Question

Liquid benzene at 298K and 1.2 bar is converted to vapor at 473K and 5 bar in a two-step process.
For compression to 5 bar by a pump, calculate the power requirement of the pump, if it is 70% efficient. Assume Cp of 105 j/mol.

Answer 1

Since, the temperature of liquid benzene is increasing from 298K to 473K and during this pressure is increasing from 1.2 bar to 5 bar.

First we assume that the process is Adiabatic compression process.

It is given in problem two-step process, there must be intermediate pressure during these stages, we need to find this pressure which will required during calculations.

The interstage pressure in a two-stage compressor is the geometric mean of the intial and final pressure. That is,

P'=

Let P₁= 1.2 bar = 1.2* 10⁵ Pa

P₂= 5 bar = 5* 10⁵ Pa and P' be the interstage pressure.

By calculation, P' = = = 244948.97 Pa

We will required n here which is ratio of Cp and Cv

So, Cp - Cv = R where R is gas constant whose value is R= 8.314 J/ (mol.K)

from above equation, Cv = 96.686 J/mol, (simply just a subsraction )

the ratio n which is exponent can be calculated as   n = = 1.085

Let the define term r = = 2.041

All of the above terms we needed to calculate the work of compresssion,the formula for evaluating the work of compression is given as W_S = , where N is the number of stages  

From calulation the work of cpmprssion we get W_S = -3635.85 J/mol (Taking T₁ as 298 K)

Negative sign indicate that the work is done on system, by sign convention any kind of work done on the system is negative.

Number of moles of the gas comprssed is given by

PV/RT = ( 1.2 * 10⁵ * 1) / (8..314 * 298 * 3600 ) = 0.0134 mol/s  (Assuming capacity of pump is 1 m³/h , not given in the problem)

the required work is

-W_S = 3635.85 * 0.0134 = 48.73 Watt

So the pump is 70% efficient, the actual power requirement will be = 70% of 3635.85 = 0.7* 3635.85 = Watt