Question 16.2: Determine the pressure ratio developed and the specific work...
Determine the pressure ratio developed and the specific work input to drive a centrifugal air compressor of an impeller diameter of 0.5 m and running at 7000 rpm. Assume zero whirl at the entry and T_{1 t}=290 \mathrm{~K} . The slip factor and power input factor to be unity, the process of compression is isentropic and for air c_{p}=1005 J/kg K, \gamma=1.4 .
The impeller tip speed
U_{2}=\frac{\pi \times 0.5 \times 7000}{60} =183.26 \mathrm{~m} / \mathrm{s}With the help of Eqs (16.6) and (16.7), we can write
\frac{T_{2 t}}{T_{1 t}}=\frac{T_{3 t}}{T_{1 t}}=1+\frac{\Psi \sigma U_{2}^{2}}{c_{p} T_{1 t}} (16.6)
\frac{p_{3 t}}{p_{1 t}}=\left(\frac{T_{3 t}^{\prime}}{T_{1 t}}\right)^{\frac{\gamma}{\gamma-1}}
=\left[1+\frac{\eta_{c}\left(T_{3 t}-T_{1 t}\right)}{T_{1 t}}\right]^{\frac{\gamma}{\gamma-1}} (16.7)
Pressure ratio =\left[1+\frac{U_{2}^{2}}{c_{p} T_{1 t}}\right]^{\frac{\gamma}{\gamma-1}}
=\left[1+\frac{(183.26)^{2}}{1005 \times 290}\right]^{\frac{1.4}{0.4}}
=1.46
From Eq (16.3),
w=\Psi \sigma U_{2}^{2} (16.3)
specific work input =U_{2}^{2}=\frac{(183.26)^{2}}{1000}=33.58 \mathrm{~kJ} / \mathrm{kg}