and substituting numerical values with Gr L = Ra L /Pr, find The appropriate empirical correlation for estimating h is given by Eq. 9.

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1 1. (Problem 9.13 in the Book) A square aluminum plate 5 mm thick and 200 mm on a side is heated while vertically suspended in quiescent air at 40 C. Determine the average heat transfer coefficient for the plate when its temperature is 15 C by two methods: using results from the similarity solution to the boundary layer equations, and using results from an empirical correlation. Schematic: Assumptions: (1) Uniform plate surface temperature, (2) Quiescent room air, (3) Surface radiation exchange with surroundings negligible, (4) Perfect gas behavior for air, β = 1/T f. Properties: Table A-4, Air (T f = (T s + T )/2 = (40 +15) C/2 = 300K, 1 atm): ν = m 2 /s, k = W/m K, α = m 2 /s, Pr = Analysis: Calculate the Rayleigh number to determine the boundary layer flow conditions, where β = 1/T f and ΔT = T - T s. Since Ra L < 10 9, the flow is laminar and the similarity solution of Section 9.4 is applicable. From Eqs and 9.20, and substituting numerical values with Gr L = Ra L /Pr, find The appropriate empirical correlation for estimating h L is given by Eq. 9.27,

2 COMMENTS: The agreement of Churchill-Chu correlation, Eq. 9.26, find h L calculated by these two methods is excellent. Using the h L = 4.87 W/m K. This relation is not the most accurate for the laminar regime, but is suitable for both laminar and turbulent regions.

3 2. (Problem 9.31 in the Book) A refrigerator door has a height and width of H = 1 m and W = 0.65 m, respectively, and is situated in a large room for which the air and walls are at T = T sur = 25 C. The door consists of a layer of polystyrene insulation (k = 0.03 W/m. K) sandwiched between thin sheets of steel (ε = 0.6) and polypropylene. Under normal operating conditions, the inner surface of the door is maintained at a fixed temperature of T s,i = 5 C. a) Estimate the heat gain through the door for the worst case condition corresponding to no insulation (L = 0). b) Compute and plot the heat gain and the outer surface temperature T s,o as a function of insulation thickness for 0 L 25 mm. Schematic: Assumptions: (1) Steady-state conditions, (2) Negligible thermal resistance of steel and polypropylene sheets, (3) Negligible contact resistance between sheets and insulation, (4) One-dimensional conduction in insulation, (5) Quiescent air. Properties: Table A.4, air (T f = 288 K): ν = m 2 /s, α = m 2 /s, k = W/m K, Pr = 0.71, β = K -1. Analysis: (a) Without insulation, T s,o = T s,i = 278 K and the heat gain is

4 (b) With the insulation, T s,o may be determined by performing an energy balance at the outer surface, where q conv + q rad = q cond, or Using the IHT First Law Model for a Nonisothermal Plane Wall with the appropriate Correlations and Properties Tool Pads and evaluating the heat gain from the following results are obtained for the effect of L on T s,o and q w. The outer surface temperature increases with increasing L, causing a reduction in the rate of heat transfer to the refrigerator compartment. For L = m, h = 2.29 W/m 2 K, h rad = 3.54 W/m 2 K, q conv = 5.16 W, q rad = 7.99 W, q w = W, and T s,o = 21.5 C. Comments: The insulation is very effective in reducing the heat load, and there would be little value to increasing L beyond 25 mm.

5 3. (Problem 9.45 in the Book) At the end of its manufacturing process, a silicon wafer of diameter D = 150 mm, thickness δ = 1 mm, and emissivity ε = 0.65 is at an initial temperature of T i = 325 C and is allowed to cool in quiescent, ambient air and large surroundings for which T = T sur = 25 C. a) What is the initial rate of cooling? b) How long does it take for the wafer to reach a temperature of 50 C? Comment on how the relative effects of convection and radiation vary with time during the cooling process. Schematic: Assumptions: (1) Negligible heat transfer from side of wafer, (2) Large surroundings, (3) Wafer may be treated as a lumped capacitance, (4) Constant properties, (5) Quiescent air. Properties: Table A-1, Silicon ( T = 187 C = 460K): ρ = 2330 kg/m 3, c p = 813 J/kg K, k = 87.8 W/m K. Table A-4, Air (T f,i = 175 C = 448K): ν = m 2 /s, k = W/m K, α = m 2 /s, Pr = 0.686, β = K -1. Solution: (a) Heat transfer is by natural convection and net radiation exchange from top and bottom surfaces. Hence, with A s = πd 2 /4 = m 2, where the radiation flux is obtained from Eq. 1.7, and with L = A s /P = m and Ra L = gβ (T i - T ) L 3 /αν = , the convection coefficients are obtained from Eqs and Hence,

6 (b) From the generalized lumped capacitance model, Eq. 5.15, Perform the integration and accounting for variations in h t and h b with T, the time t f to reach C is found to be As shown above, the rate at which the wafer temperature decays with increasing time decreases due to reductions in the convection and radiation heat fluxes. Initially, the surface radiative flux (top or bottom) exceeds the heat flux due to natural convection from the top surface, which is twice the flux due to natural convection from the bottom surface. However, because q rad and q cnv decay approximately as T 4 and T 5/4, respectively, the reduction in q rad with decreasing T is more pronounced, and at t = 181s, q rad is well below q cnv,t and only slightly larger than q cnv,b. Comments: With h, = εσ( T i + T sur )( T 2 i + T 2 sur ) = 14.7W/ m 2 K, the largest cumulative r i

7 coefficient of h tot hr, i ht, i = 26.4W/ m 2 K corresponds to the top surface. If this coefficient is used to estimate a Biot number, it follows that Bi = h tot (δ / 2) / k = and the lumped capacitance approximation is excellent.

8 4. (Problem 9.60 in the Book) A horizontal electrical cable of 25-mm diameter has a heat dissipation rate of 30 W/m. If the ambient air temperature is 27 C, estimate the surface temperature of the cable. Schematic: Assumptions: (1) Quiescent air, (2) Cable in horizontal position, (3) Negligible radiation exchange. Properties: Table A-4, Air (T f = (T s + T )/2 = 325K, based upon initial estimate for T s, 1 atm): ν = m 2 /s, k = W/m K, α = m 2 /s, Pr = Analysis: From the rate equation on a unit length basis, the surface temperature is where h is estimated by an appropriate correlation. Since such a calculation requires knowledge of T s, an iteration procedure is required. Begin by assuming T s = 77 C and calculated Ra D, For air, β = 1/T f, and substituting numerical values, Using the Churchill-Chu relation, find h Substituting numerical values into Eq. (1), the calculated value for T s is

9 This value is very close to the assumed value (77 C), but an iteration with a new value of 79 C is warranted. Using the same property values, find for this iteration: We conclude that Ts = 79 C is a good estimate for the surface temperature. Comments: Recognize that radiative exchange is likely to be significant and would have the effect of reducing the estimate for T s.

Module 1 : Conduction. Lecture 5 : 1D conduction example problems. 2D conduction

Module 1 : Conduction. Lecture 5 : 1D conduction example problems. 2D conduction Module 1 : Conduction Lecture 5 : 1D conduction example problems. 2D conduction Objectives In this class: An example of optimization for insulation thickness is solved. The 1D conduction is considered