Chapter 7, Problem 7.30P

Interpretation Introduction

Interpretation:

Shaft work of the compressor considering process is adiabatic and isentropic.

Concept Introduction:

Let us assume duration of operation is 1 hr.

It is given that this compression process is isentropic and adiabatic. Steady flow equation for compression process reduces to, $W=\dot{n}\Delta H$ after ignoring both change in kinetic and potential energy term. Here, $\Delta H$ is the change in enthalpy and $W$ is the shaft work per unit mass basis. It is to be noted that molar flow rate of air through compressor is $\dot{n}$ .

We can get enthalpy of air at the inlet to compressor as both pressure and temperature conditions are known. Compressor outlet pressure is 1000 kPa but temperature is unknown. We need to determine corresponding temperature considering process is isentropic which implies entropy at the inlet and outlet of the compressor will remain same. Specific entropy can be determined at the given compressor inlet conditions. Next we need to calculate the temperature corresponding to 1000 kPa pressure to have same specific entropy as that of the inlet condition.

Considering air is an ideal gas, specific enthalpy change is determined by the following formula:

$\begin{array}{l}\Delta H=C_p\Delta T\\ =C_p\times (T_2-T_1)\end{array}$

To determine molar flow rate through compressor, we need to calculate total change in mole in the process by the following equation:

$\begin{array}{l}\Delta n=n_2-n_1\\ =\frac{P_{2}V}{R{T}_2}-\frac{P_{1}V}{R{T}_1}\\ =\frac{V}{R}(\frac{P_2}{T_2}-\frac{P_1}{T_1})\\ \end{array}$

Then this needs to be divided by 1 hr to get rate of molar flow rate. Here compressibility factor is 1 being ideal gas.

Shaft work can be calculated by multiplying rate of molar flow with specific enthalpy.