

This thickness can be achieved by using 4 layers of type 1 bricks using the 7.5 cm dimension.

Similarly, for one-dimensional conduction through type 2 bricks

L 2 = 2 max

k q A

   

( T 12 – T cold)

L 2 = (^2)

0.85 W/(m K) 950 W/m

(870 – 200)K = 0.6 m = 60 cm

This thickness can be achieved with 8 layers of type 2 bricks using the 7.5 cm dimension.

Therefore, the most economical wall would be built using 4 layers of type 1 bricks and 8 layers of type 2 bricks with the three inch dimension of the bricks used as the thickness.

PROBLEM 1.

To measure thermal conductivity, two similar 1-cm-thick specimens are placed in an apparatus shown in the accompanying sketch. Electric current is supplied to the 6-cm by 6-cm guarded heater, and a wattmeter shows that the power dissipation is 10 watts (W). Thermocouples attached to the warmer and to the cooler surfaces show temperatures of 322 and 300 K, respectively. Calculate the thermal conductivity of the material at the mean temperature in W/(m K).

GIVEN

Thermal conductivity measurement apparatus with two samples as shown Sample thickness ( L ) = 1 cm = 0.01 cm Area = 6 cm × 6 cm = 36 cm 2 = 0.0036 m 2

Power dissipation rate of the heater ( qh ) = 10 W Surface temperatures  T hot = 322 K  T cold = 300 K

FIND

The thermal conductivity of the sample at the mean temperature in W/(m K)

ASSUMPTIONS

One dimensional, steady state conduction No heat loss from the edges of the apparatus

SKETCH

E S^ Guard Ring^ and Insulation

Heater

Wattmeter

Similar Specimen T hot T cold

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5

SOLUTION

By conservation of energy, the heat loss through the two specimens must equal the power dissipation of the heater. Therefore the heat transfer through one of the specimens is q (^) h /2.

For one dimensional, steady state conduction (from Equation (1.3)

q (^) k = k A L

Δ T = 2

q h

Solving for the thermal conductivity

k = 2

q h (^) L

A T Δ

k = (^2) (5 W)(0.01m) (0.0036 m )(322 K −300 K)

k = 0.63 W/(m K)

COMMENTS

In the construction of the apparatus care must be taken to avoid edge losses so all the heat generated will be conducted through the two specimens.

PROBLEM 1.

To determine the thermal conductivity of a structural material, a large 15 cm-thick slab of the material was subjected to a uniform heat flux of 2500 W/m 2 , while thermocouples embedded in the wall 2.5 cm apart were read over a period of time. After the system had reached equilibrium, an operator recorded the readings of the thermocouples as shown below for two different environmental conditions.

Distance from the surface (cm) Temperature (°C)

Test 1: 0 5 10 15 Test 2: 0 5 10 15

40 65 97 132

95 130 168 208

From these data, determine an approximate expression for the thermal conductivity as a function of temperature between 40 and 208°C.

GIVEN

Thermal conductivity test on a large, 15 cm slab Thermocouples are embedded in the wall, 2.5 cm apart Heat flux ( q / A ) = 2500 W/m^2 Two equilibrium conditions were recorded (shown in Table above)

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7

These points are displayed graphically.

Temperature (°C)

k (W/(mK))

(^3100 150 )

4

5

Mean Variation of with Temperature

k

We will use the best fit quadratic function to represent the relationship between thermal conductivity and temperature

k ( T ) = a + b T + c T^2

The constants a , b , and c can be found using a least squares fit.

Let the experimental thermal conductivity at data point n be designated as kn. A least squares fit of the data can be obtained as follows

The sum of the squares of the errors is

S = [ (^) n ( (^) n )]^2 N

 k^ − k T

S =  k^2 n^ − 2 a  kn − N a^2 + 2 ab  Tn − 2 b  k Tn n + 2 ac  T n^2^ + b^2^  Tn^2

• 2 2 2 3 2 4

c  k Tn n + bc  Tn + c  Tn

By setting the derivatives of S (with respect to a , b , and c ) equal to zero, the following equations result

N a + Σ T (^) n b + Σ T (^) n^2 c = Σ kn Σ T (^) n a + Σ T (^) n^2 b + Σ T (^) n^3 c = Σ kn T (^) n Σ T (^) n^2 a + Σ T (^) n^3 b + Σ T (^) n^4 c = Σ kn T (^) n^2

For this problem

Σ T (^) n = 697. Σ T (^) n^2 = 9.263 × 104 Σ T (^) n^3 = 1.3554 × 107 Σ T (^) n^4 = 2.125 × 109 Σ kn = 22. Σ kn T (^) n = 2445. Σ kn T (^) n 2 = 3.124 × 10 4

Solving for a , b , and c a = 6. b = – 4.7213 × 10 – c = 1.443 × 10 –

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8

Therefore the expression for thermal conductivity as a function of temperature between 40 and 208°C is

k ( T ) = 6.9722 – 4.7213 × 10 –2^ T + 1.443 × 10 –4^ T^2

This is plotted in the following graph

Temperature (deg C)

50 100 150 200

3

4

5

Th

er

m al Condu

ctivit

y,^

[W/mK] k

COMMENTS

Note that the derived empirical expression is only valid within the temperature range of the experi- mental data.

PROBLEM 1.

A square silicone chip 7 mm by 7 mm in size and 0.5 mm thick is mounted on a plastic substrate with its front surface cooled by a synthetic liquid flowing over it. Electronic circuits in the back of the chip generate heat at a rate of 5 watts that have to be transferred through the chip. Estimate the steady state temperature difference between the front and back surfaces of the chip. The thermal conductivity of silicone is 150 W/(m K).

GIVEN

A 0.007 m by 0.007 m silicone chip Thickness of the chip ( L ) = 0.5 mm = 0.0005 m

Heat generated at the back of the chip ( q  G^ ) = 5 W

The thermal conductivity of silicon ( k ) = 150 W/(m K)

FIND

The steady state temperature difference (Δ T )

ASSUMPTIONS

One dimensional conduction (edge effects are negligible) The thermal conductivity is constant The heat lost through the plastic substrate is negligible

SKETCH

Substrate

CNIP

7 mm

7 mm

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10

SOLUTION

The rate at which heat must be removed is equal to the rate at which heat flows into the warehouse. There will be convective resistance to heat flow on the inside and outside of the wall. To estimate the upper limit of the rate at which heat must be removed these convective resistances will be neglected. Therefore the inside and outside wall surfaces are assumed to be at the same temperature as the air inside and outside of the wall. Then the heat flow, from Equation (1.2), is

q = k A L

Δ T

q =

0.17 W/(m K) (1860 m ) 7.5 × 10 m

(32 – 4)

q = 118 kW

PROBLEM 1.

With increasing emphasis on energy conservation, the heat loss from buildings has become a major concern. For a small tract house the typical exterior surface areas and R-factors (area × thermal resistance) are listed below

Element Area (m^2 ) R -Factors = Area × Thermal Resistance [(m^2 K/W)]

Walls 150 2. Ceiling 120 2. Floor 120 2. Windows 20 0. Doors 5 0.

(a) Calculate the rate of heat loss from the house when the interior temperature is 22°C and the exterior is –5°C. (b) Suggest ways and means to reduce the heat loss and show quantitatively the effect of doubling the wall insulation and the substitution of double glazed windows (thermal resistance = 0.2 m^2 K/W) for the single glazed type in the table above.

GIVEN

Small house Areas and thermal resistances shown in the table above Interior temperature = 22°C Exterior temperature = –5°C

FIND

(a) Heat loss from the house ( q (^) a ) (b) Heat loss from the house with doubled wall insulation and double glazed windows ( qb ). Suggest improvements.

ASSUMPTIONS

All heat transfer can be treated as one dimensional Steady state has been reached The temperatures given are wall surface temperatures Infiltration is negligible The exterior temperature of the floor is the same as the rest of the house

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11

SOLUTION

(a) The rate of heat transfer through each element of the house is given by Equations (1.33) and (1.34)

q = th

T R

Δ

The total rate of heat loss from the house is simply the sum of the loss through each element:

q = Δ T wall ceiling floor windows doors

1 1 1 1 1 AR AR AR AR AR A A A A A

   (^) + + + +   ^ ^ ^ ^ ^ ^ ^ ^ ^    ^ ^ ^ ^ ^ ^ ^ ^ ^    

q = (22°C – –5°C)

2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 2.0 (m K)/W 2.8 (m K)/W 2.0 (m K)/W 0.5 (m K)/W 0.5 (m K)/W 150 m 120 m 120 m 20 m 5 m

     + + + +   ^ ^ ^ ^ ^ ^ ^ ^ ^               

q = (22°C – –5°C) (75 + 42.8 + 60 + 200 + 10) W/K

q = 10,500 W

(b) Doubling the resistance of the walls and windows and recalculating the total heat loss:

q = (22°C – –5°C)

2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 4.0 (m K)/W 2.8 (m K)/W 2.0 (m K)/W 0.2 (m K)/W 0.5 (m K)/W 150 m 120 m 120 m 20 m 5 m

     (^) + + + +              (^)                      

q = (22°C – –5°C) (37.5 + 42.8 + 60 + 100 + 10) W/K

q = 6800 W

Doubling the wall and window insulation led to a 35% reduction in the total rate of heat loss.

COMMENTS

Notice that the single glazed windows account for slightly over half of the total heat lost in case (a) and that the majority of the heat loss reduction in case (b) is due to the double glazed windows. Therefore double glazed windows are strongly suggested.

PROBLEM 1.

Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m^3 ) of 5 cm thickness and 2 m^2 area. If the hot surface is at 70°C, determine the temperature of the cooler surface.