HW24soln - PROBLEM(p.18 A 3-hp pump operating at steady...

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Unformatted text preview: PROBLEM (p.18! A 3-hp pump operating at steady state draws in liquid water at 1 atm, 60°F and delivers it at 5 atm at an elevation 20 ft above the inlet.There is no significant change in velocity between the inlet and exit, and the local acceleration of gravity is 32.2 ft/sz. Would it be possible to pump 1000 gal in 10 min or less? Explain. <cH1MA'nc &4w:~ on 7,}; W5.- l' TKOf-‘on‘h’bl VOIUM-l Sk-OWm «new: S-teady - ski-e operah'vg A‘s-0'“ ¢f€ Prov.ded ‘Fvfi A PUMP. FIND: Desi-erqu: H: it would be posrihle ‘l'opu-IMP Iooo gallon: 0F 99-40:“ in lo Wu Or- le:$_ tax—fu- Sch“... h”: is 01‘ sud“ theft. 2. For +l~ coan volume, Ktmh‘c (“(351 effect: an be ijmrcd. '5- l—Ifiu.d bunt-va3 “Md .9 «r: \rficcr.) = 0.01604» {XS/(b (7%A-Ley ANALYSK: 1', a volvmcl‘vra {4mm Lot; (AV) Qfl) (loco s.u)| meme?” I22 I gal chuncc'owd; +9 5 ms: ‘0 W" “a 7 0.2.7.15 H3" -.- 13.8%.“; ’ o-dco4-F13Hb s 69‘ 3'1 claw-14””: 6’4 WINN- pump loco 3gl w; to man“: I crave“, Woo-1U “stun-.0. k 2 l: 0.7.7.28 £2. (I) Tn +99 «bum; 0-?- n‘nhrwd irreuourub.ll.h‘*rj ‘KM power («flu-1‘ Mikael-9.4 wean—9* b.- an)?“ (an, Ark. FDUM‘KS “fussing Olaf-rum} d ’6'“... 53.6.9” urm‘ a :Juw—phm 1/3 C-W‘Am - * E vm-v-J + amt-z.) 45%”7’] Than) Wf‘l’k 0) cm)» 2 QM“ %l[(°'°"°*fie “4"”“4‘ ) + 70V é“- [u “l lfifik‘l] WP ' n I f 163.2% Lffi Ias.au£+_‘_bf 4 2o r+ lb] \ ‘ 3 “L1. lb SSDH-lbf/s To exp-ream: «We! of. saw irrevarc‘ubilJ-h'oc , suck a! fic‘ch‘m between *N Was Ju‘fimd And Paw waJ-AJ pouwr (upwf +0 +8.: r.»wa {-0 Eu («DC lay/aegi- N “Ch/04 § {Low calmd More, Rune-J Nut * 3"”? PM“ “VW’l‘w/ “Minsur'eJ Hoth Cauvo‘h Eu. ‘— 'accomuucl . PROBLEM 6.183 6.183 As shown in Fig. P6. 1 83, 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 200 ft below the intake at 19 psia, 45 ft/s, and a specific volume of 0.01602 ft3/lb. The diameter of the exit pipe is 5 ft and the local acceleration of gravity is 32.2 ft/sz. Evaluating the electricity generated at 8.5 cents per kW-h, determine the value of the power produced, in $/day, for operation at steady state and in the absence of internal irreversibilities. KNOWN: Water at specified pressure and velocity enters a turbine exiting at known elevation difference, pressure, velocity, and specific volume. FIND: The economic value of the power produced. SCHEMATIC AND GIVEN DATA: p2 = 19 psia v2 = 0.01602 fi3/lb V2 = 45 fl/s D2 = 5 ft Water ENGINEERING MODEL: 1. The control volume encloses the piping and turbine—generator as shown in the schematic. 2. The control volume operates with no internal irreversibilities at steady state. 3. Water can be treated as an incompressible substance with constant specific volume (density). ANALYSIS: For a steady-state control volume with no internal irreversibilities and a fluid with constant specific volume, Eq. 6.51a applies and can be used to determine the power (ch )2; = ’57 [V001 —P2) + 1/2 (VI2 — V22) + 8(21 —22)] The constant mass flow rate can be determined by PROBLEM 6.183 (Cowfi'vtuecl) R The cross—sectional area, A, is Mimiz. Applying the mass flow rate expression to the exit [3(5 £0145? fi3 ) =55,154.4lb/s 0.01602— lb "'1: Substituting values, applying appropriate conversion factors, and solving for power give (% )im = rev W (“filz 3 A2 * ‘ _ 55,154.4E 0.01602fi— 241b—f—19fl 144‘" + #+ 3223(20011) lbf hp kW 3 lb in2 in2 112 2 52 lb-ft fi-lbf 134111}, 32.2 2 550 S S =13,496 kW The positive sign associated with power indicates power is produced by the control volume. The generator converts turbine shaft power to electricity. The value of electricity per day is Value = (13,496 kW 8.5 Cents 24—h kW -h 100 cents day = $27,532 per day <———— The effect of irreversibilities is expected to reduce the power produced and thus the daily economic value. PROBLEM c. (84- 6.184 As shown in Figure P6.184, water flows from an elevated reservoir through a hydraulic turbine operating at steady state. Determine the maximum power output, in MW, associated with a mass flow rate of 950 kg/s. The inlet and exit diameters are equal. The water can be modeled as incompressible with v :103 m3/kg. The local acceleration of gravity is 9.8 m/sz. Fig. P6.184 KNOWN: Water flows from an elevated reservoir through a hydraulic turbine at steady state. Mass flow rate is known. FIND: Determine the maximum power output, in MW. SCHEMATIC AND GIVEN DATA: Refer to Fig. P6.184 ENGINEERING MODEL: (1) A one-inlet, one-exit control volume with inlet at 1 and exit at 2, and enclosing the turbine, is at steady state. (2) Liquid water is modeled as incompressible, with v z Vf =10"3 m3/kg. ANALYSIS: The maximum power output of the actual hydraulic turbine is limited by the power produced by an internally reversible turbine, as follows: )actual S )int rev The relationship for internally reversible work is given in Eq. 6.51a and simplified based on assumption 2, as follows: (VIZ—V22)“ (1) 2 J In Eq. (1), the kinetic energy term is eliminated because the diameter, specific volume, and mass flow rate are equal at states 1 and 2. Substituting known values: (Wt )actual S ml:_ v(p2 " p1)+ g(zl _ Zz)+ 10, N l . 3 —2 MW (W,) ,39505 — 10'3m— —0.5bar m +9.832(160m 1N 1~=154MW acma S kg lbar s 1kg-m 106 N.m 2 s J s The maximum power output of the hydraulic turbine is 1.54 MW. ’ ...
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