Question 14.1: Water flow is produced by a centrifugal pump with the follow......
Water flow is produced by a centrifugal pump with the following dimensions:
\begin{array}{lccc}{{r_{1}=6\,\mathrm{cm}}} & \beta _{1} = 33^{\circ }\\ {{r_{2}=10.5\,\mathrm{cm}}} & \beta _{2} = 21^{\circ }\\ {{L=4.75\,\mathrm{cm}}}\end{array}
At a rotational speed of 1200 rpm determine
(a) the design flow rate
(b) the power added to the flow
(c) the maximum pressure head at the pump discharge
To experience minimum losses, equation (14-11) must be satisfied; thus,
v_{r1}\,=\,r_{1}\omega\,\tan\beta_{1}
\quad =\,(0.06\,\mathrm{m})\Bigl(1200\,{\frac{\mathrm{rev}}{\mathrm{min}}}\Bigr)\Bigl({\frac{2\pi\,\mathrm{rad}}{\mathrm{rev}}}\Bigr)\Bigl({\frac{\mathrm{min}}{60\,\mathrm{s}}}\Bigr)(\mathrm{tan}\ 33^{\circ}\Bigr) \quad \quad \quad \quad (14-11)
\quad=4.896\,\mathrm{m/s}
The corresponding flow rate is
\dot{\mathrm{V}}=2\pi r_{1}L v_{r1}
\quad=2\pi(0.06\,\mathrm{m})(0.0475\,\mathrm{m})(4.896\,\mathrm{m}/s)
\quad=0.0877\,{\mathrm{m}}^{3}/{\mathrm{s}}\qquad(1390\,\mathrm{gpm})\quad \quad \quad \quad (a)
The power imparted to the flow is expressed by equation (14-10):
\dot{W}=\frac{\delta W_{\mathrm{s}}}{d t}=M_{\mathrm{z}}\omega=\rho\dot{V}r_{2}\omega\left[r_{2}\omega-\frac{\dot{ V}}{2\pi r_{2}L}\mathrm{cot}\,\beta_{2}\right] (14-10)
\dot{\mathrm{{W}}}=\rho\dot{\mathrm{{V}}}r_{2}\omega\biggl[r_{2}\omega-{\frac{\dot {\mathrm{{V}}}}{2\pi r L}}\mathrm{{cot}}\beta_{2}\biggr]
Evaluate the following:
\omega={\bigg(}1200{\frac{\mathrm{rev}}{\mathrm{min}}}{\bigg)}{\bigg(}2\pi{\frac{\mathrm{rad}}{\mathrm{rev}}}{\bigg)}{\bigg(}{\frac{\mathrm{min}}{60\,s}}{\bigg)}=125.7\,\mathrm{rad/s}
\rho \dot{V} r_{2}\omega\;=(1000\;{\mathrm{kg/m}}^{3})(0.0877\,{\mathrm{m}}^{3}/{\mathrm{s}})(0.105\,{\mathrm{m}})(125.7\,{\mathrm{rad/s}})
=\,1157\,{\mathrm{kg}}\cdot{\mathrm{m/s}}
we obtain
\dot{\mathrm{W}}=(1157\,{\mathrm{kg}}\cdot{\mathrm{m/s}})[(0.105\,{\mathrm{m}})(125.7\,{\mathrm{rad/s}})-(2.80\,{\mathrm{m/s}})({\mathrm{cot}}\,21)]
\quad=6830\,\mathrm{W}=6.83\,\mathrm{kW}\quad \quad \quad \quad (b)
Equation (14-1) expresses the net pressure head as
{\frac{P_{2}-P_{1}}{\rho g}}=-{\frac{\dot{W}}{\dot{m}g}}-h_{L} (14-1)
The maximum value, with negligible losses, will be
{\frac{P_{2}-P_{1}}{\rho g}}={\frac{6830\,\mathrm{W}}{(1000\,\mathrm{kg/m^{3}})(0.0877\,\mathrm{m^{3}}/{\mathrm{s(9.81~m/s^{2})}}}}
=7.94\operatorname*{m}\operatorname{H}_{2}\operatorname{O}\operatorname{gage}
{\mathrm{For~}}P_{1}=1{\mathrm{~atm}}=14.7{\mathrm{~psi}}=10.33{\mathrm{~m~H_{2}O}}
P_{2}=(7.94+10.33)\mathrm{m}\,\mathrm{H}_{2}\mathrm{O}=18.3\,\mathrm{m}\,\mathrm{H}_{2}\mathrm{O}\,\mathrm{(26\,psi)}\quad \quad \quad \quad (c)
The actual discharge pressure will be less than this owing to friction and other irreversible losses.