Page 455 - Advanced thermodynamics for engineers

18.4 PROBLEMS 445

If the initial and final pressures are 50 bar and 2 bar respectively and the initial temperature
is 300 K, calculate
a. the value of the Joule–Thomson coefficient at the initial state, and
b. the final temperature of the gas, given that
3
k ¼ 11.0 kJ m /(kmol) 2
c p,m ¼ 29.0 kJ/kmol K.
3
7
[3.041 10 m K/J; 298.5 K]
P18.4 A gas has the equation of state
pv m 3 2 2 2
¼ 1 þ Ap T 9:75T c T þ 9T T þ Bp T
c
<T
where A and B are positive constants and T c is the critical temperature. Determine
the maximum and minimum inversion temperatures, expressed as a multiple of T c .
½6T c ; 0:5T c Š
P18.5 A gas has the equation of state

pv m 2
¼ 1 þ Np þ Mp
<T
where N and M are functions of temperature. Show that the equation of the inversion curve is

dN dM
p ¼
dT dT
If the inversion curve is parabolic and of the form
2
ðT T 0 Þ ¼ 4a p 0 p
where T 0 , p 0 and a are constants, and if the maximum inversion temperature is five times the
T 0 2
minimum inversion temperature, show that a ¼ and give possible expressions for N and M.
9p 0
3
" #
ðT T 0 Þ
M ¼ T; N ¼ p 0 T þ þ c
12a
P18.6 In a simple Linde gas-liquefaction plant (see Fig. 18.13), air is taken in at the ambient
conditions of 1 bar and 300 K. The water-jacketed compressor delivers the air at 200 bar and
300 K and has an isothermal efficiency of 70%. There is zero temperature difference at the
warm end of the regenerative heat exchanger (i.e. T 2 ¼ T 7 ). Saturated liquid air is delivered
at a pressure of 1 bar. Heat leakages into the plant and pressure drops in the heat exchanger
and piping can be neglected.
Calculate the yield of liquid air per unit mass of air compressed, the work input per
kilogram of air liquefied, and the rational efficiency of the liquefaction process.
[7.76%; 8.39 MJ/kg; 8.84%].