IV. First Law of Thermodynamics. Cooler. IV. First Law of Thermodynamics

Transcription

1 D. Applcatons to stady flow dvcs. Hat xchangrs - xampl: Clkr coolr for cmnt kln Scondary ar 50 C, 57,000 lbm/h Clkr? C, 5 ton/h Coolr Clkr 400 C, 5 ton/h Scondary ar 0 C, 57,000 lbm/h a. Assumptons. changs ktc and potntal nrgy ar nglgbl. stady stat. constant mass flow rats v. nglgbl hat loss to surroundgs v. nglgbl nonflow work v. Evaluat hat capacts at avrag tmpraturs b. Enrgy balanc ovr clkr coolr 0 = Q W + V h + + gz out V h + + gz C a a = h 0 h out ( T T ) = C ( T T ) a aout c c cout c

2 c. Analyss Solvg for T cout gvs C = ( Ta Taout ) Tc a a T cout + mccc 57,000 lb / h(040 J / kgk ) T cout = (0 C 50 C ) C = 60 C 30,000 lb / h(0 J / kgk ) Commnt: sc th hat capacts wr valuatd at th avrag tmpraturs and th fal tmpratur of th clkr was not known, th valu of 60 C can b usd to r-valuat C c. Itrat untl T cout rachs a constant valu.. Nozzls and dffusrs (subsonc flows) a. Nozzls cras vlocty at th xpns of prssur drop drcton of flow b. Dffusrs cras prssur at th xpns of dcras vlocty A common xampl s th nozzl on a gardn or fr hos. Q CV V V 0 = + h+ h+ Nozzl Unts: kj/kg = 0 3 m /s Dffusr

3 3. Turbs and comprssors a. In a turb, a flud dos work Q on th turb blads to rotat a shaft b. In a comprssor, shaft work from th surroundgs dos work on th flud to cras Comprssor ts prssur c. A fan slghtly crass prssur ordr to mov a W flud from on locaton to anothr d. Nglct changs KE and PE. Hat transfr may b 0 = W + Q + h h nglgbl as wll. 4. Throttlg Dvcs a. Usd whn a dcras prssur s ndd wthout any work ffcts. b. Accomplshd by a partally opn valv, a porous plug, or a long capllary tub. c. Commonly usd rfrgraton whr th prssur drop s accompand by a drop tmpratur. d. No work. Nglgbl hat transfr, KE, PE. h = h (5-4) P T P T 3

4 . Exampl of throttl - rfrgraton cycl Rfrgrant 34a s throttld from th saturatd lqud stat at 700 kpa to a prssur of 0 kpa. What s th drop tmpratur? 4 Q out Condnsr Comprssor W comp Boundary of systm Throttl 3 Evaporator Q (from cold rgon). Exampl of throttl - rfrgraton cycl Rcall (5-4), h = h. From Tabl A- at 700 kpa, T = T sat = 6.69 ºC and h = h f = 88.8 kj/kg. From Tabl A- at 0 kpa, h f =.49 kj/kg and h g = kj/kg. Bcaus h = h = 88.8 kj/kg, our fal condton s a saturatd mxtur and T = T sat = -.3ºC 4

5 5. Pp Flow a. May hav sgnfcant hat transfr. b. Work trm ndd f thr s a fan or pump CV. c. Wth lquds KE usually small. May b mportant wth gass. d. PE can b larg for lqud flow, for xampl, nglctg frcton and hat transfr, W + gz gz = 0 W z z E. Unstady Flow Procsss - Exampl: rapd chargg of an vacuatd tank. Supply l s r-34a at 40 C and 0.4 MPa (absolut). If tank s tally vacuatd, fd th tmpratur th tank whn th flow stops (0.4 MPa).. Assumptons: () no hat transfr to th surroundgs durg fllg, () nglgbl changs KE and PE. r-34a 0.4 MPa 40 C 5

6 3. Enrgy and matral balanc on contnts of tank V V decv = Q + W + m h + + gz m h + + gz du CV = mh dmcv = a. Intgrat nrgy balanc wth rspct to tm du CV cv = h = h m u mu = ( m m ) h dm b. Bcaus tank s tally mpty, m = 0 and u = h c. At 40ºC, 0.4 MPa, from Tabl A-3, h = kj/kg d. Intrpolat Tabl A-3 to fd T to gv u = h : T = 69.6ºC.. Why s th fal tmpratur hghr than th lt tmpratur, T = 40ºC? 6