hmwk1_solutions_sum08

# hmwk1_solutions_sum08 - " e MEBN duh” 9:51 A l-m3...

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Unformatted text preview: /" e; MEBN duh” 9:51. A l-m3 rigidtankwith airat 1 MPa,400 Kis connectedto anairlineas shownin Fig. P3 .9. The valve is opened and air ﬂows into the tank until the pressure reaches 5 MPa, at which point the valve is closed and the temperature inside is 450K. 3.. What is the mass of air in the tank before and after the process? b. The tank eventually cools to room temperature, 300 K. What is the pressure inside the tank then? 9 Solution: Assume the air behaves as an ideal gas and P,T known at both states. PIV 1000 x1 main ' RT1 ’ 0.287 x 400 PzV 5000 x 1 main ' RT2 ‘ 0.287 x 450 Process 2 -> 3 is constant V, constant mass cooling to T3 P3 = P2 X ('1'3/1'2) = 5000 X (300/450) = 3.33 MPa = 8.711 kg = 38.715 kg A piston/cylinder arrangement, shown in Fig. P3.1 1, contains air at 250 kPa, ‘ ‘ 300°C. The SO-kg piston has a diameter of 0.1 m and initially pushes against the stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools as heat is transferred to the ambient. a. At what temperature does the piston begin to move down? b. How far has the piston dropped when the temperature reaches ambient? Solution: Piston AP =ﬁ-x012 = 0.00785 m2” Balance forces when piston ﬂoats: m g ' 50 x 9.807 AP = 100 + 0.00785 x 1000 Pﬂoat = P0 + = 162.5 kPa =P2 = P3 To find temperature at 2 assume ideal gas: VStOP T2 = Tlx%21'= 573.15 x 135265 = 372.5 K b) Process 2 -> 3 is constant pressure as piston ﬂoats to T3 = To = 293.15 K V2 = V1= AP x H = 0.00785 x025 = 0.00196 m‘ 5 1.96 L Ideal gas and P2 = P3 => V3 = V2 X'E‘é 1.96 X 239732}: = 1.54 L AH = (V2 -V3)/A = (l.96-l.54)*0.001/0.00785 = 0.053 m = 5.3 cm --_-.... A... .‘/\ I __ 3.2%. Determine the quality (if saturated) or temperature (if superheated) of the " \w“ following substances at the given two states: - Solution: a) Water, H20, use Table 8.1.1 or 3.12 1) 120°C, 1 m3/kg => v > vg superheated vapor,_ '1' = 120 °C 2) 10 MPa, 0.01 m3/kg => two-phase v < v3 2: =(0.01— 0.001452 ) / 0.01657 = 0.516 b) Nitrogen, N2, table B.6 1) 1 MPa, 0.03 m3/kg => superheated vapor since v > vg Interpolate between sat. vapor and superheated vapor 8.62: T 5 103.73 + (0.03-0.02416)*(120-103.73)/(0.03l17-0.02416) = 117 K 2) 100 K, 0.03 m3/kg => sat. liquid + vapor as two-phase v < vg v = 0.03 = 0.001452 + x x 0.029764 = x = 0.959 c) Ammonia, NH3, table 3.2 1) 400 kPa, 0.327 m3/kg => v > vg = 0.3094 m3/kg at 400 kPa Table B.2.2 superheated vapor T :- 10 °C 2) 1 MPa, 0.1 m3/kg => v < vg 2-phase roughly at 25 °c x = ( 0.1 — 0.001658 ) / 0.012647 = 0.7776 d) R-22, table 8.4 1) 130 kPa, 0.1 m3/kg => sat. liquid + vapor as v < vg vf 5 0.000716 m3/kg, vg 5 0.1684 m3/kg v = o.1= 0.000716 + x x 0.16768 :5 x = 0.592 2) 150 kPa, 0.17 m3/kg => v > vg superheated vapor, T .=. 0°C p A boiler feed pump delivers 0.05 m3/s of water at 240°C, 20 MPa. What is the mass ﬂowrate (kg/s)? What would be the percent error if the properties of saturated liquid at 240°C were used in the calculation? What if the properties of saturated liquid at 20 MPa were used? Solution: At 240°C, 20 MPa: v = 0.0012046 m3/kg (from 13.1.4) m = V/v = 0.5/0.0012046 = 41.5 kgs Vf<24ooc) = 3 m = kg/S error 2% vf (20 MP3) = 0.002036 =5 11: = 24.56 kg/s error 41% ,,_- . 3.4 A cylinder/piston arrangement contains water at 105°C, 85% quality with a volume of 1 L. The system is heated, causing the piston to rise and encounter a linear spring as shown in Fig. P3 .45. At this point the volume is 1.5 L, piston diameter IS 150 mm, and the spring constant is 100 N/mm. The heating continues, so the piston compresses the spnng. What is the cylinder temperature when the pressure reaches 200 kPa? - Solution: 1 P1 = 120.8 kPa, v1 = vf+ x vfg = 0.001047 + 0.85’1.41831 = 1.20661 = _ 0.001 _ _4 P In W vl — 1.20661- 8.288x10 kg v2 = v1 (v2 / v1) = 1.20661x 1.5 = 1.8099 200 M 3 & P=P1=_120.8kPa (T2=203.5°C) ‘ v P3 = P2 + (ks/Apz) m(V3-v2) linear spring 8 1 1.5 liters AI, = (2:14) x 0.152 = 0.01767 m2; 1ts = 100 kN/m (matches P in kPa) 200 = 120.8 + (100/001767 2 ) x 8.288x10'4(v3-1.8099) 200 = 120.8 + 265.446 (V3 — 1.8099) => v3 = 2.1083 m3/kg r3 5 600 + 100 x (2.1083 — 2.01297)/(22443-2.01297) 5 641°C ,/‘ 3&6) Refrigerant-12 in a piston/cylinder arrangement is initially at 50°C, x = 1. It is then ' expanded in a process so that P = CW1 to a pressure of 100 kPa. Find the ﬁnal temperature and speciﬁc volume. - Solution: I State 1: 50°C, x=l = P1 = 1219.3 kPa, v1= 0.01417 m3/kg Process: Pv = C = Plvl; => P2 = C/v2= Plvl/vz State 2: 100 km and v2 = lel/P2 = 0.1728 m3/kg T2 5 -13.2°C from Table B.3.2 3.60 Consider two tanks, A and B, connected by a valve, as shown in Fig. P360. Each ‘J' . has a volume of 200 L and tank A has R—lZ at 25°C, 10% liquid and 90% vapor by volume, while tank B is evacuated. The valve is now opened and saturated vapor ﬂows from A to B until the pressure in B has reached that in A, at which point the valve is closed. This process occurs slowly such that all temperatures stay at 25°C throughout the process. How much has the quality changed in tank A during the process? —-——-9—V1i 1 +-—-E—V‘v’a —--———°‘1 X 0'2 +———°'9 x 0'2 — 26 212 + 6 703 = 32 915 k mm ‘ vmoc vg m: ‘ 0.000763 0.026854 ‘ ' ' ' g - V _6.703_ , _ 13 _ 0.2 = "A1 ‘ 32915 ‘ 03°35 ’ “‘32 ‘ vg we ' 0.26854 “48 kg = noA2 = 32.915 - 7.448 = 25.467 kg .2 vA2 = Baggi= .007853 = .000763 + xA2 x 0.026091 XAZ = 0.2718 Axp= 6.82% ...
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