example 1 cengel 4 38 a piston-cylinder device initially...
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Example 1 Cengel 4_38
A piston-cylinder device initially contains air at 200 kPa, 200
deg C and 0.5 m3. At this state, a linear spring is touching the
piston but exerts no force on it. Heat is now slowly transferred
to the AIR., causing the pressure and volume to rise to 500
kPa and 0.6 m3, respectively. Show the process on a P-v
diagram and determine (a) the final temperature, (b) the work
done by the steam and (c) the total heat transferred.
Pa 1.00 bar
100.00 kN/m2
P1 2.0 bar
200 kN/m2
P2 5.0 bar
500 kN/m2
T1 200 deg C
473 K
V1 0.50 m3
V2 0.60 m3
R 0.29 kJ/kg K
Cv 0.718 kJ/kg K
m=P1*V1/(R*T1) = P2*V2/(R*T2) 0.74 kgm=P1*V1/(R*T1) = P2*V2/(R*T2) 0.74 kg
T2 = T1*P2*V2/(P1*V1) 1419.0 K
Wa Work Done by Atmosphere = Pa*(V2 - V1) 10.0 kJ
Wp Work Done by Piston = W/A *(V2 - V1) 10.0 kJ
W/A 100.0 kN/m2
Ws Work Done by Spring = 0.5*Fs*e
Fs = (P2 -P1)*A
e = (V2 -V1)/A
Ws = 0.5*Fs*e = 0.5*(P2-P1)*(V2-V1) 15.0 kJ
Total W12 = Wa + Wp + Ws 35.0 kJ
Alternatively W12 = area of P-V diagram
W12 = 0.5*(P1+P2)*(V2-V1) 35.0 kJ
dQ-dW= dU = m*(u2-u1)=m*Cv(T2 -T1) 500.3 kJ
dQ = dU +dW 535.3 kJ
Tut 5 Q2 Cengel 4_38 (STEAM)
A piston-cylinder device initially contains steam at 200 kPa,
200 deg C and 0.5 m3. At this state, a linear spring is
touching the piston but exerts no force on it. Heat is now
slowly transferred to the steam., causing the pressure and
volume to rise to 500 kPa and 0.6 m3, respectively. Show the
process on a P-v diagram with respect to the saturation lines
and determine (a) the final temperature, (b) the work done by
the steam and (c) the total heat transferred.
Pa 1.0 bar
100.0 kN/m2
P1 2.0 bar
200.0 kN/m2
P2 5.0 bar
500.0 kN/m2
V1 0.50 m3
V2 0.60 m3
Initial Condition
See A below v1 at 2 bar and 200 degC (Cengel superheated steam tables) 1.081 m3/kg
m = V1/v1 0.4625 kg
u1 2654.6 kJ/kg
Final ConditionFinal Condition
v2 = V2/m 1.2972 m3/kg
At 5 bar, vg = 1.2972 m3/kg Superheated T2 v2
1100 1.2673
1200 1.3597
See B below T2 at 5 bar (Cengel superheated steam tables) 1132.4 1.2972
deg C m3/kg
u2 v2
4259 1.2673
4470 1.3597
u2 4327.3 1.2972
kJ/kg m3/kg
Wa Work Done by Atmosphere = Pa*(V2 - V1) 10.0 kJ
Wp Work Done by Piston = W/A *(V2 - V1) 10.0 kJ
W/A 100.0 kN/m2
Ws Work Done by Spring = 0.5*Fs*e
Fs = (P2 -P1)*A
e = (V2 -V1)/A
Ws = 0.5*Fs*e = 0.5*(P2-P1)*(V2-V1) 15.0 kJ
Total W12 = Wa + Wp + Ws 35.0 kJ
Alternatively W12 = area of diagram
W12 = 0.5*(P1+P2)*(V2-V1) 35.0 kJ
dQ-dW= dU = m*(u2-u1)
dQ = dU +dW 808.7 kJ
Tut 5 Q1 (AIR)
Air is enclosed in a 6 cm diameter vertically sliding piston in a
cylinder. The pressure and temperature of the air in the
cylinder is 2 bar and 30 deg C. The pressure in the cylinder is
resisted by a spring acting on the top surface of the piston.
The height of the piston above the base of the cylinder is 4
cm. The air enclosed below the puston is heated until the
piston has moved to a height of 8 cm above the base.
Determine the work done by the air enclosed in the cylinder
and the heat transfer to the air. The spring constant is 5 kN/m.
Assume the weight of the piston is negligible
d 6.0 cm
0.06 m
A = pi*d2/4 0.0028 m2 pi 3.142
P1 = Fs1/A +Pa L1 4.0 cm
Spring Force Fs1 = (P1-Pa)*A =k*e1 0.28 kN 0.04 mSpring Force Fs1 = (P1-Pa)*A =k*e1 0.28 kN 0.04 m
L2 8.0 cm
e1 = Fs1/k 0.0566 m 0.08 m
V1 =A*L1 0.00011 m3
e2 = e1 + (L2 -L1) 0.0966 m
V2 =A*L2 0.00023 m3
Spring Force Fs2 = k*e2 0.48 kN k 5.0 kN/m
Wa Work Done by Atmosphere = Pa*(V2 - V1) 0.0113 kJ Pa 1.0 bar
100.0 kN/m2
Wp Work Done by Piston = W/A *(V2 - V1) 0.0 kJ T1 30.0 deg C
W/A = 0 303.0 K
Ws Work Done by Spring = 0.5*(Fs1 +Fs2)*(e2 - e1) 0.0153 kJ P1 2.0 bar
200.0 kN/m2
Total W12 = Wa + Wp + Ws 0.0266 kJ P2 270.7 kN/m2
P2 = Fs2/A+Pa 270.7 kN/m2 R 0.287 kJ/kg K
Cv 0.718 kJ/kg K
T2 = T1*P2*V2/(P1V1) 820.3 K
547.3 degC
m = pV1/(RT1) 0.000260 kg
Q - WD = DU
Q = WD + m*Cv*(T2 -T1) 0.1232 kJ