13.

Past ME Board Problem

A vertical furnace wall is made up of an inner wall of firebrick 20 cm thick followed by

insulating brick 15 cm thick and an outer wall of steel 1 cm thick. The surface temperature of the
wall adjacent to the combustion chamber is 1200°C while that of the outer surface to steel is
50°C. The thermal conductivities of the wall material in W/m-K are: firebrick, 10; insulating
brick, 0.26; and steel, 45; Neglecting the film resistances and contact resistance of joints,
determine the heat loss per sq.m. of wall area.
A. 1.93 W/m2 C. 1.55 W/m2

B. 2.93 w/m2 D. 2.55 W/m2

Solution

Q t 1−t 4
=
A x 12 x 23 x 34
+ +
k 12 k 23 k 34

Q 1200−50
= =1.93
A 0.20 0.15 0.01
+ +
10 0.26 45

Thus;

Q
=1.93 W /m 2 (A)
A

14. Past ME Board Problem

A composite wall is made up of an external thickness of brickwork 110 mm thick inside which is
a layer of fiber glass 75 mm thick. The fiberglass is faced internally by an insulating board 25
mm thick. The coefficient of thermal conductivity for the three is as follows:

Brickwork 1.5 W/m-K

Fiberglass 0.04 W/m-K

Insulating board 0.06 W/m-K

The surface transfer coefficients of the inside wall is 3.1 W/m 2-K while that of the outside wall is
2.5 W/m2-K. take the internal ambient temperature as 10°C and the external temperature is 27°C.
Determine the heat loss through such wall 6 m high and 10 m long.

A. 330.10 W C. 430.10 W
B. 230.10 W D.530.10 W
Solution:

A∆T
Q=
RT

Where:

1 x 12 x 23 x 34 1
RT = + + + +
hi k 12 k 23 k 34 ho

1 0.110 0.075 0.025 1

¿ + + + +
3.1 1.5 0.04 0.06 2.5

m2 ° C
RT =3.09
W

Then:

( 6 x 10 )( 27−10 )
Q=
3.09

Thus;

Q=330.10 ( A )

15. Supplementary Problems

One insulated wall of cold storage compartment is 8m long by 2.5 m high and consists of an
outer steel plate 18 mm thick. An inner wood wall 22.5 mm thick, the steel and wood are 90 mm
apart to form a cavity which is filled with cork. If the temperature drop across the extreme faces
of the composite wall is 15°C. Calculate the heat transfer per hour through the wall and the
temperature drop across the thickness of the cork. Take the coefficients of thermal conductivity
for steel, cork and wood as 45, 0.045 and 0.18 W/m-K respectively.

A. 408.24 kJ, 12.12°C C. 608.24kJ, 13.12°C

B. 708.24 kJ, 11.12°C D. 508.24kJ, 14.12°C

Solution:

A∆T x x x
Q= Wher e : R T = 12 + 23 + 34
RT k 12 k 23 k 34
 0.018 0.09 0.0225
¿ + +
45 0.045 0.18

= 2.215

Then:

( 8 x 2.5 x 15 )
Q=
2.125

=141.176 W or J/s

=508.24 kJ/hr

Thus, the heat transfer per hour is 508.24 kJ

Solving for the temperature drop across the Cork:

A∆T
Q=
x 23
k 23

20 ∆ T
141.176=
0.09
0.045

∆ T =14.12 ° C

Thus;

508.24 kJ, 14.12°C (D)

16. Supplementary Problem

A cubical tank of 2 m sides is constructed of metal plate 12 mm and contains water at 75°C. The
surrounding air temperature is 16°C. Calculate the overall heat transfer coefficient from water to
air. Take the coefficient of thermal conductivity of the metal as 48 W/m-K, the coefficient of
heat transfer of water is 2.5 kW/m2-K and the coefficient of heat transfer of the air is 16 W/m2-K.

A. 15.84 W/m2°C C. 16.84 W/m2°C

B. 14.84 W/m2°C D.13.84 W/m2°C

Solution:

Let U = overall heat transfer coefficient

 1
U=
RT

Where

1 x12 1
RT = + +
hwater k 12 h air

1 0.012 1
¿ + +
2.5 x 10 3
48 16

C
RT =0.063 m 2 °
W

Then:

1 W W
U= =15.84 2 ° C
0.063 m ° C
2
m

Thus;

U =15.84 W /m2 ° C( A)

17. Supplementary Problems

Calculate the quantity of heat conducted per minute through a duralumin circular disc 127 mm
diameter and 19 mm thick when the temperature drop across the thickness of the plate is 5°C.
Take the coefficient of thermal conductivity of duralumin as 150 W/m-K.

A. 30 kJ C. 35 kJ
B. 40 kJ D.45 kJ

Solution:

kA ∆ T
Q=
x

Q=150 x ¿ ¿

=500.04W

=30 kJ/min
Thus;

(A) The quantity of heat conducted per minute is 30 kJ

18. Supplementary Problems

A cold storage compartment is 4.5 m long by 4 m wide by 2.5 m high. The four walls, ceiling
and floor are covered to a thickness of 150 mm with insulation material which has a coefficient
of thermal conductivity of 5.8 x 10-2 W/m-K. Calculate the quantity of heat leaking through the
insulation per hour when the outside and inside face temperatures of the material is 15 °C and -5°C
respectively.

A. 2185.44 kJ C. 31854.44 kJ
B. 1185.44 kJ D. 4185.44 kJ

Solution:

kA ∆ T
Q=
x

Where:

A=2 [ ( 4.5 ) (2.5 )+ ( 4 )( 2.5 ) + ( 4.5 )( 4 ) ]

= 78.50 m2

Then:

( 5.8 x 10−2 ) ( 78.50 )( 15+5 )

Q=
0.15

Q= 607.07 W

Q= 2185.44 kJ/hr

Thus;

(A) The quantity of heat leaking through the insulation per hour is 2185.44 kJ
19. Supplementary Problem

A thin square steel plate, 10 cm on the side, is heated in a blacksmiths forge to a temperature of
800°C. if the emissivity is 0.60, what is the total rate of radiation of energy?

A. 900 Watts C. 300 Watts

B. 300 Watts D. 700 Watts

Solution:

Q= Aeσ T 4

W
Q=(2 x 0.010 m 2)(0.60)(5.67 x 10−8 )¿
m2 K 4

Thus;

(A) Q=900 Watts

20. Supplementary Problems

A furnace wall consist of 35 cm firebrick (k=1.557 W/m-K), 12 cm insulating refractory

(k=0.346) and 20 cm common brick (k=0.692) covered with 7 cm steel plate (k=45). The
temperature at the inner surface of the firebrick is 1230°C and at the outer face of the steel plate is
60°C. Atmosphere 27°C. What is the value of the combined coefficient for convection and radiation
from the outside wall?

A. 31.13 W/m2-K C. 41.3 W/m2-K

B. 30.13 W/m2-K D. 40.13 W/m2-K

Solution:

Q ∆T
=
A RT

Where:

k 12 k 23 k 34 k 45
RT = + + +
x12 x 23 x34 x 45

0.35 0.12 0.20 0.07

¿ + + +
1.557 0.346 0.692 45

m2 K
¿ 0.8621
W
Then,

Q (1230−60 ) K W
= =1357.15 2
A 2
m K m
0.8621
W

Q Q5 −0 t 5−t 0
= =
A A 1
h0

60−27
1357.15=
1
h0

W
h0 =41.13
m2 K

Thus;

(C) 41.13 W/m2-K

21. Supplementary Problem

A dry ice storage chest is a wooden box lined with glass fiber insulation 5 cm thick. The wooden
box (k=0.069) is 2 cm thick and the cubical 60 cm on an edge. The inside surface temperature is
-76°C and the outside surface temperature is 18°C, use k-0.035 for glass fiber insulation.
Determine the heat gain per day.

A. 10,211 kJ C. 12,211 kJ
B. 11,195 kJ D. 9,185 kJ

Solution:

A∆T
Q=
RT

Where:

A=6 [ ( 0.60 ) ( 0.60 ) ]

= 2.16 m2

k 12 k 23
RT = +
x12 x 23
 0.05 0.02
¿ +
0.035 0.069

=1.1718

Then,

2.16 ( 18+76 ) J
Q= =118.18
1.1718 s

J 3600 s
¿ 118.18
s( hr )( 24dayhrs )( 1000
1 kJ
J)

kJ
¿ 10,211. 092
day

Thus;

(A) The heat gain per day is 10,211.092 kJ

22. Supplementary Problem

One side of refrigerated cold chamber is 6m long by 3.7 m and consists of 168 mm thickness of
cork between outer and inner walls of wood. The outer wood wall is 30 mm thick and its outside
face temperature is 20°C, the inner wood wall is 35 mm thick and its inside face temperature is
-3°C. Taking the coefficient of thermal conductivity of cork and wood as 0.042 and 0.20 W/m-K
respectively, calculate the heat transfer per second per sq., m of surface area.

A. 5.138 J C. 6.318 J
B. 4.138 J D. 3.318 J

Solution:

Q ∆T
=
A RT

Where:

0.03 0.168 0.035

RT = + + =4.325
0.20 0.042 0.20

Then,

Q 20+3
= =5.318 W
A 4.325
Thus;

(A) The heat transfer per second per sq., m of surface area is 5.318 J

23. Supplementary Problems

Hot gasses at 280°C flow on one side of a metal plate of 50 mm thickness and air at 35°C flows
on the other side. The heat transfer coefficient of the gases is 31.5 W/m 2-K and that of the air is
32 W/m2-K. Calculate the overall transfer coefficient if the value of thermal conductivity is 0.01.

A. 15.82 W/m2-K C. 14.82 W/m2-K

B. 16.82 W/m2-K D. 17.82 W/m2-K

Solution:

1
U=
RT

Where:

1 x 12 1 1 0.050 1
RT = + + = + +
h1 k 12 h2 31.5 0.01 32

=0.0632

Thus,

1
U= =15.82
0.0632

W
( A ) U=15.82 2
m −K

24. Supplementary Problem

The surface temperature of the hot side of the furnace wall is 1200°C. it is desired to maintain
the outside of the wall at 38°C. A 152 mm of refractory silica is used adjacent to the combustion
chamber and 10 mm of steel covers the outside. What thickness of insulating bricks is necessary
between refractory and steel, if the heat loss should be kept at 788 W/m 2 ? use k=13.84 W/m-K
for refractory silica; 0.15 for insulating brick and 45 for steel.

A. 220 mm C. 260 mm
B. 240 mm D.280 mm
Solution:

x12 x 23 x34 0.152 x 23 0.010

RT = + + = + +
k 12 k 23 k 34 13.84 0.15 45

Solving for RT

∆T
788=
RT

( 1200−38 )
788=
RT

RT =1.475

Then,

0.152 x23 0.010

1.475= + +
13.84 0.15 45

Thus;

x 23=0.22 m
(A)
x 23=220 mm