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A centrifugal compressor running at 12000 rpm delivers 600$m^3$/min of free air. The air is compressed from 1 bar and $27^o$C to a pressure ratio of 4 with an isentropic efficiency of 85%.

The blades are radial at the impeller outlet and flow velocity of 60 m/s may be assumed throughout constant. The outer radius of the impeller is twice the inner one and slip factor is 0.9. calculate; i) Final temperature of air, ii) Power input to compressor, iii) Impeller diameter at inlet and outlet and iv) Width of impeller at inlet.

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Given:

$N = 12000\hspace{0.05cm}rpm,\hspace{0.25cm}P_1 = 1\hspace{0.05cm}bar,\hspace{0.25cm}T_1 = 27^\circ = 300\hspace{0.05cm}K,\hspace{0.25cm}V = 600\hspace{0.05cm}m^3/min\\ \frac{P_2}{P_1} = 4,\hspace{0.25cm}\eta_{iso} = 0.85,\hspace{0.25cm}V_f = V_{f1} = V_{f2} =60\hspace{0.05cm}m/s,\hspace{0.25cm} R_2 = 2R_1,\hspace{0.25cm}\phi_s =0.9$

To Find:

$T_2, \textit{Power}, D_1,D_2,B_1$

Solution:

$\hspace{5cm}1 - 2^1 : \textit{Isentropic Compression}\\ \frac{T_2^{'}}{T_1} = (\frac{P_2}{P_1})^\frac{r -1}{r}\\ T_2^{'}= 300\hspace{0.05cm} (4)^\frac{1.4 - 1}{1.4}\\ T_2^{'} = 445.798\hspace{0.05cm}K\\ \eta_{isen} = \frac {T_2^{'} - T_1}{T_2 - T_1}\\ 0.85 =\frac{445.798 - 300}{T_2 - 300}\\ T_2 = 471.53\hspace{0.05cm}K\\ PV = mRT\\ m = \frac{PV}{RT} = \frac{10^5\hspace{0.05cm}\times\hspace{0.05cm}600}{287\hspace{0.05cm}\times\hspace{0.05cm}300} = 696.84\hspace{0.05cm}kg/min = 11.61\hspace{0.05cm}kg/sec\\ \textit{Power} = W = mC_p(T_2 - T_1)\\ \hspace{0.8cm}=11.61\hspace{0.05cm}\times\hspace{0.05cm}1005\hspace{0.05cm}(471.53 - 300)\\ \hspace{0.8cm}=2.0014\hspace{0.05cm}MW\\ \hspace{0.5cm}\textit{W.D} = m.V_{b2}.V_{w2}\\ \textit{But} \hspace{0.05cm}\varphi_s = \frac{V_{w2}}{N_{b2}} = 0.9\\ \hspace{0.8cm} = V_{w2} = 0.9V_{b2}\\ 2.0014\hspace{0.05cm}\times\hspace{0.05cm}10^6 = 11.61\hspace{0.05cm}\times\hspace{0.05cm}0.9V_{b2}^2\\ \hspace{2cm}V_{b2} = 437.65\hspace{0.05cm}m/s\\ \hspace{2cm}V_{w2} = 393.89\hspace{0.05cm}m/s\\ \hspace{02cm}V_{b2} = \frac{\pi D_2N}{60}\\ \hspace{2cm}437.65 = \frac{\pi \hspace{0.05cm}\times\hspace{0.05cm}D_2\hspace{0.05cm}\times\hspace{0.05cm}1200}{60}\\ \hspace{02cm}D_2 = 0.697\hspace{0.05cm}m = 697\hspace{0.05cm}mm\\ \hspace{02cm}D_1 = \frac{1}{2}\hspace{0.05cm}\times\hspace{0.05cm}0.697 = 0.348\hspace{0.05cm}m = 348\hspace{0.05cm}mm\\ \textit{Volume flow rate} = \textit{Area over which air flows X Flow velocity}\\ \hspace{02cm}V = \pi \hspace{0.05cm}D_1\hspace{0.05cm}B_1\hspace{0.05cm}V_f\\ \hspace{02cm}\frac{600}{60} = 3.14\hspace{0.05cm}\times\hspace{0.05cm}0.348\hspace{0.05cm}\times\hspace{0.05cm}B_1\hspace{0.05cm}\times\hspace{0.05cm}60\\ \hspace{02cm}B_1 = 0.1525\hspace{0.05cm}m\\ \textit{Width of impeller at inlet}\hspace{0.05cm}(B_1) = 152.5\hspace{0.05cm}mm$