PROBLEM 3.136

PROBLEM 3.137

Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 280 K

to p2 = 11 bar. The initial volume is 0.262 m3. The process is described by pV1.25

= constant.

Assuming ideal gas behavior and neglecting kinetic and potential energy effects, determine the

work and heat transfer for the process, each in kJ, for (a) constant specific heats evaluated at 300

K, and (b) data from Table A-23. Compare the results and discuss.

KNOWN: Data are provided for the polytropic compression of carbon dioxide in a piston-

cylinder assembly.

FIND: Determine the work and heat transfer for the process using (a) constant specific heats, (b)

data from Table A-23.

SCHEMATIC AND GIVEN DATA:

ANALYSIS: Work is evaluated from W12 =

=

(See Example 2.1 for

details). For the polytropic process

V2 =

V1 =

(0.262 m3) = 0.02892 m

3

The work is

W =

= 53.89 kJ (in)

carbon

dioxide

(CO2)

p1 = 0.7 bar

T1 = 280 K

V1 = 0.262 m3

pV1.25

= constant

p2 = 11 bar

p

v

.

. 0.7 bar

11 bar

280 K

T2

ENGINEERING MODEL: (1) The carbon

dioxide within the piston-cylinder assembly is

a closed system. (2) The process is polytropic,

with pV1.25 = constant. (3) Volume change is

the only work mode. (4) Kinetic and potential

energy effects can be neglected. (5) The

carbon dioxide is modeled as an ideal gas with

(a) constant specific heats, and (b) with

variable specific heats.

PROBLEM 3.137 (CONTINUED)

Reducing the energy balance gives: Q12 = m(u2 – u1) + W12

The mass is

m = p1V1/RT1 =

= 0.3467 kg

and the final temperature is

T2 = p2V2/mR =

= 485.7 K

(a) From Table A-20 for CO2 at 300 K; cv = 0.657 kJ/kg∙K. Using constant specific heats, the

energy balance becomes

Q = mcv(T2 – T1) + W12 = (0.3467 kg)(0.657 kJ/kg∙K)(485.7 – 280)K + ( 53.89 kJ)

= 7.035 kJ (out)

(b) Using data from Table A-23: 1 = 6369 kJ/kmol and interpolating; 2 = 13004 kJ/kmol

Q = m

+ W12

= (0.3467)

+ ( kJ) = kJ (out)

In this case, the assumption of constant specific heat results in a value for Q that differs from the

variable specific heat value obtained using data from Table A-23 by approximately 334 %. The

assumption of constant specific heat evaluated at 300 K is questionable in this case.

PROBLEM 3.132

Revised 10-14

Revised 10-14

Problem 3.139 Air contained in a piston-cylinder assembly undergoes two processes in series ,

as shown in Fig. P3.139. Assuming ideal gas behavior for the air, the work and heat transfer for

the overall process, each in kJ/kg.

KNOWN: Air contained in a piston-cylinder assembly undergoes two processes in series.

FIND: For the overall process of the air, find the work and the heat transfer, in kJ/kg.

Using the ideal gas equation of state, the mass is

m = p1V1/RT1 =

= 0.2323 kg

So, W12/m = (40 kJ)/(0.2323 kg) = 172.2 kJ/kg

For 2-3, T is constant and p = mRT/V, so

W23 =

= mRT

= mRT2 ln (V3/V2)

0

1

2

0 0.1 0.2 0.3 0.4 0.5 0.6

ANALYSIS: For 1-2, the work determined using Eq. 2.17 is

W12 =

= p(V2 – V1) = (2 bar)(0.3 – 0.1) m

3/kg

= 40 kJ

pressure is constant

V (m3)

p

(bar)

1

3

2

ENGINEERING MODEL: (1) The air

in the piston-cylinder is the closed

system. (2) The air is modeled as an

ideal gas. (3) Kinetic and potential

energy effects are ignored.

Air

. .

.

Isothermal

process

T1 = 300 K

Revised 10-14

Revised 10-14

Problem 3.139 (Continued)

Process 1-2 is at constant pressure, so

= p1 = p2 =

→ T2 = (V2/V1)T1 = (0.3/0.1)(300 K) = 900 K

Thus

W23/m = RT2 ln (V3/V2) = (8.314/28.97 kJ/kg∙K)(900 K) ln (0.6/0.3) = 179.0 kJ/kg

The total work is W/m = W12/m + W23/m = 172.2 + 179.0 = 351.2 kJ/kg

An energy balance for the overall process is obtained as follows: ΔKE + ΔPE + ΔU = Q – W

Q = ΔU + W = m(u3 – u1) + W → Q/m = (u3 – u1) + W/m

With data from Table A-22

Q/m = (674.58 – 214.07)kJ/kg + 351.2 kJ/kg = 811.7 kJ/kg

1 Note that the heat transfers Q12 and Q23 could be determined from the respective energy

balances, as follows

Q12 = (u2 – u1) + W12/m = (674.58 – 214.07) + (172.2) = 632.7 kJ/kg

and

Q23 = (u3 – u2) + W23/m = (0) + 179.0 = 179.0 kJ/kg

So, the total for the overall process is Q = Q12 + Q23 = 632.7 + 179.0 = 811.7 kJ/kg

which is the same result as above, as expected.

1