Heat Transfer: A Practical Approach - Yunus A Cengel Assignment 11 Fall 2003 Tuesday, November 18, 2003 Chapter 11, Problem 49
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1 Heat Transer: A Practical Approach - Yunus A Cengel Assignment Fall 00 Tuesday, November 8, 00 Chapter, Problem 9 The variation o the spectral transmissivity o a 0.6- cm-thick glass window is as given in Figure P 9. Determine the average transmissivity o this window or solar radiation (T < 5800 K and radiation coming rom suraces at room temperature (T < 00 K. Also, determine the amount o solar radiation transmitted through the window or incident solar radiation o 650 W/m. Answers: 0.870, , 565. W/m Figure P.9 Chapter, Solution 9 The variation o transmissivity o a glass is given. The average transmissivity o the pane at two temperatures and the amount o solar radiation transmitted through the pane are to be determined. Analysis For T=5800 K: T = (. 0µ m(5800 K = 70 µ mk = 005. T = = ( µ m(5800k = 7,00µ mk The average transmissivity o this surace is τ T = τ( = (0.9( = ( For T=00 K: T = (. 0µ m(00 K = 90 µ mk = 00. T = ( µ m(00 K = 900 µ mk = , µm Then, τ 0.9 τ T = τ( = (0.9( = ( The amount o solar radiation transmitted through this glass is G = τ = (650 W/m tr G incident = 565. W/m Copyright 00 The McGraw-Hill Companies Inc.
2 Heat Transer: A Practical Approach - Yunus A Cengel Assignment Fall 00 Tuesday, November 8, 00 Chapter, Problem 60 The air temperature on a clear night is observed to remain at about C. Yet water is reported to have rozen that night due to radiation eect. Taking the convection heat transer coeicient to be 8 W/m C, determine the value o the maximum eective sky temperature that night. Chapter, Solution 60 Water is observed to have rozen one night while the air temperature is above reezing temperature. The eective sky temperature is to be determined. Properties The emissivity o water is ε = 0.95 (Table A-. Analysis Assuming the water temperature to be 0 C, the value o the eective sky temperature is determined rom an energy balance on water to be and air surace s sky ht ( T = εσ( T T (8 W/m C( C 0 C = 0.95( W/m K K T sky = 5.8 K T = C T sky =? [(7 T ] Thereore, the eective sky temperature must have been below 55 K. sky Water T s = 0 C ε = 0.8 Copyright 00 The McGraw-Hill Companies Inc.
3 Heat Transer: A Practical Approach - Yunus A Cengel Assignment Fall 00 Tuesday, November 8, 00 Chapter, Problem 8 The spectral absorptivity o an opaque surace is as shown on the graph. Determine the absorptivity o the surace or radiation emitted by a source at (a 000 K and (b 000 K. Figure P.8 Chapter, Solution 8 The variation o absorptivity o a surace with wavelength is given. The average absorptivity o the surace is to be determined or two source temperatures. Analysis (a T = 000 K. The average α absorptivity o the surace can be determined rom α( T = α = α 0- ( - - ( where and are blackbody radiation unctions corresponding to Tand T, determined rom T = (0. µ m(000 K = 00 µ mk T = (. µ m(000 K = 00 µ mk = 0.0 = = 0 = since 0 = 0and = since =. and, α = ( (0.8( (0.0( 0.00 = (a T = 000 K. T = (0. µ m(000 K = 900 µ mk T = (. µ m(000 K = 600 µ mk = = α = ( (0.8( (0.0( = , µm Copyright 00 The McGraw-Hill Companies Inc.
4 Heat Transer: A Practical Approach - Yunus A Cengel Assignment Fall 00 Tuesday, November 8, 00 Chapter, Problem 8 The spectral transmissivity o a glass cover used in a solar collector is given as τ = 0 or < 0. µm τ = 0.85 or 0. µm τ = 0 or > µm Solar radiation is incident at a rate o 950 W/m, and the absorber plate, which can be considered to be black, is maintained at 0 K by the cooling water. Determine (a the solar lux incident on the absorber plate, (b the transmissivity o the glass cover or radiation emitted by the absorber plate, and (c the rate o heat transer to the cooling water i the glass cover temperature is also 0 K. Chapter, Solution 8 The spectral transmissivity o a glass cover used in a solar collector is given. Solar radiation is incident on the collector. The solar lux incident on the absorber plate, the transmissivity o the glass cover or radiation emitted by the absorber plate, and the rate o heat transer to the cooling water are to be determined. Analysis (a For solar radiation, T = 5800 K. τ The average transmissivity o the surace can be determined rom τ( T = τ + τ ( + τ ( where and are blackbody radiation unctions corresponding to T and T. These unctions are determined rom Table - to be T = (0. µ m(5800 K = 70 µ mk T = ( µ m(5800 K = 7,00 µ mk and τ = = 0.05 = ( 0.0( (0.9( (0.0( = 0.85 Since the absorber plate is black, all o the radiation transmitted through the glass cover will be absorbed by the absorber plate and thereore, the solar lux incident on the absorber plate is same as the radiation absorbed by the absorber plate: E abs. plate = τi = 0.85(950 W/m = W/m (b For radiation emitted by the absorber plate, we take T = 00 K, and calculate the transmissivity as ollows: T = (0. µ m(0 K = 0 µ mk = 0.0 T = ( µ m(0 K = 00 µ mk = , µm Copyright 00 The McGraw-Hill Companies Inc.
5 Heat Transer: A Practical Approach - Yunus A Cengel 5 Assignment Fall 00 Tuesday, November 8, 00 τ = ( 0.0(0.0 + (0.9( (0.0( = (c The rate o heat transer to the cooling water is the dierence between the radiation absorbed by the absorber plate and the radiation emitted by the absorber plate, and it is determined rom Energy balance: Q & τ E = Q& since the cover is also at 0 K and there is no interaction absorbed room emitted water E emitted = εσt = ( (5.67e 8(0 = 757 W/m Then Copyright 00 The McGraw-Hill Companies Inc.
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