HCB-3 Edition. Solutions Chapter 12 Problems. SOLUTION: Refer to saturated steam table (Table A3-SI) and superheated steam table (Table A4-SI)

Size: px

Start display at page:

Download "HCB-3 Edition. Solutions Chapter 12 Problems. SOLUTION: Refer to saturated steam table (Table A3-SI) and superheated steam table (Table A4-SI)"

Darcy O’Brien’
5 years ago
Views:

1 HCB- Editin 12.1 Slutins Chapter 12 Prbles GIVEN: Fllwing table fr water: T (C p (kpa v ( /kg Phase 60 (1.25 (2 ( 175 (4 Saturated vapr (5 ( (7 (8 (9 ( Saturated vapr (11 (12 FIND: Cplete the table (spaes labeled (1 (12. Refer t saturated stea table (Table A-SI and superheated stea table (Table A4-SI (1 At T = 60.06C fr saturated water, vf = and vg = /kg fr Table A (SI. The value fr v given lies between vf and vg, s it is a saturated liquid-vapr ixture and p = 20 kpa. (2 As desribed in (1, it is saturated liquid-vapr ixture. The quality an be fund fr: v = vliq (1-x + x. vvap (eq /kg = /kg (1-x + x (7.649 /kg x = 0.42 ( Sine it is given that it is saturated vapr, lk at the stea table fr the saturatin teperature at a saturatin pressure f 175 kpa. T = C (4 Lk fr vg at saturatin pressure 175 kpa. vg = /kg (5 At a saturatin pressure f 00 kpa, the saturatin teperature T = 1.55C. The given teperature f 00C is greater than the saturatin teperature at the given pressure. Therefre, it is superheated stea. Fr Table A4, rrespnding t superheated stea at 0. MPa and 00 C, v = /kg (6 Superheated stea as desribed in (5. 1

2 12.1 ntinued (7 At a saturatin pressure f 10 kpa, the saturatin teperature T = 99.6C. The given teperature f 100C is greater than the saturatin teperature at the given pressure. Therefre, it is superheated vapr. Fr Table A5, fr p= 0.01 MPa and T = 99.6 C, v = /kg (8 Superheated vapr as desribed in (7. (9 Sine it is knwn t be saturated liquid, find vf = /kg n Table A. T = C (10 P=550 kpa at vf = /kg. (11 The given teperature (1000C is greater than the saturatin teperature at the given pressure (10kPa. Fr Table A5, fr p = 0.01 MPa and T = 1000 C. v = /kg (12 Superheated stea as desribed in (11. 2

3 12. GIVEN: Fllwing table fr R-22: T (F p (psia v (ft /lb Phase (1 ( Saturated vapr ( (4 10 (5 0.5 (6 ( (8 Saturated liquid FIND: Cplete the table (spaes labeled (1 (8. Refer t Table A5 fr saturated nditin and Table A7 fr superheated (1 Sine the phase is saturated vapr, find vg = ft /lb fr R-22 Table A5 (IP. Read T and P. T = 40F (2 p = 8.28 psia (see (1. ( Fr R-22 table A5(IP, at T = 45F, p = psia; therefre it is saturated. Hwever, ne has n infratin t deterine the quality. All that an be said is that: vf v vg 1/( lb/ft v ft /lb ft /lb v ft /lb (4 Saturated vapr f unknwn quality. (see (. (5 At 10F, vg = ft /lb and vf = 1/82.54 ft /lb = ft /lb. The v falls between vg and vf, s it is saturated and p = psia (6 The R-22 is saturated. The quality an be fund fr: v = vf(1-x + xvg (eq ft /lb = ft /lb (1-x + x ( ft /lb x = 0.44 (7 Sine it is knwn t be saturated liquid, T an be read fr the R-22 table A5 at p = psia. T = 40F (8 Sine it is saturated liquid v = vf = 1/79.07 ft /lb. v = ft /lb

4 12.5 GIVEN: Vtank = 5, water = 10 kg saturated liquid vapr at 75C. Heat added until all liquid pletely evaprated. FIND: The teperature after evapratin. ASSUMPTIONS: The tank is lsed and n vapr is lst ut f the tank g 10kg kg when the liquid is entirely evaprated. Fr saturated stea table A (SI, vg = 0.5 when T = 141C 4

5 12.8 GIVEN: Water p = 40 psia, (a x = 70%, (b x = 90% FIND: h, v, T Read values fr saturated stea table A (IP. hf = /lb, hfg = 9.8 /lb h = hf + xhfg (eq fr h (a h = /lb (9.8 /lb = /lb ( kj/kg (b h = /lb (9.8 /lb = /lb ( kj/kg vf = ft /lb, vg = ft /lb v = vf,(1-x + x. vg (eq (a v = ft /lb( ( ft /lb = 7.56 ft /lb (0.459 /kg (b v = ft /lb( ( ft /lb = 9.45 ft /lb (0.590 /kg T = Tsat = F (10.7 C 5

6 12.10 GIVEN: Refrigerant 14a ndensed fr saturated vapr t saturated liquid at T = 100F; = 500 lb/. Twi = 60F, = 1500 lb/. FIND: Tw, Q r ASSUMPTIONS: Water and refrigerant flws are steady. Open syste first law f therdynais: 2 2 vi v i ( gzi hi ( gz h W (eq Assuing the height, z, is nstant, the flw is steady, and n wrk is dne, this redues t: ihi h i h h ( i On the refrigerant side, lb 500 ( h hi Saturated liquid: h = /lb = hf (Table A6-IP sat. vapr at 100F Saturated vapr hi = /lb = hg lb 500 ( ,611.5 lb lb (The negative sign signifies heat is reved fr the refrigerant. On the water side, the heat reved fr the refrigerant is gained by the water. h h ( T T (eq w ( i w p i lb 5, (1.0 / lb F( Tw 60 F Tw = 8.7F r 6

7 12.12 GIVEN: Refrigerant 14a tttled fr a saturated liquid at 70F t 6.1 psia. FIND: T, x2, and v2 ASSUMPTIONS: Adiabati press, state 2 is saturated. State 1: T1 = 70F, saturated liquid h1 = hf = 4.58 /lb fr R-14a Table A6-IP. State 2: In the tttling press, h2 = h1 h2 = h1 = 4.58 /lb p2 = 6.1 psia; hf2 = /lb, hg2 = /lb T find the quality, x, slve eq , h = hf(1-x + xhg 4.58 /lb = /lb(1-x + x( /lb x = 18.1% v = vf(1-x + xvg (eq v = 1/( lb/ft ( (1.292 ft /lb v = 0.24 ft /lb T = 70F - 24F T = 46F where 24F is the teperature f saturated Refrigerant 14a at 6.1 psia pressure. 7

8 12.1 GIVEN: Air is heated by passing ver an eletri il. Tai = 15C, Va = 00 /in Eletri heater input = Qh=0 kw and heat lss Qlss= 0. kw FIND: Ta, ASSUMPTIONS: Cnstant air density = 1.2 kg/ and speifi heat = 1.00 kj/kg.k Heat balane:. Q V ( T T (eq a And kW i.e., ai 00 in kg / 1.00kJ /( kg. K ( T 60 in/ s Or T 15 C 4.95 C 19. C a 95 a 15 C 8

9 12.14 GIVEN: Air is heated by passing ver a stea il. TAi = 90F, pai = 14.7 psia, TA = 140F; Tstea,i = 400F, pstea,i = 0 psia, = 0 lb/in; Tstea, = 200F, pstea, = 25 psia, s = 0 lb/in s FIND: a ASSUMPTIONS: There are n ptential r kii energy effets. The pen syste first law f therdynais, 2 2 vi v i ( gzi hi ( gz h W 2 2 Fr the assuptins it redues t h h ( i (eq. 2.2 On the stea side: State 1 is superheated stea: hsi = /lb at 0 psia and 400F (HCB sftware r therdynais text State 2(p = 25 psia, T = 200F is subled liquid, s hs is a funtin f teperature nly: hs = hf = /lb at T = 19.19F hs = hf = /lb at T = F interplating fr T = 200F, hs = /lb at T = 200F lb 0 ( in lb ,146 lb in On the air side: The heat lst by the stea il is gained by the air. h h ( T T (eq ( i p i 2,146 a (0.24 (140 F 90 F in lb F lb a 2679 = lb/s = 20. kg/s in 9

10 12.15 GIVEN: V a = 50 ft /in, Tai = 85F R22, pri = 69.6 psia, xi = 0.0; ri =.5 lb/in R22 leaves as saturated vapr. FIND: Ta, Q air ASSUMPTIONS: Steady air and refrigerant flw. a = lb/ft. Q h h (fr eq. 2.2 with vi = v, zi = z, W =0 ( i On the refrigerant side: At p = 69.6 psia, hf = /lb, hg = /lb hi = xhg + (1-xhf (eq hi = (0.( /lb + (0.7( /lb = /lb h = hg = /lb lb.5 ( in lb lb in On the air side: The heat lst by the stea il is gained by the air. Q T T (eq and 2.2 p ( i ft lb (0.075 (0.24 ( T 85 F in in ft lb F (Q is negative beause the heat is reved fr the air. T 50.6 F air 217 in 10

11 12.16 Air led in a ling il GIVEN: R22: r =100 lb/in, p1 = 76 psia, and quality xin = 0.5,, xut =0 ( sat. liquid Air: =10,000 f Ti = 8F FIND: T,Q a ASSUMPTIONS: Steady air and refrigerant flws. Cnstant air density = 1.2 kg/ and speifi heat = 1.00 kj/kg.k Refrigerant: refer t Table A5 (saturated tables fr R22 hi = x ( = 50.9 /lb h = hf = /lb Q ( h hi (fr eq. 2.2 with vi = v, zi = z, W =0 lb 100 ( , 687 lb lb On the air side: The heat lst by air is gained by refrigerant: 5,687 10,000f 0.075lb / ft 0.24 /( lb. F (8 T T F a 4 a 11

12 12.18 GIVEN: R14a: T1 = 140F, p1 = 120 psia, =800 lb/, p2 = 120 psia sat. liquid Water: T1 = 85F, p = 0 psia, T2 = 104F FIND: w ASSUMPTIONS: Steady water and refrigerant flws. Refrigerant: State 1: T1 > Tsat at 120 psia, therefre it is superheated. hi = /lb, (Table A6 fr R14a h = hf = 41.0 /lb Q ( h hi (fr eq. 2.2 with vi = v, zi = z, W =0 lb 800 ( ,400 lb lb On the air side: The prble states that the water is pressed (r subled in whih ase enthalpy is a funtin f teperature nly and an be read fr the Table A-IP at the given teperatures: h1 = hf at 85F = 5.08 /lb h = hf at 104F = 72.0 /lb The heat lst by the R14a is gained by the water: 68,400 w ( lb lb lb w

13 12.19 GIVEN: FIGURE: 12. FIND: COPl, COP Q L = 1200 kj/in; Q H L l (eq W W = 10 kw 1200kJ / in(1in/ 60s COP l 10kW 2.0 COP l H W H 0 kw L kj 1in 10kW 1200 ( in 60s 1

14 12.20 GIVEN: FIND: P r Q envelp= 60 kw; Q int = 20 kw ; COPl = 2.4 W Ttal lad n ling syste = = 80 kw L COPl (eq W W Q / COP 80kW / 2.4. kw L Eletri resistane heating vs heat pup GIVEN: Eele.heat = 2000 kwh/; unit st f eletriity = $0.09/kWh ; COPhp = 2.4 FIND: Savings S Assue that COPhp = 1.0. Utility bill fr eletri resistane heater = 2000 kwh x $0.09/kWh = $180 Utility bill fr heat pup = (2000/2.4 kwh x $0.09/kWh = $75 Savings S fr the nth = $180 - $75 = $105 14

15 12.22 Huse heated by heat pup GIVEN: T1 = 41F; T2 = 7F; COPhp =.0; FIGURE: 12.2 = 0.5 /lbf W FIND: Tie required t raise teperature fr T1 t T2. COP h l (eq W h 17, , COP l = 17,750 /; = 100 lb; t ( T 2 T (eq. 2.15, 2.26 Q h ( 1 t t 100lb (0.5 (7 lb F 5, in F 41 F 15

16 12.2 GIVEN: TL = 68F, TH = 95F, FIGURE: 12. FIND: Q gains W =.4 kw ASSUMPTIONS: Reversible Carnt yle. Q gains are all sensible gains. COP W L TL T T ( R COP (95 68 R COP 19.6 H L (eq Q L W COP kW 66. 5kW 16

17 12.24 GIVEN: TL = 72F, TH = 92F, FIGURE: 12. FIND: W, pwer input t keep building at 72F ASSUMPTIONS: Reversible Carnt yle. Q gains = 8000 /in /in Q gains are all sensible gains. Q L Q gains 10,000 in L TL COP W T T ( R COP (92 72 R COP 26.6 H L (eq W L COP 10,000 / in 26.6 W kw in 17

18 12.26 GIVEN: FIGURE: 12.2 Q h = 0,000 /in; Th,ideal = 72F, Tl,ideal = 40F; Real heat pup requires 10F teperature differene t transfer heat fr the ideal reservirs. FIND: COPhp,ideal, COPhp,real TL COP hp, ideal 1 (eq and 12.5 T T H L ( R (72 40 R COP hp, ideal COP hp, ideal In the real ase, the heat transfer penalties require that Th =72F+10F= 82F and TL = 40F-10F= 0F (0 460 R (82 0 R COP hp, real COP hp, real

19 12.28 GIVEN: Duble pipe, unterflw heat exhanger with water as the wrking fluids; V shell = 200 gal/in, V tube = 200 gal/in; Ti = 50ºF; T = 100ºF; Thi = 115ºF FIND: UA and ASSUMPTIONS: Jaket lsses are negligible. LOOKUP VALUES: p,water = 1.0 /lb-ºf water = 62. lb/ft (Table A2, Prperties f Water at an average tep. f 75ºF gal 60 in 1 ft lb lb 200 ( ( (62. 99, 9 in 7.481gal ft h lb C 1.0 (99,9 99, lb F ft 9 F C h C C in C ax Q C ( T T i 99,9 (50 F 100 F 4,996, 658 F Q ( C in ( Thi Ti (eq ( C in ( Thi Ti 5,068,850 / (101,77 / F(115 F 50 F 0.77 C in 1.0 C ax Fr figure with = 0.77, NTU.5 U A NTU.5 (eq C U A in C in ( NTU 101,77 (.5 54,800 = 187. kw/ C F F 19

20 12.0 GIVEN: Duble pipe, unterflw heat exhanger with the ld strea % prpylene glyl and the ht strea water; V V h = 200 gal/in; Ti = 50ºF; Thi = 115ºF; p,glyl = 0.95 /lb-ºf, glyl = 64.0 lb/ft FIND: UA and ASSUMPTIONS: Jaket lsses are negligible. gal 60 in 1 ft lb lb 200 ( ( ( , 674 in 7.481gal ft h 101, 77 (see slutin C h 101, 77 (see slutin F lb C p glyl lb F 540, 102, , F C in 97, C 101,77 ax Fr slutin fr tw water streas, UA = 54,820 /-ºF. 90% f resistane is attributed t nvetin, equally split between the tw water t tube surfaes (given. The equivalent iruit f UA is: Rw + Rpipe + Rw R = 1/UA (eq. s 2.40,2.41, 2.42 R = 2.8 x ºF/ Rw = 0.45R = 1. x ºF/ Rpipe = 0.10R = 0. x ºF/ Replae ne Rw with Rglyl: hn,w = 1/Rw = 7.7 x 10 5 /-ºF hn,g = Fghn,w (eq Fg = 0.5 (Fig. 2.15b at 75ºF and % hn,g = 0.5(7.7 x 10 5 /-ºF =.85 x 10 5 /-ºF Rglyl = 1 / hn,g = 2.6 x ºF/ Calulating the new R: R = Rw + Rpipe + Rglyl R = (1. x x x 10-6 = 4.2 x ºF/ UA = 1/ R = 2.4 x 10 5 /-ºF 20

21 12.0 ntinued U A NTU (eq C in / F NTU ,540 / F C in Using Figure at and NTU = 2.5, = 0.72 C ax 21

22 12.2 GIVEN: Duble pipe, unter-flw heat exhanger between tw water streas; V ld = 00 gp, V ht = 500 gp, Ti = 50ºF, T = 100ºF, Thi = 115ºF FIND: UA, LMTD, and A heat balane n the tw streas, ld ( T Ti ht ( Thi Th ; p ld p ht,, 00gp(100 F 50 F 500gp(115 F Th T h 85 F T T T 115 F 100 F 15 F (eq T 1 hi 2 Th Ti 85 F 50 F 5 F T1 T LMTD 21.6 F (eq ln( T1 / T2 ln(15/ 0 p T V n the ld side the average, = x 62. lb/ft = 62.1 lb/ft (fr Table A2 prp. f water at 75ºF p = 1.0 /lb-ºf lb gal 1 ft 62.1 (00 ( (1.0 (100 F 50 F ft in 7.481gal lb F k M in 7.47M / U A (eq.12.0 LMTD 21.6 F k U A 46 F (eq ( p in ( Thi Ti V p ( T Ti (100 F 50 F V ( T T (115 F 50 F p 0.77 hi i 22

23 12.5 GIVEN: Cunter-flw heat exhanger between tw water steas; Thi = 60ºC, kg/s, Ti = 8ºC, T = 42ºC, = 15 kg/s FIND: UA, ( T Ti h ( Thi Th ; p, p, h kg kg 15 (42 C 8 C 12 (60 C Th s s 17.5 C T h T T T 60 C 42 C 18 C (eq T 1 hi 2 Th Ti 17.5 C 8 C 9.5 T1 T LMTD 1. C (eq ln( T1 / T2 ln(18/ 9.5 p T kg kj 12 (4.186 (60 C 17.5 C s kg C kW 214.9kW U A (eq LMTD 1. C kw U A C h p ( Thi Th ( ( T T h p p in hi ( p in (60 C 17.5 C (60 C 8 C 0.82 i C h = 12 2

24 12.6 GIVEN: Cunter-flw heat exhanger between tw water steas; Thi = 60ºC, kg/s, Ti = 8ºC, = 15 kg/s, UA = 0.8(117.7 kw/ºc h = 12 FIND: and Q. By hw uh des the ht fluid inlet teperature have t be inreased t ahieve the sae heat transfer rate in kw kw U A ( ( C C U A LMTD Q ( T Ti h ( Thi Th ; p, p, h kg kg 15 ( T 8 C 12 (60 C Th s s 12 T (60 C Th 8 C 15 T 1 T hi T 60 C T T2 Th Ti Th 8 C T1 T2 LMTD (eq ln( T1 / T2 12 plugging in T (60 C Th 8 C int the LMTD equatin will result in ne 15 equatin and ne unknwn. (60 (0.8(60 Th 8 ( Th 8 LMTD 60 (0.8(60 Th 8 ln ( Th 8 (12 0.2T h LMTD 4 0.8T h ln ( Th 8 U A LMTD ( T Th h p hi kg kj 12 (4.186 (60 C Th 014kW (50.2T h kw s kg C 24

25 12.26 ntinued (12 0.2Th 11.6 (014 (50.2T h / 4 0.8T h ln ( Th 8 by trial and errr, Th = 19.7ºC kg kj 12 (4.186 (60 C 19.7 C s kg C ( Thi Th ( ( T T (60 8 hi i Fr slutin 12.5, lean 215kW ( Thi T Sine, 0.78 ( T T hi h i 2024 kw Thi Th 0.78( Thi Ti 215kW h p ( Thi Th h p (0.78( Thi Ti kg kj 215kW 12 (4.186 (0.78( Thi 8 C s kg C T 62.5 t ahieve the sae utput as the lean heat exhanger. hi 25

th th th The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,

th th th The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, Cheical Reactins 14-14 rpane is burned wi 75 percent excess during a cbustin prcess. The AF rati is t be deterined. Assuptins 1 Cbustin is cplete. The cbustin prducts cntain CO, H O, O, and N nly. rperties