Modern Engineering Thermodynamics

1Q 2 − 1 W 2 = mðu 2 − u 1 Þ + KE 2 − KE 1 + PE 2 − PE 1
(Continued )
5.7 Piston-Cylinder Devices 155
and since we know that, for the argon, p 1 = 0:0100 MPa = 10:0 kPa = 10:0 kN/m 2 , T 1 = 20:0°C, and V 1
= 0:0100 m 3 , we can
find the value of R from Table C.13b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics as
R = 0:208 kJ/kg⋅K; then, determine the mass as
m =
This equation for T 2 can be solved to give
ð10:0 kN/m 2 Þð0:0100 m 3 Þ
= 0:00164 kg
ð0:208 kJ/kg⋅KÞð20:0 + 273:15 KÞ
T 2 = 20:0°C +
0:100 kJ
ð0:00164 kgÞð0:315 kJ/kg⋅KÞ = 214°C
where we use the value of c v = 0:315 kJ/kg⋅K for argon found in Table C.13b. Note that, if you express T 1 in °C, then T 2 is
in °C, but if you express T 1 in K, then T 2 is in K also. Finally, we can compute the final pressure from the ideal gas equation
of state as
p 2 = mRT 2
V 2
=
ð0:00164 kgÞð0:208 kJ/kg⋅KÞð214 + 273:15 KÞ
0:0400 m 3 = 4:15 kPa
where in the final state the argon fills the entire volume of the box.
Exercises
4. Suppose the balloon in Example 5.5 is designed to burst after absorbing 0.200 kJ of radiation heat transfer. What would
the final temperature and pressure be inside the box after the balloon burst? Answer: T 2 = 407°C and p 2 = 5:8 kPa.
5. If air were substituted for the argon in Example 5.5 with no changes in the remaining parameters, what would the final
temperature and pressure be inside the box after the balloon burst? Answer: T 2 = 137°C, p 2 = 3:5 kPa.
6. Determine the radiation heat transfer in Example 5.5 that would produce a final pressure of 20.0 kPa in the box after
the balloon bursts. What would be the corresponding temperature in the box? Answers: 1 Q 2 = 1:06 kJ, and T 2 = 2070°C:
5.7 PISTON-CYLINDER DEVICES
One of the oldest pieces of effective technology is the piston in a cylinder apparatus. It was used in early Roman
pumps, and its use in the steam engine of the 18th century brought about the Industrial Revolution. It is still
commonly used today in piston-type pumps and compressors and in a wide variety of internal and external
combustion engines. The following example illustrates its use in a refrigeration process.
EXAMPLE 5.6
You have 0.100 lbm of Refrigerant-134a initially at 180.°F and 100. psia in a cylinder with a movable piston that undergoes
the following two-part process. First, the refrigerant is expanded adiabatically to 30.0 psia and 120.°F, then it is compressed
isobarically (i.e., at constant pressure) to half its initial volume.
Determine
a. The work transport of energy during the adiabatic expansion.
b. The heat transport of energy during the isobaric compression.
c. The final temperature at the end of the isobaric compression.
Solution (This is an example of a multiple-part solution process)
First, draw a sketch of the system (Figure 5.6).
The unknowns are (a) 1 W 2 , (b) 2 Q 3 , and (c) T 3 . The system is closed, consisting of the R-134a in the cylinder.
Because this is a two-part process, three states are involved, as follows.
Process: Adiabatic
State 1 ƒƒƒƒƒƒƒƒƒ ƒ!
expansion
State 2
Process: Isobaric
ƒƒƒƒƒƒƒƒƒƒ!
compression
State 3
p 1 = 100: psia p 2 = 30:0 psia p 3 = p 2 = 30:0 psia
T 1 = 180:°F T 2 = 120:°F v 3 = v 1 /2
The basic energy balance equations for these two processes are