Problem 1.:A double glazed window is made of 2 glass panes of 6 mm
thick each with an airgap of 6 mm between them. Assuming that the air
layer is stagnant and only conduction isinvolved, determine the thermal
resistance and overall heat transfer coefficient. The inside isexposed
to convection with h = 1.5 W/m2K and the outside to 9 W/m2K. Compare
the valueswith that of a single glass of 12 mm thickness. The
conductivity of the glass = 1.4 W/mK andthat for air is 0.025 W/mK.
2
h1 = 15 W/m K
Glass
Air gap
Glass
k = 1.4 W/mK
k = 0.025 W/mK
Q
2
h2 = 9 W/m K
Q
T1
6 mm
6 mm
T2
RC1
6 mm
(a)
R1
R2
R3
RC2
(b)
Solution: Considering unit area
Double glazing: (Fig. P. 2.3)
Total resistance
1 0.006 0.006 0.006 1
+
+
+
+
15
1.4
0.025
1.4
9
= 0.067 + 0.0043 + 0.24 + 0.0043 + 0.11 = 0.426°Cm2/W
=
1
, A = 1 here
ΣR
1
1
=
U=
= 2.35 W/m2°C
ΣR 0.426
1 0.012 1
+
+
U = 1/
= 5.37 W/m2°C
15
1.4
9
UA =
Single glass,
LM
N
OP
Q
The heat flow will be almost doubled for the same temperature drop. The resistance for
some unit of thickness like inch or cm is often used to compare insulating materials.
Problem 2: A composite wall is made up of 3 layers of thicknesses 25 cm, 10 cm and 15 cm
with thermal conductivities of 1.7, kB and 9.5 W/mK. The outside surface is exposed to air at
20°C with convection coefficient of 15 W/m2K and the inside is exposed to gases at 1200°C
witha convection coefficient of 28 W/m2 K and the inside surface is at 1080°C. Determine the
unknownthermal conductivity, all surface temperatures, resistances of each layer and the
over all heattransfer coefficient. Compare the temperature gradients in the three layers.
Solution: As the gas temperature, inside surface temperature and convection coefficients are
known, the heat flow can be found. Assuming unit area,
Q = 28 (1200 – 1080) = 3360 W
The heat flow is the same allthrough and is equal to Total temperature deep/Total
resistance
1200 − 20
1180
=
1 0.25 0.1 0.15
1
0.1
0.2652 +
+
+
+
+
28
17
9.5
15
.
kB
kB
3360 =
∴
1200°C
kA =
1.7 W/mK
kC = 9.5 W/mK
1080°C
T1
2
h1 = 28 W/m K
2
h2 = 15 W/m K
T2
120°C
Q
20°C
A
B
C
T3
20°C
T2
1080
T2
T1
T2
T3
1200°C
0.25 m
0.1 m
20°C
RC1
0.15 m
RA
(a)
891.07 +
Total resistance
RB
RC
RC2
(b)
336
= 1180
kB
=
∴ kB = 1.163 W/mK
1 0.25
0.1
0.15
1
+
+
+
+
= 0.3512°Cm2/W
28
17
1.163
9.5
15
.
1
= 2.85 W/m2K
R
To determine the surface temperatures, the heat flow is equated to the (temperature
drop/resistance) of each layer
Overall heat transfer coefficient = U =
3360 =
1080 − T1
0.25 / 1.7
3360 =
T1 − T2
585.9 − T2
=
01
. / 1163
.
01
. / 1163
.
297 − T3
T2 − T3
=
015
. / 9.5 015
. / 9.5
Check using outside convection
3360 =
∴
T1 = 585.9°C
T2 = 297°C
∴
T3 = 243.95°C
243.95 − 20
= 3359.25 W
1 / 15
Resistance of layers including convection are: 0.0357, 0.147, 0.086, 0.0158, 0.067°C
Q=
m2/W
Temperature gradient =
T2 − T1
x 2 − x1
1st layer = 585.9 – 1080 = – 494.1°C. gradient =
∆T – 494.1
=
= – 1976.4°C/m
L
0.25
2nd layer = 297 – 585.9 = – 288.9°C. gradient = – 2889°C/m
3rd layer = 243.95 – 297 = – 53.05°C. gradient = – 353.7°C/m
Higher the resistance larger the gradient.
Problem 3: A composite wall is made of 12 mm and 18 mm layers of materials of thermal
conductivity 12.5 and 22.5 W/mK. The contact resistance between surfaces is 5 × 10–4
m2 °C/W.
2
The hot side is exposed to fluid at 400°C with h = 75 W/m K and the cold side is exposed to
fluids at 60°C with h = 400 W/m2K. Determine the heat flow, temperature drop at various
resistances and overall heat transfer coefficient. Comment on the contribution of contact
resistance.
Solution: The specified data are shown in Fig. P. 2.5. Assuming unit area
Q
k1 = 12.5 W/m K
k2 = 22.5 W/m K
400°C
T1
T1
2
h1 = 75 W/m K
T2
T2
T2
T3
T3
400°C
2
h2 = 400 W/m K
T4
60°C
T5
400°C
T2
T1
T2
T3
60°C
60°C
0.012 m 0.018 m
Total resistance
=
RC1
R1
RS
L
L
1
1
+ 1 + Rcontact + 2 +
h1 k1
k2 h2
1 0.012
0.018
1
+
+ 5 × 10 −4 +
+
75
12.5
22.5
400
= 0.018093333 m2 K/W
=
1
= 55.27 W/m2K
R
400 − 60
∆T
=
Q=
= 18791.45 W
R
0.018093333
U=
Temperature drop:
1st convection layer: Q = h1(T∞ – T1)
∴ T1 = 149.447°C
18791.45 = 75(400 – T1) ∆T1 = 250.553°C
149.447 − T2
∆T
=
L / k ( 0.012 / 12.5)
T2 = 131.407°C ∆T2 = 18.04°C
1st conduction layer Q =
∴
Contact surface drop: ∆T3 = Q.R 5 × 10–4 × 18791.45 = 9.4°C
∴
T21 = 122°C
R2
RC2
2nd conduction layer
L
∆T
18791.45 × 0.018
∆T4 = Q × 2 =
L2 / K 2
K2
22.5
∆T4 = 15.03°C T3 = 122 – 15.03 = 106.97°C
Check: using last convection layer
Q = h(∆T5) = 400(106.97 – 60) = 18788 W checks
The contribution of contact resistance is small as this involves a temperature drop of
9.4°C out of 340°C. This is the general order of contact resistance the heat flow calculated
neglecting contact resistance is 19325.50 W. Which is higher by 2.76%. This is less than errors
due to uncertainty in values of k. However, whenever possible the contact resistance should be
taken into account.
Q=
Problem 4:A composite slab is made of three layers 15 cm, 10 cm and 12 cm thickness. The
first layer is of material with thermal conductivity 1.45 for 60% of the area and the rest is of
material with conductivity of 2.5 W/mK. The second layer is made of material with
conductivityof 12.5 W/mK for 50% area and of material with conductivity 18.5 W/mK is used
for the other50%. The third layer is of single material of thermal conductivity 0.76 W/mK.
The slab isexposed on one side to warm air at 26°C and to cold air at – 20°C on the other side.
The convectioncoefficients are 15 and 20 W/m2K on the inside and outside respectively.
Determine the heatflow and interface temperatures.
Solution: The thermal resistances of the five material section are found assuming a total area
of 1 m2. Refer Fig. P. 2.6.
Q
k = 2.5
k = 12.5 k = 0.76
RB
RC
RC1
RC2
RE
B, 40%
26°C
C, 50%
– 20°C
RA
h1 = 15 W/m K
A, 60%
26°C
k = 1.45
2
1
0.15 m
RD
h2 = 20 W/m2K
E
D, 50%
k = 18.5
(b)
– 20°C
2
3
4
0.1 m
0.12 m
(a)
Q
26°C
– 20°C
RC1
Req1
Req2
R0
RC2
(c)
∴
Rc1 =
1
0.15
0.15
= 0.06667, RA =
, RB =
1 × 15
1.45 × 0.6
0.4 × 2.5
∴
RC =
0.1
0.1
0.12
, RD =
, RE =
= 0.1579
12.5 × 0.5
18.5 × 0.5
0.76 × 1
Rc2 =
1
= 0.05
1 × 20
The equivalent resistance for the parallel paths are found using
1
1
+
where R1 and R2 are the resistances in parallel
R1 R2
1/Req1 = [1/(0.15/1.45 × 0.6)] + [1/(0.15/0.4 × 2.5)]
1/Req =
∴
∴
∴
∴
∴
Req1 = 0.0802 m2C/W
1/Req2 = [1/(0.1/18.5 × 0.5)] + [1/(0.1/12.5 × 0.5)]
Req2 = 0.006452 m2C/W
ΣR = 0.06667 + 0.0802 + 0.006452 + 0.1579 + 0.05 = 0.3612
Q=
Interface temperatures:
face 1:
∴
face 2:
∴
face 3:
∆T 26 − ( − 20)
=
= 127.35 W/m2
ΣR
0.3612
∆T2 26 − T1
=
= 127.35
1 / 15
R1
∆T1 = 8.49
∴
Q=
Q=
T1 = 17.51°C
∆T2
∆T2
=
= 127.35
Req1 0.0802
∆T2 = 10.21°C
∴ T2 = 7.3°C
∆T3
∆T3
=
Q=
= 127.35
Req2 0.006452
∆T3 = 0.82°C
∴ T3 = 6.48°C
∆T4
∆T4
=
face 4:
Q=
= 127.35
0.12 / 0.76
R4
∴
∆T4 = 20.11°C
∴ T4 = – 13.63°C
Check: heat flow by convection on the cooler side:
h {– 13.63 – (– 20)} = 20 × 6.372 = 127.44 W
It may be noted that some heat flow occurs between A and B as well as between C and D
at their interface and hence the temperature variation in individual materials cannot be
established by this analysis.
∴
Problem 5: A heat flux of 1500 W/m2 is incident on the surface of a slab 10 cm thick with
thermal conductivity of 7.5 W/mK. The hot side is found to be at 120°C. On the otherside, the
heat is passed on to the surroundings at 30°C both by convection and radiation. It radiation
isideal, determine the convection coefficient and also the share of heat flow between the two
processes.
Solution: The specified data are shown in Fig. P. 2.7. Looking at the equivalent circuit,
Fig P. 2.7(b), the value of h is identified as the unknown. The value of h can be found if the
surface temp. is determined. Using the conduction layer only
Q = 1500 =
120 − T2
. / 7.5
01
∴
T2 = 100°C
To find hr, temperatures should be in absolute units
T2 = 273 + 100 = 373 K, Ts = 273 + 30 = 303 K
k = 7.5 W/mK
Ts = 30°C
120°C
h=?
2
Q = 1500 W/m
T2
1/hr
Qr
Ts = 30°C
T1 = 120°C
Qcv
L/k
T = 30°C
1/h
0.1 m
(a)
1/hr =
(b)
1
2
2
σ(T2 + Ts ) (T2 + Ts )
=
1
5.67 × 10
−8
(373 + 303) × (3732 + 303 2 )
= 0.112973 m2°C/W
100 − 30
= 1500
R
Using the parallel circuit (Fig. 2.7 b)
Q=
∴
1
= hr + h
R
h = 12.58 W/m2K
∴
R = 0.0466°Cm2/W
∴
h = (1/0.0466) – (1/0.112973)
Qr = σ (T24 – Ts4) = 5.67 (3.734 – 3.034) = 619.62 W 41.3%
Qcv = 12.58 (100 – 30) = 880.6 W. 58.7% Total: 1500.72 W checks.
Problem 6 :A 2 kW heater element of area 0.04 m2 is protected on the backside with
insulation50 mm thick of k = 1.4 W/mK and on the front side by a plate 10 mm thick with
thermalconductivity of 45 W/mK. The backside is exposed to air at 5°C with convection
coefficient of10W/m2 K and the front is exposed to air at 15°C with convection coefficient
including radiationof 250 W/m2K. Determine the heater element temperature and the heat
flow into the roomunder steady conditions.
Solution: The equivalent circuit can be drawn as in Fig. 2.8(b).
Q1 + Q2 = 2000 W
Q2 =
Q1 =
To − 5
To − 15
= 9.4737 (To – 15)
0.01
1
+
45 × 0.04 250 × 0.04
0.05
1
+
1.4 × 0.04 10 × 0.04
= 0.29473 (To – 5)
Q1 + Q2 = 9.4737 (To – 15) + 0.29473 (To – 5) = 2000
k = 1.4 W/mK
2
h = 250 W/m K
T0
2
h = 10 W/m K
T1
15°C
5°C
Q
T2
Q2
Heater
Q2
T2
5°C
Q1
Q1
RC2
50 mm
T0
T1
R2
R1
(a)
(b)
9.7684 To – 143.58 = 2000
Room side
Back side
Q2 =
Surface temperature
∴
RC1
Q
10 mm
Q1 =
15°C
∴
To = 219.44°C
(219.44 − 15)
= 1936.8 W, 96.84%
0.01
1
+
45 × 0.04 250 × 0.04
(219.44 − 5)
= 63.2 W, 3.16%.
1
0.05
+
10 × 0.04 1.4 × 0.04
Room side
∆T1 = Q1 ×
1
1936.8
= 193.68°C
=
h1 A 250 × 0.04
T1 = 193.68 + 15 = 208.68°C
Back side
∆T2 = Q2 ×
1
1
= 158°C
= 63.2 ×
h2 A
10 × 0.04
T2 = 158 + 5 = 163°C.
Problem 7:To reduce frosting it is desired to keep the outside surface of a glazed window at4
°C. The outside is at – 10°C and the convection coefficient is 60 W/m2K. In order to maintain
the conditions a uniform heat flux is provided at the inner surface which is in contact with
room air at 22°C with a convection coefficient of 12 W/m2K. The glass is 7 mm thick and has
athermal conductivity of 1.4 W/mK. Determine the heating required per m2area
The data are shown is Fig. P. 2.9.
Solution: The heat flow through the barrier
= heat convected on the outside
= h (T – T∞1)
= 60 (4 – (– 10)) = 840 W/m2
The heat flow through the barrier is the same
840 =
T1 − 4
∆T
=
0.007 / 1.4
R
∴
T1 = 8.2°C
The heat flux + heat received by convection from room = heat flow through barrier
heat flux = heat flow through glass barrier – heat convected from inside
= 840 – 12 (22 – 8.2) = 840 – 165.6 = 674.4 W
k = 1.4 W/mK
2
T1
h = 60 W/m K
2
h = 12 W/m K
Q2
– 10°C
T2
22°C
Heater
4°C
T1
22°C
– 10°C
Q1
– 10°C
RC2
R1
0.007 mm
Q
(a)
(b)
RC1
If it is desired that the inside well temperature and room temperatures should be equal
for comfort, determine the heat flux. In this case T1 = 22°C and T2 is not known
But heat conducted = heat convected
22 − T2
= 60 (T2 – (– 10))
0.007 / 1.4
solving
22 – T = 0.3 T + 3,T2 = 14.62°C
Q=
Check
This is almost double.
22 − 14.62
= 1477 W
0.007 / 1.4
Q = h(T – T∞) = 60 × 24.62 = 1477.2 W