Question 11.2: Figure 11.21 shows a centrifugal compressor stage, which is ......
Figure 11.21 shows a centrifugal compressor stage, which is equipped with a totally radial impeller (i.e., one with no appreciable axial-velocity component anywhere). The stage geometrical data and operating conditions are as follows:
• Mass-flow rate (\dot m) = 2.5 kg/s
• Shaft speed (N) = 22,000 rpm
• Impeller-inlet total pressure (p_{t\ 1}) = 1.09 bars
• Impeller-inlet total temperature (T_{t\ 1}) = 300 K
• Sidewall spacing (b) = 2.2 cm
• Impeller-inlet blade angle (β^{\prime}{}_{1}) = −52°
• Impeller-exit blade angle (β^{\prime}{}_{2}) = −55°
• Impeller-inlet flow is assumed incompressible
• Impeller-inlet velocity is totally radial (i.e., α_{1} = 0)
• Impeller-inlet radius (r_{1}) = 8.0 cm
• Impeller-exit radius (r_{2}) = 20 cm
• Impeller total-to-total pressure ratio (p_{t\ 2}/p_{t\ 1}) = 3.6
• Impeller total-to-total efficiency (η_{t−t }) = 78%
Assuming a specific-heat ratio (γ ) of 1.4, calculate:
a) The incidence-caused total-pressure loss using Fig. 9.15;
b) The blade-exit deviation angle (\epsilon_{2});
c) The specific speed (N_{s});
d) The torque (\tau) transmitted to the impeller;
e) The exit/inlet total relative temperature ratio ( T_{t\ r\ 2}/T_{t\ r\ 1}).
Part a: Let us begin by computing the impeller-inlet thermophysical properties, noting that the inlet flow is stated to be incompressible:
\rho_{1}\approx\rho_{t\ 1}={\frac{p_{t\ 1}}{R T_{t\ 1}}}=1.27\,\mathrm{kg/m}^{3}
V_{1}=V_{r\ 1}=\frac{\dot{m}}{\rho_{1}(2\pi r_{1}b)}=178.0\,\mathrm{m/s}
U_{1}=\omega r_{1}=184.3\,\mathrm{m/s}
W_{1}=\sqrt{V_{r\ 1}{}^{2}+U_{1}{}^{2}}=256.2\,{\mathrm{m/s}}
\beta_{1}=\tan^{-1}\left({\frac{U_{1}}{V_{r\ 1}}}\right)=46.0^{\circ}
i_{i m p.}=\beta_{1}-\beta_{1}{}^{\prime}=-6.0^{\circ}
Using the graph in Figure 9.15, we get
(\Delta p_{t})_{i n c i d.}=0.022p_{t\ 1}=0.024\,\mathrm{bars}
Part b:
U_{2}=\omega r_{2}=460.0\,\mathrm{m/s}
T_{t\ 2}=T_{t\ 1}\Biggl\{1+\frac{1}{\eta_{t-t}}\biggl[\biggl(\frac{p_{t\ 2}}{p_{t\ 1}}\biggr)^{\frac{\gamma-1}{\gamma}}-1\biggr]\biggr\}=470.0\,\mathrm{K}
V_{\theta\ 2}=\frac{c_{p}}{U_{2}}(T_{t\ 2}-T_{t\ 1})=370.5\,\mathrm{m/s}
\rho_{t\ 2}={\frac{p_{t\ 2}}{R T_{t\ 2}}}=2.9\,\mathrm{kg/m^{3}}
V_{c r\ 2}={\sqrt{\left({\frac{2\gamma}{\gamma+1}}\right)R T_{t\ 2}}}=396.7\,{\mathrm{m/s}}
In order to compute V_{r\ 2}, we need to apply the continuity equation at the impeller exit station. However, the application of this equation requires knowledge of the static density ρ_{2}. This, in turn, requires knowledge of the impeller-exit critical Mach number (M_{cr\ 2}), a step we cannot conduct until we determine V_{r\ 2}, which (at this point) is unknown.
The computational difficulty just outlined is precisely the same as that encountered in Example 1. We will therefore follow the same iterative procedure we executed then. The final results come out to be as follows:
V_{r\ 2}=46.3\,\mathrm{m/s}
V_{2}={\sqrt{V_{r\ 2}{}^{2}+V_{\theta\ 2}{}^{2}}}=373.4\,{\mathrm{m/s}}
M_{c r\ 2}=0.941
\rho_{2}=1.95\,\mathrm{kg/m^{3}}
It follows that
\beta_{2}=\tan^{-1}\left({\frac{W_{\theta\ {2}}}{V_{r\ 2}}}\right)=-62.5^{\circ}
\mathrm{Deviation~angle}\,\epsilon_{i m p.}=\beta_{2}{}^{\prime}-\beta_{2}=-55^{\circ}-(-62.5^{\circ})=7.8^{\circ}
Part c: The specific speed can now be easily computed as follows:
N_{s}=\frac{N\left(\frac{2\pi}{60}\right)\sqrt{\frac{\dot m}{\rho_{2}}}}{\left[\eta_{C}c_{p}(T_{t\ 2}-T_{t\ 1})\right]^{\frac{3}{4}}}=0.37\,\mathrm{radians}
Referring to Figure 5.8, we see that this magnitude places the stage well within the centrifugal-stage range.
{\mathsf{P a r t{\ d}}}\colon {\mathrm{Supplied\;torque}}\;(\tau)=r_{2}V_{\theta\ 2}=74.1{\mathrm{~N/m}}\;(\mathrm{zero ~inlet ~swirl})Part e: The exit/inlet total relative temperature ratio can be computed as follows:
T_{t\ r\ 1}=T_{t\ 1}+\left({\frac{W_{1}{}^{2}-V_{1}{}^{2}}{\displaystyle2c_{p}}}\right)=316.9\,\mathrm{K}
T_{t\ r\ 2}=T_{t\ 2}+\left({\frac{W_{2}{}^{2}-V_{2}{}^{2}}{2c_{p}}}\right)=405.7\,\mathrm{K}
{\frac{T_{t\ r\ 2}}{T_{t\ r\ 1}}}=1.28
