Application of Steady-State Heat Transfer
Steady-state heat transfer Temperature in a system remains constant with time. Temperature varies with location.
Conductive heat transfer in a rectangular slab
Example For the stainless steel plate 1 cm thick is maintained at 110C, while the other face is at 90 C. Calculate temperature at 0.5 cm from the 110C-temperature face. Given : heat flux = 34,000 W/m2 thermal conductivity of stainless steel = 17 W/m C
Conductive Heat Transfer through a Tubular Pipe Consider a long hollow cylinder l r ro dr Ti ri To T
Conductive Heat Transfer through a Tubular Pipe Consider a long hollow cylinder
Example A 2 cm thick steel pipe (k= 43 W/mC) 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 115C, and the outside pipe surface temperature is 90C. Under steady state conditions, calculate total heat loss to the surrounding.
Heat conduction in multilayered systems
Composite rectangular wall (in series) x3 k3 T3 x2 k2 T2 x1 k1 T1 q R3 R2 R1
= composite thermal resistance
Composite rectangular wall (in parallel) q R3 R2 R1 x1 k1 T1 x2 k2 T2 x3 k3 T3 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))
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:
Example A cold storage wall (3m X 6m) is constructed of a 15 cm thick concrete (k = 1.37 W/mC). 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/mC. The outside surface temperature of the wall is 38C and the inside wall temperature is 5C.
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.
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 k (insulation) = 0.20 J/(s-m-C) k (wood) = 0.80 J/(s-m-C)
Series and parallel one-dimensional heat transfer through a composite wall and electrical analog 1 3 2 5 4 B A D C E G F RB RE RA RD RC RG RF
Composite cylindrical tube (in series)
Example A stainless steel pipe (k= 17 W/mC) is being used to convey heated oil. The inside surface temperature is 130C. 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/mC). The outer insulation temperature is 25C. Calculate the temperature of interface between steel and isulation. Assume steady-state conditions.
THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE Heat transfer through a slab k = k0(1+T) km is thermal conductivity at T =
THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE Heat transfer through a cylindrical tube
Problem 1. Find the heat transfer per unit area through the composite wall. Assume one-dimensional heat flow. Given: kA = 150 W/mC kB = 30 W/mC kC = 50 W/mC kD = 70 W/mC AB = AD T = 370C C A D B q AA = AC = 0.1 m2 T = 66C 2.5 cm 7.5 cm 5.0 cm
Problem 2. One side of a copper block 5 cm thick is maintained at 260C. The other side is covered with a layer of fiber glass 2.5 cm thick. The outside of the fiber glass is maintained at 38C, 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 500C.
Problem 4. A certain material has a thickness of 30 cm and a thermal conductivity of 0.04 W/mC. 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 35C and the other at 95C. The temperature at the center plane of the material is 62C, 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.