PROBLEM 3.9

COMMENT: As the pressure increases, the difference in specific volume between saturated

vapor and saturated liquid decreases. At the critical pressure, the two states

coincide and the difference is zero.

PROBLEM 3.37

PROBLEM 3.55

PROBLEM 3.63

PROBLEM 3.85

PROBLEM 3.98

Five kg of butane (C4H10) in a piston-cylinder assembly undergo a process from p1 = 5 MPa, T1 =

500 K to p2 = 3 MPa during which the relationship between pressure and specific volume is pv =

constant. Determine the work, in kJ.

KNOWN: Five kg of C4H10 undergo a process for which pv = constant in a piston-cylinder

assembly between two states.

FIND: Determine the work.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL: 1. The Butane is a closed system. 2. The process is polytropic with

pv = constant. 3. Volume change is the only work mode.

ANALYSIS: The work is given by W = m

. With pv = constant = p1v1

W = m

= (p1v1) m ln

(*)

To evaluate W requires v1 and v2. The compressibility chart can be used to obtain v1: From

Table A-1; pc = 38 bar, Tc = 425 K, M = 58.12 kg/kmol.

pR1 = p1/pc = (50)/(38) = 1.32

Fig. A-2: Z1 ≈ 0.67

TR1 = T/Tc = (500)/(425) = 1.18

Accordingly, with v1 = Z1 (RT1/p1), we get

v1 = (0.67)

= 0.0096 m

3/kg

Now, with pv = constant,

= [(5 MPa)(0.0096 m3)]/(3 MPa) = 0.016 m

3

Now, inserting values into (*), we get

Butane

(C4H10)

m = 5 kg

p1 = 5 MPa = 50 bar

T1 = 500 K p1 = 3 MPa = 30 bar →

pv = constant

PROBLEM 3.98 (CONTINUED) – PAGE 2

W = (5 MPa)(0.0096 m3/kg)(5 kg) ln [(0.016)/(0.0096)]

= 122.6 kJ (out)

Alternative Evaluation of

pR1 = p1/pc = (50)/(38) = 1.32

Fig. A-2: ≈ 0.6

TR1 = T/Tc = (500)/(425) = 1.18

v1 =

= (0.6)

= 0.0096 m

3/g

PROBLEM 3.105

PROBLEM 1.333

Two kg of nitrogen (N2) gas are contained in a closed, rigid tank surrounded by a 10-kg water

bath, as shown in Fig. P1.133. Data for the initial states of the nitrogen and water are shown on

the figure. The entire unit is well-insulated, and the nitrogen and water interact until thermal

equilibrium is achieved. The measured final temperature is 34.1oC. The water can be modeled

as an incompressible substance, with c = 4.179 kJ/kg∙K, and the nitrogen is an ideal gas with

constant cv. From the measured data, determine the average value of the specific heat cv, in

kJ/kg∙K.

KNOWN: A tank of nitrogen gas is surrounded by a well-insulated water bath. The gas and

water are initially at different temperatures and they interact until equilibrium is achieved.

FIND: Determine the average values of the specific heat cv for the nitrogen.

SCHEMATIC AND GIVEN DATA:

ANALYSIS: Reducing the energy balance

ΔKE + ΔPE + ΔUw + Δ = Q – W → ΔUw + Δ

= 0

The final equilibrium temperature is T2. With ΔUw = mwcw (T2 – T1,w) and Δ =

cv(T2 - )

mw(T2 – T1,w) + cv(T2 -

) = 0

Solving for cv

cv =

=

= 0.741 kJ/kg∙K

Nitrogen

10 kg

water bath

2 kg

Nitrogen

= 50

oC

Water

T1, w = 20oC

ENGINEERING MODEL: (1) As

shown on the accompanying sketch, the

closed system consists of the nitrogen

tank and the water bath. (2) W = 0 and

Q = 0. (3) Kinetic and potential energy

effects can be neglected. (4) The water

is an incompressible substance with cw =

4.179 kJ/kg∙K. (5) The nitrogen is

modeled as an ideal gas with constant cv.

PROBLEM 3.135

PROBLEM 3.142

PROBLEM 3.142 (CONTINUED)