Question

An ideal compressor compresses 12 kg/min of air isothermically from 99 kPa and a specific volume of 0.81 m3 /kg to a final pressure of 600 kPa. Determine the work in kW

Answer 1

At first, since the temperature is constant, for an ideal gas we have the following:

$p_1v'_1=RT=p_2v'_2; v'_1=v_1/n; v'_2=v_2/n \\ \cfrac{v'_2}{v'_1} = \cfrac{p_1}{p_2}$

Then, we proceed to find the work for this type of process, where n^* is the rate of gas that is being compressed on the device ($n^*=12\frac{kg}{min}\times\frac{1\,min}{60\,s}=0.2\frac{kg}{s}$), and the relation between RT and p₁v'₁ to find:

$W = \int p\,{dv}=\int (\frac{n^*RT}{v})\,{dv}=n^*RT\int \cfrac{dv}{v} \\ W=n^*RT\ln{ \cfrac{v'_2}{v'_1}} = n^*p_1v'_1\ln{\cfrac{p_1}{p_2}}$

Now, we proceed to substitute and we calculate the rate of work done as:

$W = (0.2\frac{kg}{s})(99\,{kPa})(0.81\frac{m^3}{kg})\ln{ \bigg(\cfrac{99\,{kPa}}{600\,{kPa}} \bigg)} \\ W = -28.8974\frac{kJ}{s}=-28.8974\,{kW}$

In conclusion, we were able to find that W = -28.8974 kW(W < 0 because the gas is being compressed and the work is being performed on it).

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.