Question 11.5: The geometrical details of a centrifugal-compressor stage ar......

Part a: Let us calculate the impeller-inlet static density, assuming M_{cr\ 1} to be sufficiently small to justify an incompressible flow at this flow station:

\rho_{1}\approx\rho_{t\ 1}={\frac{P_{t\ 1}}{R T_{t\ 1}}}=1.71\,\mathrm{kg/m}^{3}

V_{1}=V_{r\ 1}=\frac{\dot{m}}{\rho_{1}\pi(r_{0}{}^{2}-r_{i}{}^{2})}=190.7\,\mathrm{m/s}

M_{c r\ 1}=0.54

This critical Mach number magnitude is disappointing, for it is too high to be consistent with our impeller-inlet incompressible-flow assumption (above). To overcome this difficulty, we take the newly computed critical Mach number (above) as an initial guess, re-calculate the static density, re-apply the continuity equation to find a new critical Mach number, repeating the entire procedure until convergence is attained. In the end, the critical Mach number came out to be:

M_{c r\ 1}=0.65

It follows that

V_{1}=V_{r\ 1}=228.9\,\mathrm{m/s}

\rho_{1}=1.425\,\mathrm{kg/m}^{3}

Part b:

U_{2}=\omega r_{2}=380.3\,\mathrm{m/s}

Knowing that the impeller dynamic head is 57,921 J/kg (given), we can proceed to calculate the impeller-exit critical Mach number and the impeller total-to-total efficiency as follows:

V_{2}={\sqrt{V_{1}{}^{2}+2(\Delta h)_{d y n.}}}=410.2\,{\mathrm{m/s}}

V_{\theta\ 2}=V_{2}\sin\alpha_{2}=380.3\,\mathrm{m/s}

W_{\theta\ 2}=V_{\theta\ 2}-U_{2}=0

This means that the exit relative velocity (W_{2}) is totally in the radial direction:

W_{2}=V_{r\ 2}=V_{2}\cos\alpha_{2}=153.7\,\mathrm{m/s}

With V_{θ\ 1} being zero, the impeller-exit total temperature can now be calculated:

T_{t\ 2}=T_{t\ 1}+{\frac{U_{2}V_{\theta\ 2}}{c_{p}}}=509.9\,\mathrm{K}

M_{c r\ 2}={\frac{V_{2}}{V_{c r\ 2}}}=0.99 (subsonic, as it should be)

p_{t\ 2}=\frac{p_{2}}{\left[1-\left(\frac{\gamma-1}{\gamma+1}\right)M_{c r\ {2}}{}^{2}\right]^{\frac{\gamma}{\gamma-1}}}=4.78\,\mathrm{bars}

\eta_{C}=\frac{\left(\frac{p_{t\ 2}}{p_{t\ 1}}\right)^{\frac{\gamma-1}{\gamma}}-1}{\left(\frac{T_{t\ 2}}{T_{t\ 1}}\right)-1}=81.8%

Part c:

\rho_{2}=\rho_{t\ 2}\biggl[1-\biggl(\frac{\gamma-1}{\gamma+1}\biggr)M_{cr\ 2}{}^{2}\biggr]^{\frac{1}{\gamma-1}}=2.09\,\mathrm{kg/m}^{3}

b_{2}=\frac{\dot m}{\rho_{2}V_{r\ 2}(2\pi r_{2})}=1.15\,\mathrm{cm}

Part d: Let us first calculate the impeller-exit total relative pressure (p_{t\ r\ 2}):

T_{t\ r\ 2}=\,T_{t\ 2}-\left(\frac{V_{2}{}^{2}-W_{2}{}^{2}}{2c_{p}}\right)=437.9\,\mathrm{K}

p_{t\ r\ 2}=p_{t\ 2}\biggl(\frac{T_{t\ r\ 2}}{T_{t\ 2}}\biggr)^{\frac{\gamma}{\gamma-1}}=2.81\,\mathrm{bars}

As for the impeller inlet station, we have

W_{\theta\ 1}=V_{\theta\ 1}-U_{1}=0-(\omega r_{1m})=-134.2\,\mathrm{m/s}

W_{z1}=V_{z1}=V_{1}=228.9\,\mathrm{m/s}

T_{t\ r\ 1}=T_{t\ 1}-\left({\frac{V_{1}{}^{2}-W_{1}{}^{2}}{2c_{p}}}\right)=375.0\,\mathrm{m/s}

p_{t\ r\ 1}=p_{t\ 1}\!\left({\frac{T_{t\ r\ 1}}{T_{t\ 1}}}\right)^{\frac{\gamma}{\gamma-1}}=1.96\,\mathrm{bars}

Now, we can calculate the percentage of the impeller-wise increase in total relative pressure as follows:

{\frac{\Delta\,p_{t\ r}}{p_{t\ r\ 1}}}={\frac{(p_{t\ r\ 2}-p_{t\ r\ 1})}{p_{t\ r\ 1}}}=43.4%