Engineering Optimization: Theory and Practice, Fourth Edition

2.2 Single-Variable Optimization 67

Example 2.2 In a two-stage compressor, the working gas leaving the first stage of
compression is cooled (by passing it through a heat exchanger) before it enters the
second stage of compression to increase the efficiency [2.13]. The total work input to
a compressor(W )for an ideal gas, for isentropic compression, is given by

W=cpT 1

[(

p 2

p 1

)(k − 1 )/k

+

(

p 3

p 2

)(k − 1 )/k

− 2

]

wherecpis the specific heat of the gas at constant pressure,kis the ratio of specific
heat at constant pressure to that at constant volume of the gas, andT 1 is the temperature
at which the gas enters the compressor. Find the pressure,p 2 , at which intercooling
should be done to minimize the work input to the compressor. Also determine the
minimum work done on the compressor.

SOLUTION The necessary condition for minimizing the work done on the compres-
sor is

dW

dp 2

=cpT 1

k

k− 1

[(

1

p 1

)(k − 1 )/k

k− 1

k

(p 2 )−^1 /k

+ p( 3 )(k^ −^1 )/k

−k+ 1

k

(p 2 )(^1 −^2 k)/k

]

= 0

which yields
p 2 = (p 1 p 3 )^1 /^2

The second derivative ofWwith respect top 2 gives

d^2 W

dp^22

=cpT 1

[

−

(

1

P 1

)(k − 1 )/k

1

k

(p 2 )−^ (^1 +k)/k

− p( 3 )(k^ −^1 )/k

1 − 2 k

k

(p 2 )(^1 −^3 k)/k

]

(

d^2 W

dp^22

)

p 2 =(p 1 p 2 )^1 /^2

=

2 cpT 1

k− 1

k

p 1 (^3 k−^1 )/^2 kp 3 (k+^1 )/^2 k

Since the ratio of specific heatskis greater than 1, we get

d^2 W

dp^22

> 0 at p 2 = (p 1 p 3 )^1 /^2

and hence the solution corresponds to a relative minimum. The minimum work done
is given by

Wmin= 2 cpT 1

k

k− 1

[(

p 3

p 1

)(k − 1 )/ 2 k

− 1

]