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Application of Steady-State Heat Transfer. Steady-state heat transfer Temperature in a system remains constant with time. Temperature varies with location.

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Presentation on theme: "Application of Steady-State Heat Transfer. Steady-state heat transfer Temperature in a system remains constant with time. Temperature varies with location."— Presentation transcript:

1 Application of Steady-State Heat Transfer

2 Steady-state heat transfer Temperature in a system remains constant with time. Temperature varies with location.

3 Conductive heat transfer in a rectangular slab T 1 > T 2 T1T1 T2T2

4 Example For the stainless steel plate 1 cm thick is maintained at 110  C, while the other face is at 90  C. Calculate temperature at 0.5 cm from the 110  C-temperature face. Given : heat flux = 34,000 W/m 2 thermal conductivity of stainless steel = 17 W/m  C

5 Conductive Heat Transfer through a Tubular Pipe Consider a long hollow cylinder l r roro dr TiTi riri ToTo T

6 Conductive Heat Transfer through a Tubular Pipe Consider a long hollow cylinder

7 Example A 2 cm thick steel pipe (k= 43 W/m  C) with 6 cm inside diameter is being used to convey steam from a boiler to process equipment for a distance of 40 m. The inside pipe surface temperature is 115  C, and the outside pipe surface temperature is 90  C. Under steady state conditions, calculate total heat loss to the surrounding.

8 Heat conduction in multilayered systems

9 Composite rectangular wall (in series) x3k3T3x3k3T3 x2k2T2x2k2T2 x1k1T1x1k1T1 q q R3 R3 R2 R2 R1 R1

10

11 = composite thermal resistance

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13 q q R3 R3 R2 R2 R1 R1 x 1 k 1  T 1 x 2 k 2  T 2 x 3 k 3  T 3 q = A  T k / x = A  T / (x/k)  T 1 +  T 2 +  T 3 =  T R = Resistance = x/k = 1/C 1/R T = 1/R 1 +1/ R 2 +1/ R 3 = (1/(x 1 / k 1 ))+ (1/(x 2 / k 2 ))+ (1/(x 3 / k 3 )) Composite rectangular wall (in parallel)

14 and it is resistance which is additive when several conducting layers lie between the hot and cool regions, because A and Q are the same for all layers. In a multilayer partition, the total conductance is related to the conductance of its layers by: So, when dealing with a multilayer partition, the following formula is usually used:

15 Example A cold storage wall (3m X 6m) is constructed of a 15 cm thick concrete (k = 1.37 W/m  C). Insulation must be provided to maintain a heat transfer rate through the wall at or below 500 W. If k of insulation is 0.04 W/m  C. The outside surface temperature of the wall is 38  C and the inside wall temperature is 5  C.

16 Example How many joules of thermal energy flow through the wall per second? ------------------------------------------- Heat is like a fluid: whatever flows through the insulation must also flow through the wood.

17 k (insulation) = 0.20 J/(s-m-C) k (wood) = 0.80 J/(s-m-C) Across insulation: Hins = (0.20)(40)(25 - T)/0.076 = 2631.6 -105.3 T Across wood: Hwood = (0.80)(40)(T - 4)/0.019 = 1684.2 T - 6736.8 Heat is like a fluid: whatever flows through the insulation must also flow through the wood: Hwood = Hins 1684.2 T - 6736.8 = 2631.6 -105.3 T 1789.5 T = 9368.4 T = 5.235 C H=Hwood=Hins H= 1684.2 (5.235) - 6736.8 = 2080 J/s H= 2631.6 - 105.3 (5.235) = 2080 J/s

18 13254 B A D CE G F RB RB RE RE RA RA RD RD RC RC RG RG RF RF Series and parallel one-dimensional heat transfer through a composite wall and electrical analog

19 Composite cylindrical tube (in series) r1r1 r2r2 r3r3

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21 Example A stainless steel pipe (k= 17 W/m  C) is being used to convey heated oil. The inside surface temperature is 130  C. The pipe is 2 cm thick with an inside diameter of 8 cm. The pipe is insulated with 0.04 m thick insulation (k= 0.035 W/m  C). The outer insulation temperature is 25  C. Calculate the temperature of interface between steel and isulation. Assume steady-state conditions.

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23 THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE k = k 0 (1+  T) k m is thermal conductivity at T = Heat transfer through a slab 

24 THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE Heat transfer through a cylindrical tube

25 Problem 1. Find the heat transfer per unit area through the composite wall. Assume one-dimensional heat flow. Given: k A = 150 W/m  C k B = 30 W/m  C k C = 50 W/m  C k D = 70 W/m  C A B = A D T = 370  C C A D B q A A = A C = 0.1 m 2 T = 66  C 2.5 cm 7.5 cm 5.0 cm

26 Problem 2. One side of a copper block 5 cm thick is maintained at 260  C. The other side is covered with a layer of fiber glass 2.5 cm thick. The outside of the fiber glass is maintained at 38  C, and the total heat flow through the copper-fiber-glass combination is 44 kW. What is the area of the slab? 3. A wall is constructed of 2.0 cm of copper, 3.0 mm of asbestos, and 6.0 cm of fiber glass. Calculate the heat flow per unit area for an overall temperature difference of 500  C.

27 Problem 4. A certain material has a thickness of 30 cm and a thermal conductivity of 0.04 W/m  C. At a particular instant in time the temperature distribution with x, the distance from the left face, is T = 150x 2 - 30x, where x is in meters. Calculate the heat flow rates at x = 0 and x = 30 cm. Is the solid heating up or cooling down? 5. A certain material 2.5 cm thick, with a cross-sectional area of 0.1 m 2, has one side maintained at 35  C and the other at 95  C. The temperature at the center plane of the material is 62  C, and the heat flow through the material is 1 kW. Obtain an expression for the thermal conductivity of the material as a function of temperature.


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