1 / 27

Application of Steady-State Heat Transfer

Application of Steady-State Heat Transfer. Steady-state heat transfer. Temperature in a system remains constant with time. Temperature varies with location. T 1. T 2. T 1 > T 2. Conductive heat transfer in a rectangular slab. Example.

annis
Download Presentation

Application of Steady-State Heat Transfer

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

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. T1 T2 T1 > T2 Conductive heat transfer in a rectangular slab

  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/m2 thermal conductivity of stainless steel = 17 W/m C

  5. To dr ro T ri Ti r l ConductiveHeatTransferthroughaTubularPipe • Consider a long hollow cylinder

  6. ConductiveHeatTransferthroughaTubularPipe • 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. x1 k1 T1 x2 k2 T2 x3 k3 T3 q q R1 R2 R3 Composite rectangular wall (in series)

  10. = composite thermal resistance

  11. x1 k1 T1 q x2 k2 T2 q x3 k3 T3 R1 R2 R3 Composite rectangular wall (in parallel) q = A T k / x = A T / (x/k)  T1 + T2 + T3= T R = Resistance = x/k = 1/C 1/RT = 1/R1+1/ R2+1/ R3 = (1/(x1 / k 1))+ (1/(x2 / k 2))+ (1/(x3 / k 3))

  12. 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:

  13. 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.

  14. 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.

  15. 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 - 6736.8Heat 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     k (insulation) = 0.20 J/(s-m-C)k (wood)      = 0.80 J/(s-m-C)

  16. B F C E A G D 1 2 3 4 5 RB RF RE RA RC RG RD Series and parallel one-dimensional heat transfer through a composite wall and electrical analog

  17. Composite cylindrical tube(in series) r1 r3 r2

  18. 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.

  19. k = k0(1+T) km isthermal conductivity at T = THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE Heat transfer through a slab 

  20. THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE Heat transfer through a cylindrical tube

  21. B AA = AC = 0.1 m2 q A C D T = 370C T = 66C 2.5 cm 7.5 cm 5.0 cm Problem 1. Find the heat transfer per unit area through the composite wall. Assume one-dimensional heat flow. Given: kA = 150 W/mC kB = 30 W/mC kC = 50 W/mC kD = 70 W/mC AB = AD

  22. 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.

  23. 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 = 150x2 - 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 m2, 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.

More Related