Find out the amount of heat transfer for a material in which thermal
conductivity is 0.24 Wim K.The cross-sectional area is 150 square meter
and a thickness of 100 cm. The hot temperature is 300 °C, where the cold
temperature is 25°C.
Answers
Answer:
Example 3.20 Determine the heat transfer rate across a composite slab which is made of different materials with top and bottom as shown in fig. 3.16. The entire left-hand face is held at the temperature T1 while the entire right hand face is at the temperature T2. The conductivities of the two different materials are stated as ka and kb, and their areas as viewed in the direction of slab thickness δ are Aa and Ab respectively. Steady state exists, there is no heat generation and the slab is so thin that any edge effects can be neglected. Interpret the result in terms of an electrical circuit.
Solution: Applying Fourier law of heat conduction separately to materials a and b, we obtain
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The desired quantity of heat transfer through the slab equal the sum of Qa and Qb
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Apparently, the two thermal resistances
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Appear in the same way as two electrical resistors in parallel. Accordingly the electrical circuit for the heat transfer through the given composite slab will be as indicated in fig. 3.16.
Example 3.20 Find the heat flow rate through the composite wall as shown in fig.3.17. Assume one-dimensional flow and take
Ka =150 W/m-deg; kb = 30 W/m-deg; kc = 65 W/m-deg; kd = 50 W/m-deg
Solution: The equivalent thermal circuit for heat flow in the composite system has been shown in fig. 3.18.
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The resistances Rb and Rc are in parallel and their equivalent resistance Req is
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The equivalent resistance is now in series with resistance Ra and Rd. The total thermal resistance for the entire circuit then becomes
∑ Rt=Ra + Req + Rb
=0.02+0.1469+0.1 =0.2669°C/W
Hence heat transfer rate through the system is
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Example 3.24 Two slabs, each 200 mm thick and made of materials with thermal conductivities of 16 W/m-deg and 1600 W/m-deg, are placed in contact which is not perfect. Due to roughness of surfaces, only 40% of area is in contact and air fills 0.02 mm thick gap in the remaining area. If the extreme surfaces of the arrangement are at temperatures of 250° C and 30° C, determine the heat flow through the composite system, the contact resistance and temperature drop in contact.
Take thermal conductivity of air as 0.032 W/m-deg and assume that half of the contact (of the contact area) is due to either metal.
Solution: Refer fig. 3.22 for the composite system and its equivalent thermal resistance
The various thermal resistances to flow of heat are:
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The resistances , and are in parallel and their equivalent resistance (Rt)eq is
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This equivalent resistance is now in series with resistance and . The total thermal resistance for the entire circuit then becomes
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Hence, heat transfer rate through the system is
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Example 3.34 A glazed window, made of 8mm thick glass of thermal conductivity 1.5 W/mK, has its outside surface maintained at 5°C so that frosting is reduced. The surroundings are at-10°C with convective coefficient 55 W/m2K.the desired condition is attained by providing a uniform heat flux at the inner surface of the window which is fitted into a room where the air temperature is 25° with convection of 12.5 W/m2K. Make calculations for the heating required per m2 area.
Solution: Refer fig.3.35 for the window fixture with specified data and thermal circuit for the resistance involved.
Let t1 be the temperature at the heater. Under steady state conditions heat conducted through the glass barrier equals the heat convected through the outde film. That is
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Considering unit area,
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That gives:
(b) From energy balance
Heat flow (Q) + heat received by convection fram room (Q2)
= heat conducted through the glass barrier (Q1)
or heat flux Q = Q1 – hi A (ti –t1)
=825-12.51(25-9.4)= 630 W
Thus, the heat required per m2 area is 630W.
Example 3.34 A square plane heater of 0.8 kW rating and measuring 15cm15cm is placed between two slabs A and B and the following data refers to these slabs:
Slab A is 1.8 cm thick with k = 55 W/m-deg
Slab B is 1cm thick with k = 0.2 W/m-deg
The outside heat transfer coefficients on the side of plate A and B are 200 W/m2-deg and 45 W/m2-deg respectively. If the surrounding environment is at 27°C temperature, make calculations for the maximum temperature of the system and outside surface temperature of both slabs.
Solution: Refer fig. 3.36 for the arrangement and thermal resistance network for the system.
The individual resistances are evaluated as:
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These resistances are in series and accordingly for slab A (left branch of the circuit)
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These resistances are in series and accordingly for slab B (right branch of the circuit)
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(a) Rating of Heater, Q = QA+ QB
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Maximum temperature in the system,
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(b) Considering left side branch of the circuit (slab A)
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If T1 is the temperature at exposed surface of slab A, then
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Considering right side branch of the circuit (slab B)
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