Solved Problems

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SOLVED PROBLEMS SOLVED PROBLEMS

Combined Convection and Radiation Combined Convection and Radiation Problem 1:

Problem 1:

_{A surface is}

_{A surface is at 200°C and}

_{at 200°C and is exposed to surroundings at}

_{is exposed to surroundings at 60°C}

_60°C

and convects and

radiates heat to the surroundings. The convection coefficient is 80W

radiates heat to the surroundings. The convection coefficient is 80W //m

m

22

_K.

The radiation factor is one. If the heat is conducted to the surface through

a solid of conductivity 12 W/mK, determine the temperature gradient at

the surface in the solid.

Solution: Solution:

Heat convected + heat radiated = heat conducted considering 1

Heat convected + heat radiated = heat conducted considering 1m

m

22

_,,

h

h((T

T

11

–

– T

T

22

) +

σσ

((T

T

1144

–

– T

T

2244

) = –

) = – kdT

kdT //dx

dx

Therefore, 80(200 – 60) + 5.67 {[(200 +

Therefore, 80(200 – 60) + 5.67 {[(200 + 273)/100]

273)/100]

44

_{– [(60 + 273)/100]}

44

_}

= – 12

= – 12 dT/dx

dT/dx

Therefore

Therefore dT/dx

dT/dx =

= – (11200

– (11200 + 2140.9)/12

+ 2140.9)/12 = –

= –

1111.7°C/m.1111.7°C/m. Problem 2:

Problem 2:

_{Heat is conducted through a material with a temperature}

gradient of – 9000 °C

gradient of – 9000 °C//m. The conductivity of the material is 25W/mK. If

m. The conductivity of the material is 25W/mK. If

th

this

is he

heat

at is

is co

conv

nvec

ecte

ted

d to

to su

surr

rrou

ound

ndin

ings

gs at

at 30

30°C

°C wi

with

th a

a co

conv

nvec

ecti

tion

on

coefficient of 345W/m2K, determine the surface temperature.

If the heat is radiated to the surroundings at 30°C determine the surface

temperature.

Solution:

_{In this case only convection and conduction are}

_{In this case only convection and conduction are involved.}

_involved.

–

– kAdT

kAdT //dx

dx =

= hA

hA((T

T

11

–

– T

T

22

). Considering unit area,

– 25 × 1 × (– 9000) = 345 × 1 (

– 25 × 1 × (– 9000) = 345 × 1 ( T

T

11

– 30)

Therefore,

TT11 = 682.17°C= 682.17°C

In this case conduction and radiation are

In this case conduction and radiation are involved.

involved.

Heat conducted = Heat radiated

– 25 × 1 × (– 9000) = 5.67 [(

– 25 × 1 × (– 9000) = 5.67 [(T

T

11

/100)

44

– (303/100)

44

]]

Therefore,

TT11 = 1412.14K = 1139°C.= 1412.14K = 1139°C.

Problem 3:

_{There is a heat flux through a wall of 2250W/m}

22

_{. The same is}

dissipated to the

surroundings by convection and radiation. The surroundings is at 30°C.

The convection

coefficient has a value of 75W/m

22

_{K. For radiation F = 1. Determine the}

wall surface temperature.

Solution:

For the specified condition, Consider unit area.

The heat conducted = heat conv

The heat conducted = heat convected + heat radiated

ected + heat radiated

Using the rate equations, with absolute

Using the rate equations, with absolute temperature

temperature

2250 = [(T

22

– 303)/(1/75 × 1] + 5.67 × 1[(

– 303)/(1/75 × 1] + 5.67 × 1[(T

T

22

/100)

44

– (303/100)

44

]]

= 75

= 75T

T

22

– 22725 + 5.67(

– 22725 + 5.67(T

T

22

/100)

44

– 477.92

or, (

or, (T

T

22

/100)

44

+ 13.2275

+ 13.2275T

T

22

– 4489.05 = 0.

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This equation can be solved only by trial. It may be noted that the

contribution of (T

2

/100)

4

is small and so the first choice of T

2

can be a little

less than 4489/13.227 = 340K. The values of the reminder for T

2

= 300,

310, 320, 330 are given below:

Assumed value of T 2

300 310 320 330 330.4 330.3

Remainder – 439.80 – 296.2 – 15.1 – 5.38 0.484 – 0.98

So, the temperature T

2

is near 330K. By one more trial T

2

is obtained

as

330.4K or 57.4°C.

Check:

_{Q = 75(330.4 – 303) + 5.69(3.304}

4

_{– 3.03}

4

₎

= 2047.5 + 206 = 2253.5 W.

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PROBLEM 1.1

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 (

 Ti) = 20°C

Temperature of the outer surface (

 To) = –5°C

FIND

The heat loss through the wall (

 qk )

ASSUMPTIONS

One dimensional heat flow 

The system has reached steady state 

SKETCH

SOLUTION

The rate of heat loss through the wall is qk = (AK/L)

×

(∆T )

qk = [(10m) (3m)

(

1.2W/(mK)

)/

0.2m]

×

(20°C – (–5°C)) qk = 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.

PROBLEM 1.9

Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m3_{) of 5 cm thickness and 2 m}2 _{area. If the hot surface is at 70°C, determine the}

temperature of the cooler surface. GIVEN

Glass wool insulation with a density (

 ρ) = 100 kg/m3

Thickness (

 L) = 5 cm = 0.05 m

Area (

 A) = 2 m2

Temperature of the hot surface (

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Rate of heat transfer (

 qk ) = 0.1 kW= 100 W

FIND

The temperature of the cooler surface (

 T C)

ASSUMPTIONS

One dimensional, steady state conduction 

Constant thermal conductivity 

SKETCH

PROPERTIES AND CONSTANTS

From Appendix 2, Table 11

The thermal conductivity of glass wool at 20°C ( k ) = 0.036 W/(m K) SOLUTION

For one dimensional, steady state conduction, the rate of heat transfer, is qk = (Ak/L)

×

(T h – T c)

Solving this for Tc Tc = T h – qk L/A k

T c= 70°C – [(100W) (0.05m)]/[(2 m2)( 0.036W/mK)]

T c= 0.6°C

PROBLEM 1.10

A heat flux meter at the outer (cold) wall of a concrete building indicates that the heat loss through a wall of 10 cm thickness is 20 W/m2_{. If a thermocouple at the inner surface}

of the wall indicates a temperature of 22°C while another at the outer surface shows 6°C, calculate the thermal conductivity of the concrete and compare your result with the value in Appendix 2, Table 11.

GIVEN Concrete wall  Thickness (  L) = 100 cm = 0.1 m Heat loss (  q/ A) = 20 W/m2 Surface temperature  Inner (  T i) = 22°C Outer (  T o) = 6°C FIND

The thermal conductivity (

 k ) and compare it to the tabulated value

ASSUMPTIONS

One dimensional heat flow through the wall 

Steady state conditions exist 

SKETCH

SOLUTION

The rate of heat transfer for steady state, one dimensional conduction, is qk = (k A/L)

×

(T hot– T cold)

Solving for the thermal conductivity k = (qk / A)

×

L/(T i – T o )

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This result is very close to the tabulated value in Appendix 2 , Table 11 where the thermal conductivity of concrete is given as 0.128 W/(m K).

PROBLEM 1.11

Calculate the heat loss through a 1-m by 3-m glass window 7 mm thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer.

GIVEN Window: 1 m by 3 m  Thickness (  L) = 7 mm = 0.007 m Surface temperature  Inner (  T i) = 20°C and outer (To) = 17°C FIND

The rate of heat loss through the window (

 q)

ASSUMPTIONS

One dimensional, steady state conduction through the glass 

Constant thermal conductivity 

SKETCH

PROPERTIES AND CONSTANTS From Appendix 2, Table 11

Thermal conductivity of glass (k ) = 0.81 W/(m K) SOLUTION

The heat loss by conduction through the window is qk = (k A/L)

×

(T hot– T cold)

qk = [[(0.81W/(mK))(1m) (3m)]/(0.007m)]

×

(20°C – 17°C)

qk = 1040 W

COMMENTS

Window glass is transparent to certain wavelengths of radiation, therefore some heat may be 

lost by radiation through the glass.

During the day sunlight may pass through the glass creating a net heat gain through the 

window.

PROBLEM 1.12

If in Problem 1.11 the outer air temperature is –2°C, calculate the convective heat transfer coefficient between the outer surface of the window and the air assuming radiation is negligible.

Problem 1.11: Calculate the heat loss through a 1 m by 3 m glass window 7 mm thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C.

Comment on the possible effect of radiation on your answer. GIVEN Window: 1 m by 3 m  Thickness (  L) = 7 mm = 0.007 m Surface temperatures  Inner (  T i) = 20°C and outer (T o) = 17°C

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The rate of heat loss = 1040W (from the solution to Problem 1.11) 

The outside air temperature = –2°C 

FIND

The convective heat transfer coefficient at the outer surface of the window (

 hc )

ASSUMPTIONS

The system is in steady state and 

radiative loss through the window is negligible

SKETCH

SOLUTION

For steady state the rate of heat transfer by convection (Equation (1.10)) from the outer surface must be the same as the rate of heat transfer by conduction through the glass

qc= hcA ∆T = qk Solving for hc hc = qk / A (T o – T ∞ ) hc= 1040W / (1m)(3m)( 17°C – – 2°C) hc= 18.2 W/(m2K) COMMENTS

This value for the convective heat transfer coefficient falls within the range given for the 

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PROBLEM

The wall of an industrial furnace is constructed from 0.15-m-thick fireclay

brick having a thermal conductivity of

1.7 W/m. K.

Measurements made

during steady-state operation reveal temperatures of 1400 and 1150 K at

the inner and outer surfaces, respectively. What is the rate of heat

transfer through a wall that is 0.5 m by 1.2 m on a side?

Solution

Known: Steady-state conditions with prescribed wall thickness, area, thermal conductivity, and surface temperatures.

Find:Heat transfer rate through the wall. Schematic and Given Data:

Assumptions:

1.Steady-state conditions.

2.One-dimensional conduction through the wall. 3. Constant thermal conductivity.

Analysis: Since heat transfer through the wall is by conduction, the heat flux may be determined from Fourier’s law.

q x= k ∆T /L = 1.7 W/m. K × 250 K / 0.15 m = 2833 W/m2

The heat flux represents the rate of heat transfer through a section of unit area, and it is uniform across the surface of the wall.

The heat rate through the wall of area A = H × W is then Q x= ( HW ) q x= (0.5 m × 1.2 m) 2833 W/m2= 1700 W