Solutions Manual
Principles of Heat Transfer 7th Edition SI
By
Frank Kreith,
Raj M. Manglik,
Mark S. Bohn
( All Chapters Included - 100% Verified Solutions )
1
, Chapter 1
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
L = 0.2 m
m
10 H=3m
=
L
qk
Ti = 20°C
To = – 5°C
SOLUTION
The rate of heat loss through the wall is given by Equation (1.2)
AK
qk = (ΔT)
L
(10 m) (3m) (1.2 W/(m K) )
qk = (20°C – (–5°C))
0.2 m
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.
1
© 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 1.2
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 (Rk), from Equation (1.13) is
L
Rk =
Ak
where
L = the thickness of the wall
A = the area of the wall
The weight of the wall (w) is
w =ρAL
Solving this for L
w
L =
ρA
Substituting this expression for L into the equation for the resistance
w
Rk =
ρ k A2
∴ w = ρ k Rk A2
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 1.3
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/m2 for each square
foot of area, determine the most economical arrangements for the available bricks.
2
© 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.
3
, GIVEN
Furnace wall made of 22.5 cm × 11 cm × 7.5 cm bricks of two types
Type 1 bricks Maximum useful temperature (T1, max) = 1040°C
Thermal conductivity (k1) = 1.7 W/(m K)
Type 2 bricks Maximum useful temperature (T2, max) = 870°C
Thermal conductivity (k2) = 0.85 W/(m K)
Bricks cost the same
Wall hot side temperature (Thot) = 1040°C and wall cold side temperature (Tcold) = 200°C
Maximum permissible heat transfer (qmax/A) = 950 W/m2
FIND
The most economical arrangement for the bricks
ASSUMPTIONS
One-dimensional, steady state heat transfer conditions
Constant thermal conductivities
The contact resistance between the bricks is negligible
SKETCH
Type 1 Bricks Type 2 Bricks
Tmax = 1040°C Tmax = 200°C
T12 £ 870°C
SOLUTION
Since the type 1 bricks have a higher thermal conductivity at the same cost as the type 2 bricks, the
most economical wall would use as few type 1 bricks as possible. However, there should be thick
enough layer of type 1 bricks to keep the type 2 bricks at 870°C or less.
For one-dimensional conduction through type 1 bricks (from Equation 1.2)
kA
qk = ΔT
L
qmax k
= 1 (Thot – T12)
A L1
where L1 is the minimum thickness of the type 1 bricks.
Solving for L1
k1
L1 = (Thot – T12)
qmax
A
1.7 W/(m K)
L1 = (1040 – 870)K = 0.3042 m = 30.42 cm
950 W/m 2
3
© 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.
4
Principles of Heat Transfer 7th Edition SI
By
Frank Kreith,
Raj M. Manglik,
Mark S. Bohn
( All Chapters Included - 100% Verified Solutions )
1
, Chapter 1
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
L = 0.2 m
m
10 H=3m
=
L
qk
Ti = 20°C
To = – 5°C
SOLUTION
The rate of heat loss through the wall is given by Equation (1.2)
AK
qk = (ΔT)
L
(10 m) (3m) (1.2 W/(m K) )
qk = (20°C – (–5°C))
0.2 m
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.
1
© 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 1.2
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 (Rk), from Equation (1.13) is
L
Rk =
Ak
where
L = the thickness of the wall
A = the area of the wall
The weight of the wall (w) is
w =ρAL
Solving this for L
w
L =
ρA
Substituting this expression for L into the equation for the resistance
w
Rk =
ρ k A2
∴ w = ρ k Rk A2
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 1.3
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/m2 for each square
foot of area, determine the most economical arrangements for the available bricks.
2
© 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.
3
, GIVEN
Furnace wall made of 22.5 cm × 11 cm × 7.5 cm bricks of two types
Type 1 bricks Maximum useful temperature (T1, max) = 1040°C
Thermal conductivity (k1) = 1.7 W/(m K)
Type 2 bricks Maximum useful temperature (T2, max) = 870°C
Thermal conductivity (k2) = 0.85 W/(m K)
Bricks cost the same
Wall hot side temperature (Thot) = 1040°C and wall cold side temperature (Tcold) = 200°C
Maximum permissible heat transfer (qmax/A) = 950 W/m2
FIND
The most economical arrangement for the bricks
ASSUMPTIONS
One-dimensional, steady state heat transfer conditions
Constant thermal conductivities
The contact resistance between the bricks is negligible
SKETCH
Type 1 Bricks Type 2 Bricks
Tmax = 1040°C Tmax = 200°C
T12 £ 870°C
SOLUTION
Since the type 1 bricks have a higher thermal conductivity at the same cost as the type 2 bricks, the
most economical wall would use as few type 1 bricks as possible. However, there should be thick
enough layer of type 1 bricks to keep the type 2 bricks at 870°C or less.
For one-dimensional conduction through type 1 bricks (from Equation 1.2)
kA
qk = ΔT
L
qmax k
= 1 (Thot – T12)
A L1
where L1 is the minimum thickness of the type 1 bricks.
Solving for L1
k1
L1 = (Thot – T12)
qmax
A
1.7 W/(m K)
L1 = (1040 – 870)K = 0.3042 m = 30.42 cm
950 W/m 2
3
© 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.
4
