Question

Problem 6.123 As shown in the figure below, water behind a dam enters an intake pipe at a pressure of 24 psia and velocity of 5 ft/s, flows through a hydraulic turbine-generator, and exits at a point 21 = 140ft below the intake at 19 psia, 45 ft/s, and a specific volume of 0.01602 ft/lb. The diameter of the exit pipe is 5 ft and the local acceleration of gravity is 32.2 ft/s2. Water Dam P1 = 24 psia Vi = 5 ft/s Generator Turbine z P2 = 19 psia 2n=0.01602 ft3/1b V2 = 45 ft/s Water ID=5 ft Evaluating the electricity generated at 8.5 cents per kWh, determine the value of the power produced, in $/day, for operation at steady state and in the absence of internal irreversibilities. Value = $. /day

Answer 1

Answer from given data, Pi = ?A psia, Vi = 5 ft/s Z,-Z₂ = 140 ft P2 = 19 psía, V2= 45 ft/s D2 = 54t, Az = 0.01602 fts/e6, 9= 32-2 ft/s? and 8.5 cents/kw.h from energy balance, equation, V2 ห้า -Sud v dp +967,-72) + 2 Wew De = int Yer 2 = -OCP2 - P.)+96Z1-Zz) + Substitute values + = -0.01.02 (19-24) dbkx 144 in? 32.2 et 140 ft. Elbet in? ft2 t 32.2 lbm. ft sz ] 52 452 ft² albf 32.2 lbm .ft 82 =120.478 Dbfoft abm =120.478 lbf.ft ebm IBTU 778ebf.ft بے بن = 0.154.85 BTU abm

Mass How rate (m) = A2V2 og - 2 (02)) v₂ 3.14 (52)(45) = A 0.01602 = 55126.4 lbmls from eq A wer =0.15485 m = 0.15485 Wer 55126.4 3600 sec X ükr = 8536.323 BTU/sec ::8536.323 BTU 1 kwn 3413 BTU х Ihr sec = 9004.033 kW .. The value of power required is = werx8.5 = 9004.033 * 8.5 = 76534 • 282 cents/no = 765.34 $/nr. so per day it is = 765.34 X 24 = 183688227 $/dag An aus Good Luck