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

See A here

2 bar = 0.20MPa

See B here

5 bar = 0.50MPa

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