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Unformatted text preview: Solution 1.5 A ﬂuid at 0.7 bar occupying 0.09 To3 is compressed reversibly to a pressure of 3.5 bar
according to a law p1:I‘I = constant. The ﬂuid is then heated reversibly at constant volume I
until the pressure is 4 bar; the Speciﬁc volume is then 0.5 n13i ,r’ [(3. A reversible expansion 
according to a law p03 = constant restores the ﬂuid to its initial state. Sketch the I
cycle on a p—u diagram and calculate:
(i) the mass of ﬂuid present;
(ii) the value of II in the ﬁrst process;
(iii) the net work of the cycle. (i) v: = v: = 0.5 m3/kg
w” : 4>co.En‘/o.‘lr
is v: = 1195 mlikg
9150, V1 : 009 m3 (given), therefore, Mass of fluid = 0.09/1.195 = 0.0753 kg ‘. (ii)
For the process 1 to 2:
3.5/0.7 = (1.195/0.5)ﬂ is n : 1.84? (iii)
For the process 1 to 2; work input : 0.0755x105{,(_§._5xg.5}_; (O.7x1.195)}
1.34?  1 ' = 8121 N m FOr the process .5 to 1: work input : 0.0753x105{£9.?x1_.1951  (4x0.5)_]
 1 ='8761Nm ‘ Nat. work input 2 8121  81"61 =  640 H m is Not work output = + 640 N m P
o '5"
0‘ Pressure/(bar) 0.1 5pm“; Vllklnl (n3; kg'l \/ 1.6 A ﬂuid is heated reversibly at a constant pressure of 1.05 but until it has a speciﬁc Solution 1.6 volume of 0.1 m3 / kg. It is then compressed reversibly according to a law pi) = constant
to a pressure of 4.2 bar, then allowed to expand reversibly according to a law
p1)” = constant, and is ﬁnally heated at constant volume back to the initial conditions.
The work done in the constant pressure process is —515 Nm, and the mass of ﬂuid present is 0.2 kg. Calculate the net work of the cycle and sketch the cycle on a 12—1)
diagram. For the process 1 to 2: Work input = — 515 N m = — 515/0.2
= — 2575 N m/kg
ie —2575 = 1.05x105x(V1 e 0.1) v1 : 0.0755 m3/kg : V4
For the process 2 to :5:
Work input = 1.05x105x0.Lx1n(1.05/4.2) = + 14 556 N m/«
Also, ‘ v: = 1.0‘5x0.1/4.2 = 0.025 m3/kg For the process 3 to 4: p4 = 4.2/(0.O755/0_025)17 : o_642 bar Work input
7 105((0.642x0.0755)  (4.2x0.025)} 1.71 = — 8075.6 N ME;
Then, I Net work input = O.2x(14 556 ‘ 18075.6  2575) =781Nm M1 Pressure“ be!) 1.05 V; v“ u; n.i Spam Vomitﬁring: —____________________ "In Ml“:_ﬁl_._i.;'_ ._ _'
U \.. 3
If'e Solution 2.3 14 Sciatica Steam at 7 bar and 250°C enters a pipeline and ﬂows along it at constant pressure.
If the steam rejects heat steadily to the surroundings. at what temperature will droplets
of water begin to form in the vapour? Using the steadyﬂow energy equation, and neglecting changes in velocity of the steam, calculate the heat rejected per kilogram of
steam ﬂowing. water droplets will begin to farm at the saturation temperature correSpending to 7 bar is t = 165 ‘C
From superheat tables. in : 2955 kJ/kg, and for dry
saturated steam at ‘3 bar. h2 = 2?:54 kJ/kg. then. D = 2?64  2955 = h 191 kJ/kg ie Heat rejected = 191 kJ/kg 0.05 kg of steam at 15 bar is contained in a rigid vessel of volume 0.0076n13. What
is the temperature of the steam? If the vessel is cooled, at what temperature will the
steam be just dry saturated? Cooling is continued until the pressure in the vessel is ll bar: calculate the ﬁnal dryness fraction of the steam. and the heat rejected between
the initial and the ﬁnal states. v = 0.00?6/0.05 = 0.152 m3/kg
Hence the steam is superheated since v > w ; from superheat tables at p = 15 bar and v = 0.152 m3/kg
t = 250 'C. when cooling takes place at constant volume the
steam is dry saturated when u : Vg : 0.152 m3/kg. ane i nterpoleti ng , t = 191.6  (9.1520  0,;512)(191.6 l38) (01652  0.1512) 191.4 °C At 11 bar and v = 0.152 m3/kg the steam is net with
e dryness fraction of. x = o.152/0.1?74 = 0.35?. and therefore,
uz : ?80 + 0.357(2536  780) = 2527.4 k3/kg
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This note was uploaded on 07/26/2009 for the course ENG sem314 taught by Professor Dunno during the Spring '09 term at A.T. Still University.
 Spring '09
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