MAE 110A. Homework 4: Solutions 10/27/2017

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1 MAE 0A Homwork 4: Solution 0/27/207

MS 4.20: Th figur blow provid tady-tat data for watr vapor flowing through a piping configuration. At ach xit, th volumtric flow rat, prur, and tmpratur ar qual.

8 bar Uniform proprti (D flow) T 2 = T 3 = 320 A 2 V 2 = A 3 V 3 m =?, m =? Govrning Equation i i m m = ρ(v n)da Eq. 4.3 Dtrmin th ma flow rat of th inlt.

= m i i m m 3 Sinc tat 2 and tat 3 ar qual v 2 = v 3, and inc A 2 V 2 = A 3 V 3 it can b n m 2 = m 3 = m 0 = m 2m m = m 2 = 5.76 m = 5.

2 MS 4.20: Th figur blow provid tady-tat data for watr vapor flowing through a piping configuration. At ach xit, th volumtric flow rat, prur, and tmpratur ar qual. Dtrmin th ma flow rat at th inlt and xit, ach in /. Givn Data Modl Diagram V = 30 m/, A = 0.2 m 2 Control volum analyi P = 5 bar, T = 360 Stady tat (d/ = 0) P 2 = P 3 = 4.8 bar Uniform proprti (D flow) T 2 = T 3 = 320 A 2 V 2 = A 3 V 3 m =?, m =? Govrning Equation i i m m = ρ(v n)da Eq. 4.3 Dtrmin th ma flow rat of th inlt. m = ρ (V n )da m = ρ A V = A V v Th pcific volum i fixd by P and T in uprhatd vapor tabl v = m3 m = A V v = (0.2 m2 )(30 m/) m 3 / = m = 0.4 Dtrmin th ma flow rat at ach xit. = m i i m m 3 Sinc tat 2 and tat 3 ar qual v 2 = v 3, and inc A 2 V 2 = A 3 V 3 it can b n m 2 = m 3 = m 0 = m 2m m = m 2 = 5.76 m = 5.2 Dicuion In homwork -3 w primarily conidrd clod ytm (aka a control ma) in which w idntifid a fixd quantity of mattr and applid conrvation law. Hr w hav lctd a control volum (aka opn ytm) in which w idntify a rgion of pac and mattr can pa frly through th control urfac. Th am conrvation law apply but mut b adaptd to a control volum analyi (uing Rynold Tranport Thorm for tho intrtd) to account for ma flux through th control volum. In thi problm w ud conrvation of ma which tat th rat of chang of ma inid th control volum i qual to th nt ma flux through th control urfac. Auming tady tat and uniform proprti at ach cro ction rducd th problm to a impl D flow.

Givn Data Modl Diagram P = 8 kpa Control volum (diffur) T = 26 K Stady tat (d/ = 0) V = 265 m/ Uniform proprti at and 2 T 2 = 250 K Adiabatic (Q in out = 0) V 2 =?

3 MS 4.37: A hown blow, air ntr th diffur of a jt ngin oprating at tady tat at 8 kpa, 26 K and a vlocity of 265 m/, all data corrponding to high-altitud flight. Th air flow adiabatically through th diffur and achiv a tmpratur of 250 K at th diffur xit. Uing th idal ga modl for air, dtrmin th vlocity of th air at th diffur xit, in m/. Givn Data Modl Diagram P = 8 kpa Control volum (diffur) T = 26 K Stady tat (d/ = 0) V = 265 m/ Uniform proprti at and 2 T 2 = 250 K Adiabatic (Q in out = 0) V 2 =? No work ( in = out = 0) Idal ga Nglct lvation chang (z z 2 ) Govrning Equation = m i i m m (h + i Eq. 4.5 = m i i m Contant ma flow rat for tady, on inlt, on outlt flow m (h + i 0 = m (h + 2 V 2 ) m 2 (h V 2 2 ) h V 2 2 = h + 2 V 2 V 2 = 2(h h 2 ) + V 2 Th nthalpy valu can b found from th idal ga proprti of air (Tabl A22). Sinc it i an idal ga h = h(t). h = / and h 2 = / V 2 = 2 (25.97 V 2 = 45.4 m/ N m ) [000 ] [ m ] + (265 m N 2 )2 Dicuion A diffur i ud to dclrat a flow through ara variation. No work i don xcpt th comprion work includd in h and chang in lvation ar ngligibl. A xpctd th vlocity at th diffur xit i rducd. Th chang in kintic nrgy of th fluid go into raiing it nthalpy and tmpratur. Th txt driv quation for variou dvic including diffur. In fact quation 4.2 match th abov analyi. Whil on could jump to th final rult givn in th txt, thi approach i not rcommndd. Only thr fundamntal phyical law (ma, nrgy, and 2 nd law of thrmodynamic in ch. 5/6) ar ud in thi cla and larning how to apply thm in thir gnral form i far mor important than mmorizing dozn of quation with limitd utility. 2

MS 4.43: Air xpand through a turbin from 8 bar, 960 K to bar, 450 K. Th inlt vlocity i mall compard to th xit vlocity of 90 m/. Th turbin oprat at tady tat and dvlop a powr output of 2500 kw.

4 MS 4.43: Air xpand through a turbin from 8 bar, 960 K to bar, 450 K. Th inlt vlocity i mall compard to th xit vlocity of 90 m/. Th turbin oprat at tady tat and dvlop a powr output of 2500 kw. Hat tranfr btwn th turbin and it urrounding and potntial nrgy ffct ar ngligibl. Modling air a an idal ga, calculat th ma flow rat of air, in /, and th xit ara, in m 2. Givn Data Modl Diagram P = 8 bar P 2 = bar Control volum T = 960 K T 2 = 450 K Stady tat V 0 V 2 = 90 m/ Adiabatic (Q in out = 0) out = 2500 kw Nglct lvation chang (z z 2 ) m =? A 2 =? Idal ga V V 2 Govrning Equation i i m m = ρ(v n)da Eq. 4.3 m (h + i Eq. 4.5 P = ρrt Eq = m i i m Contant ma flow rat for tady, on inlt, on outlt flow m (h + i 0 = out + m h m 2 (h V 2 2 ) 0 = out + m (h h V 2 2 ) m = out h h V 2 2 Enthalpy valu can b found from th idal ga proprti of air (Tabl A22). Sinc it i an idal ga h = h(t). m = out h h V 2 2 = m = 4.53 / Calculat th xit ara in m 2, m 2 = ρ 2 (V 2 n 2 )da = ρ 2 A 2 V 2 A 2 = m 2 ρ 2 V 2 P 2 = ρ 2 RT 2 or ρ 2 = P 2 RT 2 = (2500 kw)[ /kw ] = / ( ) (45.80 )+ 2 (90m )2 [ N 2 m ][ 000 N m ] / ( bar) N/m2 [05 )(450 K) bar (287 J K A 2 = m 2 = = ρ 2 V 2 ( m2 m3)(90 m/) A 2 = m 2 ] [ J N m ] = Dicuion: Turbin gnrat powr. High prur and tmpratur fluid ntr th turbin. Enrgy i rmovd from th fluid a th fluid prform work in rotating th turbin blad. Th lowr nrgy fluid xit th turbin at a lowr prur and tmpratur than th inlt. m 3 3

Givn Data Modl Diagram P = MPa Control volum T = 20 = 253 K Stady tat P 2 = 6 MPa Adiabatic (Q in out = 0) m = 0.02 / Nglct lvation chang (z z 2 ) in = 2 kw Nglct KE chang (V V 2 ) M = 44.

5 H4.2: Carbon dioxid ntr an adiabatic compror at MPa, 20 and xit at 6 MPa. Th ma flow rat i 0.02 / and th powr input i 2 kw. Aum tady tat opration, ngligibl kintic and potntial nrgy ffct, and idal ga bhavior. Dtrmin th xit tmpratur [ ]. Givn Data Modl Diagram P = MPa Control volum T = 20 = 253 K Stady tat P 2 = 6 MPa Adiabatic (Q in out = 0) m = 0.02 / Nglct lvation chang (z z 2 ) in = 2 kw Nglct KE chang (V V 2 ) M = 44.0 kmol T 2 =? Idal ga (CO 2) Govrning Equation = m i i m m (h + i Eq. 4.5 = m i i m m (h + i 0 = in + m (h h 2 ) h 2 = h + in M m Convrt to /kmol for conitnt unit h 2 = ( 2 kw ) (44.0 kmol h 2 = 233 kmol T K = / ) [ kmol kw ] = kmol Dicuion Conrvation of nrgy wa applid to dtrmin h 2 which wa rlatd to T 2 uing th idal ga proprti of carbon dioxid (Tabl A23). Both tat ar dfind by known prur and tmpratur. A xpctd th compror rducd th pcific volum of air it wa comprd. Enrgy addd to th ga by work input can b n in th incrad nthalpy, tmpratur, and prur. 4

H4.3: Conidr a valv in a rfrigration ytm which u ammonia a th rfrigrant. Th ammonia ntr th valv at 36 and.5 MPa and xit at 300 kpa.

tmpratur drop acro th valv [ ] Givn Data Modl Diagram T = 36 Control Volum P =.5 MPa Stady tat P 2 = 300 kpa Adiabatic (Q in out = 0) x 2 =?, ΔT =?

3 h = h f + x(h g h f ) Eq. 3.7 = m i i m m (h + i 0 = m (h h 2 ) h 2 = h Looking at th aturation data corrponding to T = 36 it can b n that th ammonia i a comprd liquid at tat (P 2 > P at (T 2 )).

3896 MPa] [ MPa m 3] h 2 = h = 35.8784 Looking at aturation data corrponding to P = 300 kpa it can b n that h f < h 2 < h g o th ammonia i a aturatd liquid vapor mixtur.

6 H4.3: Conidr a valv in a rfrigration ytm which u ammonia a th rfrigrant. Th ammonia ntr th valv at 36 and.5 MPa and xit at 300 kpa. Aum tady tat condition, ngligibl kintic and potntial nrgy chang, and ngligibl hat tranfr. Dtrmin: a. quality of th ammonia at th valv xit b. tmpratur drop acro th valv [ ] Givn Data Modl Diagram T = 36 Control Volum P =.5 MPa Stady tat P 2 = 300 kpa Adiabatic (Q in out = 0) x 2 =?, ΔT =? Nglct lvation chang (z z 2 ) Nglct KE chang (V V 2 ) Govrning Equation m (h + i Eq. 4.5 i i m h(t, P) h f (T) + v f (T)[P P at (T)] Eq. 3.3 h = h f + x(h g h f ) Eq. 3.7 = m i i m m (h + i 0 = m (h h 2 ) h 2 = h Looking at th aturation data corrponding to T = 36 it can b n that th ammonia i a comprd liquid at tat (P 2 > P at (T 2 )). Sinc comprd liquid data i unavailabl w will approximat th nthalpy uing aturation data (aturatd liquid approximation). h(t, P) h f (T) + v f (T)[P P at (T)] h = (35.69 m3 000 ) + ( ) [.5 MPa.3896 MPa] [ MPa m 3] h 2 = h = Looking at aturation data corrponding to P = 300 kpa it can b n that h f < h 2 < h g o th ammonia i a aturatd liquid vapor mixtur. Inid th vapor dom tmpratur and prur ar dpndnt proprti. T 2 = T at = 9.24 ΔT = T 2 T = h = h f + x(h g h f ) = h f + xh fg / / x = h h f = = h fg / x = 6.57% Dicuion: Valv will dcra th prur of th fluid, P 2 < P. Thi i don through a throttling proc, and h = h 2, a n from conrvation of nrgy. Whil undrgoing thi proc 6.57% of th ammonia chang pha from liquid to vapor and thr i a tmpratur drop of 45. Thi nabl th fluid to aborb nrgy at th vaporator of th rfrigration cycl. 5

Aum tady tat opration, ngligibl chang in kintic nrgy, and contant watr tmpratur. Dtrmin th rquird pumping powr [kw].

7 H4.4: A mall pump i locatd down in a wll at a dpth of 5 m, drawing in liquid watr at 0, 90 kpa (ab) at a rat of.5 /. Th xit lin of th pump i a pip that go up to a rcivr tank (at z = 0) maintaind at a gag prur of 400 kpa. Conidr th liquid watr a an incompribl ubtanc. Aum tady tat opration, ngligibl chang in kintic nrgy, and contant watr tmpratur. Dtrmin th rquird pumping powr [kw]. Givn Data Modl Diagram z = 5 m, z 2 = 0 m Control Volum T = T 2 = 0 = 283 K Stady tat P = 90 kpa, P 2 = 400 kpa (gag) Adiabatic (Q in out = 0) m =.5 / Nglct KE chang (V V 2 ) in =? Incompribl ubtanc Govrning Equation h = u + Pv Eq. 3.4 du = cdt Eq. 3.8 = m i i m m (h + i Eq. 4.5 = m i i m m (h + i 0 = in + m (h h 2 + gz gz 2 ) in = m (h 2 h + gz 2 gz ) Conidr th diffrntial of h, dh = du + vdp + Pdv dv = 0 (incompribl) dh = du + vdp dh = cdt + vdp dt = 0 (T = contant) dh = vdp 2 dh = vdp h 2 h = v(p 2 P ) v = contant (incompribl) Fluid proprti can b approximatd by aturation valu, v(t, P) = v f (T) 3 = m 3. P 2,ab = P 2,gag + P atm = 400 kpa + 0 kpa = 50 kpa h 2 h = ( m 3) (50 kpa 90 kpa) [ kpa m 3] = 0.4 in = (.5 ) {0.4 + (9.8 m N 2 2) (5 m) [ ] [ kw ]} [ ] = kw m 000 N m / in = 0.84 kw Dicuion: Th pump mut provid th nrgy to lift th watr 5 m a wll a incra th prur. Hr it wa aumd th watr wa an incompribl ubtanc, which i oftn a good approximation for liquid and olid, and implifi proprty valuation. 6

8 H4.5: Stam ntr th condnr of a tam powr plant at 20 kpa and a quality of 95% with a ma flow rat of 20,000 /hr. It i to b coold by watr from a narby rivr by circulating th watr through th condnr (imilar to Fig. P4.78). To prvnt thrmal pollution, th rivr watr i not allowd to xprinc a tmpratur ri abov 0. Aum tady tat opration and trat th rivr watr a an incompribl ubtanc with contant pcific hat. If th tam i to lav th condnr a aturatd liquid at 20 kpa, dtrmin th ma flow rat of th cooling watr rquird [/]. Givn Data Modl Diagram P = 20 kpa Control volum x = 95% Stady tat m hot = 20,000 = Rivr i incompribl hr P 2 = 20 kpa ` ubtanc with contant c Stat 2 aturatd liquid Prur rmain ΔT cold = 0. contant for ach tram m cold =? Govrning Equation h = u + Pv Eq. 3.4 du = cdt Eq. 3.8 = m i i m m (h + i Eq. 4.5 Conidr two ytm, on ncloing th hot watr from th powr plant and th othr ncloing th cool rivr watr. Apply conrvation of ma to ach ytm. Sytm A (Hot Stram) = m i i m Sytm B (Cold Stram) = m i i m 0 = m 3 m 4 hot = contant 7 m 3 = m 4 = m cold = contant Th two fluid tram rmain parat and conrvation of ma can b applid to ach tram paratly. A n th flow rat for ach tram rmain contant. Apply conrvation of nrgy to ach ytm: Sytm A (Hot Stram) Sytm B (Cold Stram) 0 = Q out + m hot(h h 2 ) 0 in + m cold(h 3 h 4 ) Q out = m hot(h h 2 ) Q in = m cold(h 4 h 3 ) Sinc all th hat i tranfrrd from ytm A to ytm B Q out in m hot(h h 2 ) = m cold(h 4 h 3 ) m cold = m hot ( h h 2 h 4 h 3 ) m (h + i Altrnativly, you could apply an nrgy balanc to th combind Sytm (C). Hat tranfr btwn hot and cold watr till occur but inc it i intrnal to ytm it i not includd in th analyi. 0 = m hot (h h 2 ) + m cold(h 3 h 4 ) m cold = m hot ( h h 2 h 4 h 3 )

Two proprti ar ndd to fix a tat. Stat i dfind by x = 0.95 and P = 20 kpa. Th fluid i inid th vapor dom.

. x = 0) and P = 20 kpa h 2 = h f = 25.4 Th fluid condn at contant prur and tmpratur until it rach aturation at tat 2. Dtrmin th nthalpy chang of rivr watr h 4 h 3 uing th incompribl ubtanc modl.

556 cδt m cold = 297.8 contant c No tmpratur wa givn, aum T = 300 K for typical day ) (249.8 4.79 K 25.4 ) = 297.8 {0 K} Dicuion Hat xhangr ar common dvic in mchanical nginring.

9 Two proprti ar ndd to fix a tat. Stat i dfind by x = 0.95 and P = 20 kpa. Th fluid i inid th vapor dom. Th nthalpy at tat can b calculatd from th quality 5% i liquid and 95% i vapor h = ( x)h f + xh g = 0.05 (25.4 ) ( ) = Stat 2 i dfind a aturatd liquid (i.. x = 0) and P = 20 kpa h 2 = h f = 25.4 Th fluid condn at contant prur and tmpratur until it rach aturation at tat 2. Dtrmin th nthalpy chang of rivr watr h 4 h 3 uing th incompribl ubtanc modl. h = u + Pv dh = du + vdp + Pdv dv = 0 (incompribl) dh = du + vdp dp = 0 (modl) dh = du = cdt dh = cdt h 4 h 3 = c(t 4 T 3 ) = cδt c = 4.79 / K m cold = m hot [ h h 2 ] = (5.556 cδt m cold = contant c No tmpratur wa givn, aum T = 300 K for typical day ) ( K 25.4 ) = {0 K} Dicuion Hat xhangr ar common dvic in mchanical nginring. Oftn th objctiv i to cool a hot objct/ytm by contact with a coolr ytm. Applying th conrvation law dpnd on th way you dfin your ytm. To olv th problm you may nd to conidr multipl control volum. Chooing a particular control volum can implify your analyi a n in th altrnativ olution. Rgardl of how you lct your ytm you hould gt th am olution. 8

ME 300 Exam 1 October 9, :30 p.m. to 7:30 p.m.

ME 300 Exam 1 October 9, :30 p.m. to 7:30 p.m. CIRCLE YOUR LECTURE BELOW: First Na Last Na 10:0 a.. 1:0 p.. Naik Gor ME 00 Exa 1 Octobr 9, 014 6:0 p.. to 7:0 p.. INSTRUCTIONS 1. This is a closd book and closd nots xaination. You ar providd with an