Question 12.1: A two-stage centrifugal compressor, as shown in Figure 12.17......
A two-stage centrifugal compressor, as shown in Figure 12.17, has an overall pressure ratio of \pi_{\text{c}} . The isentropic efficiencies of both stages are \eta_1 \text{ and }\eta_2 , respectively. If no losses are assumed in the passage between the two stages, derive a mathematical expression for T_{03} in terms of \pi_{\text{c}},T_{01},T_{06},\eta_1,\text{ and }\eta_2.
Next, if \pi_0=11,T_{01}=290 \text{ K},T_{06}=660 \text{ K}, \eta_1=\eta_2=0.8, then calculate T_{03} .
The two-stage centrifugal compressor has an overall pressure ratio, \pi_{\text{c}} , which is defined as
\pi_{\text{c}}=\frac{P_{06}}{P_{01}} \text{ and }\frac{P_{04}}{P_{03}} =1 \\ \frac{P_{06}}{P_{01}} =\frac{P_{03}}{P_{01}} \times \frac{P_{06}}{P_{04}} \\ \frac{P_{03}}{P_{01}} =\left(1+ \eta_1\frac{\Delta T_{0\text{s}1}}{T_{01}} \right)^{\gamma/(\gamma-1)} \\ \Delta T_{0\text{s}1}=T_{03}-T_{01} \\ \frac{P_{03}}{P_{01}}\left(1+\left[\eta_1\frac{T_{03}}{T_{01}}-1 \right] \right) ^{\gamma/(\gamma/1)} \\ \frac{P_{06}}{P_{04}} =\left(1+ \eta_2\frac{\Delta _{0\text{s}2}}{T_{04}} \right) ^{\gamma/(\gamma-1)} \\ T_{04}\equiv T_{03} \\ \Delta T_{0\text{s}2}=T_{06}-T_{04}\equiv T_{06}-T_{03} \\ \frac{P_{06}}{P_{04}} =\left(1+\eta_2\left[\frac{T_{06}}{T_{03}}-1 \right] \right) ^{\gamma/(\gamma-1)} \\ \pi_{\text{c}}=\frac{P_{06}}{P_{01}} =\left(1+ \eta_1\left[\frac{T_{03}}{T_{01}} -1\right] \right) ^{\gamma/(\gamma-1)} \times \left(1+ \eta_2\left[\frac{T_{06}}{T_{03}}-1 \right] \right) ^{\gamma/(\gamma-1)} \\ \pi_{\text{c}}^{(\gamma-1)/\gamma}=\left(1+\frac{T_{03}}{T_{01}}\eta_1 -\eta_1 \right) \left(1+\frac{T_{06}}{T_{03}}\eta_2 -\eta_2 \right) \\ \pi_{\text{c}}^{(\gamma-1)/\gamma}= 1+\frac{T_{06}}{T_{03}} \eta_2 -\eta_2 +\frac{T_{03}}{T_{01}} \eta_1 +\frac{T_{06}}{T_{01}} \eta_1 \eta_2-\frac{T_{03}}{T_{01}} \eta_1 \eta_2-\eta_1 -\frac{T_{06}}{T_{03}} \eta_1 \eta_2 + \eta_1 \eta_2 \\ \pi_{\text{c}}^{(\gamma-1)/\gamma}= T_{03}\left(\frac{\eta_1}{T_{01}}-\frac{\eta_1 \eta_2}{T_{01}} \right) +\frac{1}{T_{03}} \left(T_{06}\eta_2-T_{06}\eta_1 \eta_2\right) +1- \eta_1-\eta_2+\eta_1 \eta_2\left(\frac{T_{06}}{T_{01}}+1 \right) \\ \pi_{\text{c}}^{(\gamma-1)/\gamma}= A_{1}T_{03}+\frac{A_2}{T_{03}} +A_3,
where
A_1=\frac{\eta_1}{T_{01}} -\frac{\eta_1 \eta_2}{T_{01}} \\ A_2=T_{06}\eta_2-T_{06} \eta_1 \eta_2 \\ A_3=1- \eta_1 -\eta_2+\eta_1 \eta_2\left(\frac{T_{06}}{T_{01}}+1 \right) \\ \pi_{\text{c}}^{(\gamma-1/\gamma)}=\frac{A_1 T^2_{03}+A_{2}+A_3T_{03}}{T_{03}} \\ A_1T^2_{03}+A_2+A_3T_{03}=\pi_{\text{c}}^{(\gamma-1)/\gamma}T_{03} \\ T^2_{03}+\left(\frac{A_3-\pi_{\text{c}}^{(\gamma-1)/\gamma}}{A_1} \right) T_{03}+\frac{A_2}{A_1} =0.
Define
\lambda_1=\frac{A_3-\pi_{\text{c}}^{(\gamma-1)/\gamma}}{A_1} \text{ and } \lambda_2=\frac{A_2}{A_1} \\ T_{03}=\frac{-\lambda_1 \pm \sqrt{ \lambda_1^2-4 \lambda_2}}{2} \quad \quad \quad (\text{a})
Now substitute the given data in Equation (a) to get the value of T_{03}
\pi_{\text{c}}=11 , \quad T_{01}=290 \text{ K}, \quad T_{06}=660 \text{ K}, \quad \eta_1=\eta_2=0.8
The values of the constants are then
A_1=5.517 \times 10^{-4} , \ \ A_2=105.6 , \ \ A_3=1.497 , \quad \text{ and } \quad \pi_{\mathrm{c}}^{(\gamma-1)/\gamma}=1.984 \\ \frac{\left(A_{3}-\pi_{\mathrm{c}}^{(\gamma-1/\gamma)}\right)}{A_{1}}=-882.726 \quad \text{ and } \quad \frac{A_2}{A_1} =19.141 \times 10^4
From Equation (a)
T_{03}=\frac{+88.726 \pm \ \sqrt{882.726^2 -4\times 19.141 \times 10^4}}{2} \\ T_{03}=441.363 \pm 58.235 \\ T_{03}=499.592 \text{ K} \quad \quad \text{or}\quad \quad T_{03}=383.128 \text{ K}