Question 12.2: Figure 12.14 shows a single-stage axial-flow compressor toge......

Part a: Because the stator flow is given as isentropic, it follows that

p_{t\ 2}=p_{t\ 1}=0.25\;\mathrm{bars}

T_{t\ 2}=T_{t\ 1}=200\,\mathrm{K}

Let us now locate the compressor point of operation on the map by computing the following variables:

{\dot{m}}_{C2}={\frac{{\dot{m}}{\sqrt{\theta_{2}}}}{\delta_{2}}}=5.0\ \mathrm{kg/s}

\pi_{C}={\frac{p_{t\ 3}}{p_{t\ 2}}}=1.4

Referring to the compressor map in Figure 12.14, we get

N_{C2}=34,000\,\mathrm{rpm}

\eta_{C}=80.0%

We can also calculate the “physical” variables (flow rate and speed) as follows:

{\dot{m}}={\frac{\delta_{2}}{\sqrt{\theta_{2}}}}{\dot{m}}_{C2}=1.5~\mathrm{kg/s}

N={\sqrt{\theta_{2}}}N_{C}=28,333\,\mathrm{rpm}

Applying the continuity equation at the stator-exit station, we have

\frac{\dot m\sqrt{T_{t\ 2}}}{p_{t\ 2}[\pi\left(r_{t\ 2}{}^{2}-r_{h2}{}^{2}\right)\cos\alpha_{2}]} = \sqrt{\frac{2\gamma}{(\gamma+1)R}}M_{c r\ 2}\biggl[1-\biggl(\frac{\gamma-1}{\gamma+1}\biggr)M_{c r\ 2}{}^{2}\biggr]^{\frac{1}{\gamma-1}}

which, upon substitution, yields

\alpha_{2}=56.0^{\circ}

Making use of the preceding results and the given data, we get

V_{2}=M_{c r\ 2}V_{c r\ 2}=0.62\sqrt{\left(\frac{2\gamma}{\gamma+1}\right)R T_{t\ 2}}=160.4\,\mathrm{m/s}

U_{m}=\omega r_{m}=356.0~\mathrm{m/s}

V_{z2}=V_{z3}=V_{2}\cos{\alpha_{2}}=89.7~\mathrm{m/s}

V_{\theta\ 2}=V_{2}\sin{\alpha_{2}}=133.0\ \mathrm{m/s}

W_{\theta\,2}=V_{\theta\,2}-U_{m}=-223.0\,\mathrm{m/s}

W_{2}=\sqrt{W_{\theta\ 2}{}^{2}-V_{z}{}^{2}}=240.4\;\mathrm{m/s}

T_{t\ r\ 2}=\,T_{t\ 2}-\left(\frac{V_{2}{}^{2}-W_{2}{}^{2}}{2c_{p}}\right)=216.0\,\mathrm{m/s}

W_{c r\,2}=\sqrt{\left(\frac{2\gamma}{\gamma+1}\right)R T_{t\ r\,2}}=269.0\;\mathrm{m/s}

\frac{W_{2}}{W_{c r\ 2}}=0.894

Part b: To calculate the stage-exit static pressure, we proceed as follows:

T_{t\ 3}=\,T_{t\ 2}\Biggl\{1+\frac{1}{\eta_C}\Biggl[\left(\frac{p_{t\ 3}}{p_{t\ 2}}\right)^{\frac{\gamma-1}{\gamma}}-1\Biggr]\Biggr\}=225.2\,{\mathrm K}

V_{\theta\ 3}=V_{\theta\ 2}-\frac{c_{p}}{U_{m}}(T_{t\ 3}-T_{t\ 2})=204.1\,\mathrm{m/s}

V_{3}={\sqrt{V_{\theta\ 3}{}^{2}+V_{z3}{}^{2}}}=222.9\;\mathrm{m/s}

M_{c r\ 3}=\frac{V_{3}}{V_{c r\ 3}}=\frac{V_{3}}{\sqrt{\left(\frac{2\gamma}{\gamma+1}\right)R T_{t\ 3}}}=0.812

p_{3}=p_{t\ 3}\bigg[1-\bigg(\frac{\gamma-1}{\gamma+1}\bigg)M_{c r}{}^{2}\bigg]^{\frac{\gamma}{\gamma-1}}=0.233\,\mathrm{bars}