This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PROBLEM (p.18! A 3hp 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 VOIUMl SkOWm «new: Steady  skie operah'vg A‘s0'“ ¢f€ Prov.ded ‘Fvﬁ A PUMP. FIND: Desierqu: H: it would be
posrihle ‘l'opuIMP Iooo gallon: 0F 9940:“ 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 buntva3 “Md .9 «r: \rﬁccr.) = 0.01604» {XS/(b (7%ALey 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%.“;
’ odco4F13Hb s 69‘ 3'1 claw14””: 6’4 WINN pump loco 3gl w; to man“: I crave“, Woo1U “stun.0. k 2
l: 0.7.7.28 £2. (I) Tn +99 «bum; 0? n‘nhrwd irreuourub.ll.h‘*rj ‘KM power («ﬂu1‘ Mikael9.4
wean—9* b. an)?“ (an, Ark. FDUM‘KS “fussing Olafrum} d ’6'“... 53.6.9” urm‘ a :Juw—phm 1/3 CW‘Am  * E vmvJ + amtz.) 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 SSDHlbf/s To expream: «We! of. saw irrevarc‘ubilJh'oc , suck a! fic‘ch‘m between *N Was Ju‘ﬁmd And Paw waJAJ pouwr (upwf +0 +8.: r.»wa {0 Eu («DC lay/aegi N “Ch/04 § {Low calmd More, RuneJ 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 turbinegenerator, 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 kWh, 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 steadystate 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 lbft ﬁlbf 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 oneinlet, oneexit 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 1kgm 106 N.m
2
s J s The maximum power output of the hydraulic turbine is 1.54 MW. ’ ...
View
Full
Document
This note was uploaded on 12/19/2011 for the course ME 302 taught by Professor Staff during the Spring '06 term at Cal Poly.
 Spring '06
 staff

Click to edit the document details