Question 11.4: Figure 11.23 shows a centrifugal compressor stage that is ge......
Figure 11.23 shows a centrifugal compressor stage that is geometrically identical to that in Example 3. The impeller is radial, meaning there is no appreciable axial velocity component anywhere within this subdomain. The stage operating conditions are as follows:
• Inlet total pressure (p_{t\ 0}) = 11.4 bars
• Inlet total temperature (T_{t\ 0}) = 649.5 K
• Inlet critical Mach number (M_{cr\ 0}) = 0.46
• Zero swirl velocity in the inlet duct (α_{0} = α_{1} = 0)
• Impeller-inlet relative flow angle (β_{1}) = −54°
• Impeller-exit relative flow angle (β_{2}) = −46°
• Impeller-exit total pressure (p_{t\ 2}) = 18.2 bars
• Impeller total-to-total efficiency (η_{t−t}) = 78%
The following simplifications are also valid:
• Isentropic flow in the unbladed inlet duct
• Adiabatic flow throughout the stage
• A constant specific-heat ratio (γ) of 1.4
With the preceding data, calculate:
a) The mass-flow rate (\dot m);
b) The impeller-inlet critical Mach number (M_{cr\ 1});
c) The impeller “physical” speed (N);
d) The impeller-exit absolute flow angle (α_{2});
e) The stage specific speed (N_{s}).
Part a: Knowing that the stage-inlet swirl angle (α_{0}) is zero, we can apply the continuity equation there, with the result being
{\dot{m}}=12.08\,\mathrm{kg/s}
This mass-flow rate is on the high side for a centrifugal stage. However, the specificspeed magnitude also depends on the ideally supplied shaft work. The latter will have to be sufficiently large to justify a centrifugal-stage choice.
Part b: Let us reapply the continuity equation, but at the impeller-inlet station this time:
\frac{\dot m\sqrt{T_{t\ 1}}}{p_{t\ 1}(2\pi r_{1}b_{1})}=\sqrt{\frac{2\gamma}{(\gamma+1)R}}M_{c r\ 1}\biggl[1-\biggl(\frac{\gamma-1}{\gamma+1}\biggr)M_{c r\ 1}{}^{2}\biggr]^{\frac{1}{\gamma-1}}
where
T_{t\ 1}=T_{t\ 0} (adiabatic flow through the inlet passage)
p_{t\ 1}=p_{t\ 0} (isentropic flow through the same passage)
Upon substitution, the preceding continuity equation can be rewritten in the following compact form:
M_{c r\ 1}(1-0.1667M_{c r\ 1}{}^{2})=0.561
This nonlinear equation is very similar to that in Example 3. A similar trial-and-error procedure was performed in this case, and the final result is
M_{c r\ 1}=0.69
Part c:
V_{r\ 1}=V_{1}=M_{c r\ 1}V_{c r\ 1}=321.7\,\mathrm{m/s}
U_{1}=\omega r_{1}=V_{r\ 1}\tan\beta_{1}=442.8\,\mathrm{m/s}
\omega={\frac{U_{1}}{r_{1}}}=5534.8\,\mathrm{radians/s}
N=\left({\frac{60}{2\pi}}\right)\omega=52,853\,\mathrm{rpm}
Part d:
T_{t\ 2}=T_{t\ 1}\left\{1+\frac{1}{\eta_{t-t}}\left[\left(\frac{p_{t\ 2}}{p_{t\ 1}}\right)^{\frac{\gamma-1}{\gamma}}-1\right]\right\}=768.6\,\mathrm{K}
V_{\theta\ 2}=\frac{c_{p}}{U_{2}}(T_{t\ 2}-T_{t\ 1})=200.1\,\mathrm{m/s}
W_{\theta\,2}=V_{\theta\,2}-U_{2}=-397.7\,\mathrm{m/s}
V_{r\ 2}=W_{r\ 2}={\frac{W_{\theta\ 2}}{\tan\beta_{2}}}=384.1\,\mathrm{m/s}
\alpha_{2}=\tan^{-1}\left({\frac{V_{\theta\ 2}}{V_{r\ 2}}}\right)=27.5^{\circ}
Part e: To calculate the stage specific speed, we proceed as follows:
V_{2}={\sqrt{V_{\theta\ 2}{}^{2}+V_{r\ 2}{}^{2}}}=433.1\,{\mathrm{m/s}}
M_{c r\ 2}=\frac{V_{2}}{V_{c r\ 2}}=\frac{V_{2}}{\sqrt{\left(\frac{2\gamma}{\gamma+1}\right)R T_{t\ 2}}}=0.854
\rho_{2}=\left(\frac{p_{t\ 2}}{R T_{t\ 2}}\right)\biggl[1-\biggl(\frac{\gamma-1}{\gamma+1}\biggr)M_{cr\ 2}{}^{2}\biggr]^{\frac{1}{\gamma-1}}=5.97\,\mathrm{kg/m^{3}}
N_{s}=\frac{\omega\sqrt{\frac{\dot m}{\rho_{2}}}}{[\eta_{C}c_{p}(T_{t\ 2}-T_{t\ 1})]^{\frac{3}{4}}}=1.47\,\mathrm{radians}