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4 Pages

### Prob0703

Course: MAE 4263, Fall 2008
School: Oklahoma State
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Word Count: 1282

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7-3: Problem The air enters with a dry-bulb temperature of 50 o F and, at 50% relative humidity, with a wetbulb temperature of 42 o F, according to the Psychrometric Chart (page 821). The main evaporative process follows a constant-enthalpy line, which is nearly parallel to the 42-degree constant-wet-bulb-temperature line in the Psychrometric Chart, ending up at a dry-bulb temperature of 42 degrees when it reaches...

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7-3: Problem The air enters with a dry-bulb temperature of 50 o F and, at 50% relative humidity, with a wetbulb temperature of 42 o F, according to the Psychrometric Chart (page 821). The main evaporative process follows a constant-enthalpy line, which is nearly parallel to the 42-degree constant-wet-bulb-temperature line in the Psychrometric Chart, ending up at a dry-bulb temperature of 42 degrees when it reaches 100% relative humidity. This process would evaporate enough water to increase the humidity ratio from 0.0038 to 0.0055; evaporating 0.0017 lbm of water per lbm of dry air. Taking hf g 1065 BTU/lbmH2 0 , this means = that 1.8 BTU of latent heat would be carried away with each lbm of dry air. However, there is also a heat transfer taking place, tempering the air. For example, taking cp 0.24 = BTU/lbmair -o F, an eight-degree temperature rise back to 50 o F would carry away 1.9 BTU of sensible heat with each lbm of air. The higher temperature would then also provide more evaporation. Would this be enough to drive the natural convection? Equation 7-2 (page 276) suggests that we need a density difference between the environmental air and the departing air of = p gc H g sec2 32.174 lbm-ft 0.015 lbf 144 in2 = ft2 450 ft 32 ft in2 lbf-sec2 = 0.0048 lbm/ft3 according to the Psychrometric Chart (and neglecting the small correction that wet air has 1 + times the mass of dry air), the environmental air has a density of 1/12.93 = 0.077 lbm/ft3 , so the departing air has 0.072 = 1/13.8 taking us to saturated air at 74 o F. Therefore the air flow is less than we might have hoped for with a large cooling tower, and the final temperature is higher than we might have wished for to obtain maximum thermal efficiency of the power-plant Rankine cycle. The actual process evaporates enough water to increase the humidity ratio from 0.0038 to 0.0182; evaporating 0.0144 lbm of water per lbm of dry air. Taking hf g 1065 BTU/lbmH2 0 , this means that 15 = BTU of latent heat are carried away with each lbm of dry air. The temperature rise would carry away about 7 BTU of sensible heat with each lbm of air. The total could be read from an expanded Psychrometric Chart in other texts. Since about 70% of the heat is carried away by evaporation, we are about half-way between typical hotand cold-weather operation (page 268), and expect an intermediate result of about 1.2% in evaporation in Table 7-2 (page 271). Problem 7-12: Indirect dry cooling towers are described on pages 291--297. The heat Q transferred and rejected is known from the air flow and temperature rise Q = mair hair = mair cp Tair 500 106 lbm 0.240 BTU (90 - 60) o F = lbm-o F hour = 3.6 109 BTU/hour For the given flow mCW = 72.1 106 lbm/hour of circulating cooling water, leaving the cooling tower and entering the condenser at TCW in = 70 o F, this amount of heat requires a temperature rise in the condenser and corresponding temperature drop in the cooling tower (TCW out - TCW in ) = Q TCW out mCW cp lbm-o F hour 3.6 109 BTU o = 72.1 106 lbm 0.998 BTU = 50.0 F hour = 120.0 o F For the surface condenser, shown in Figure 7-17, where the circulating cooling water leaves the condenser at To = 8 o F below the temperature of the condensing steam, the condition in the condenser is T2 = 128.0 1 o F, corresponding to a saturation pressure p2 = 2.12 psia. We are now ready to proceed with a cycle calculation, similar to Problem 6-6 P psi 103 1.0 1.0 103 T o F 545 102 102 x % 100 68 0 h Btu/lb 1192.4 775.8 69.7 -3.0 4 72.7 1119.7 # 1 2 3 s /o R 1.39 1.39 w Btu/lb 416.6 q Btu/lb eff % m lb/hr W MW 606 Q 109 Btu/hr -706.1 5106 -3.53 37 5.60 but with a higher condenser pressure (leading to lower efficiency), and with a condenser heat transfer Q23 = 3.6 109 BTU/hour.For the surface condenser: # 1 2 3 4 P psi 103 2.12 1.1 2.12 103 T o F 545 128 128 x % 100 70 0 h Btu/lb 1192.4 811.0 96.0 -3.0 99.0 1093.4 35 5.5 s o / R 1.39 1.39 -715.0 5.0106 -3.6 w Btu/lb 381.4 Btu/lb q eff % m lb/hr W MW 558 Q 9 10 Btu/hr For the direct-contact condenser, shown in Firgure 7-17, circulates condensate from the cycle itself through the cooling tower, and then sprays the cooled water at TCW in = 70 o F back into the condenser to condense steam at TCW out = T2 = 120.0 o F, corresponding to a saturation pressure p2 = 1.695 psia. We repeat the cycle calculation with this condenser pressure, and find that the efficiency is slightly better because heat is rejected from the cycle at a temperature which is 8 o F lower. P psi 103 1.7 1.7 103 T o F 545 120 120 x % 100 69 0 h Btu/lb 1192.4 798.5 88.0 -3.0 4 91.0 1101.4 35.5 5.6 s /o R 1.39 1.39 -710.5 3 5.1106 -3.6 w Btu/lb 393.9 2 q Btu/lb eff % m lb/hr # 1 W MW 580 Q 109 Btu/hr Problem 7-14: Cooling lakes are described on p. 300ff; the heat to be rejected is obtainable from the efficiency as on p. 260f 1 Q = - 1 W net Rej th est 1 - 1 200 106 W = 0.39 = (1.564) 682.6 106 BTU/hour = 1.068 109 BTU/hour 2 letting us calculate from Equation 7-13 1.068 109 Q = = 53.4 106 BTU/hour-o F mcp = (TH - TC ) (95 - 75) From Equation 7-12 we obtain e-UA/mcp -U A mcp = TC - TE 75 - 65 = 0.333 = TH - TE 95 - 65 = ln (0.333) = -1.0986 mcp A = 1.0986 U 1.0986 53.4 106 BTU hour-ft2 -o F = 3.5 BTU hour-o F = 16.76 106 ft2 = 385 acres which is about six-tenths of a square mile, in the same order-of-magnitude as Boomer Lake. Problem 7-15: Spray ponds are described on pp. 303--307ff. As in Problem 7-14 above, which has the same output and efficiency, the heat to be rejected is Q = 1068 106 BTU/hour The empirical heat-transfer Equation 7-14 (page 303) for spray canals is -N (1 - f ) r NTU b TC - Twb = exp TH - Twb cp TC - Twb -N (1 - f ) r NTU b ln = TH - Twb cp where we are given the condenser cooling-water outlet temperature TH = 90 o F Also given are the environmental conditions Tdb = 70 o F at 60% relative humidity; from the psychometric chart in Appendix M, Figure M-1 on page 821, we find = 76 grains/lbm = 0.011 lbm/lbm, and the wet-bulb temperature Twb = 61 o F According to Figure 7-27, a well-designed spray pond with Twb = 61 o F and TH = 90 o F should achieve a circulating-water temperature range of TH - TC TC = 15 o F = 75 o F This is somewhat less than the 20 o F range in Problem 7-14 above, and requires a greater water flow Q = 71.2 106 lbm/hour = 142, 50...

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