If the rotational speed of the impeller of a centrifugal pump is doubled without

Question

If the rotational speed of the impeller of a centrifugal pump is doubled without changing the impeller diameter, then how does the capacity change? New flow rate is four time less than the old flow rate New flow rate is two times more than the old flow rate New rate is four times more than the old flow rate New rate is two times less than the old flow rate The specifies speed for a pump operating at 2000 rpm delivering 15000 gaVmin of water at a total head of 400 ft is given by 2659.43 rpm 2738.6 rpm 2854.9 rpm none of the above Determine the NPSH available when a pump draws gasoline at 30 degree C (density = 650 kg/m^3) from a tank whose level is 1.7 m below the pump inlet. The energy losses in the suction line total 1.8 m and the atmospheric pressure is 99.2 kPa absolute vapor pressure of gasoline is shown in the figure to the right. Assume the velocity in the pipe to be negligible 15.56 m -2.32 m 1.18 m 2.06 m. When the condenser pressure is reduced without changing the evaporator pressure the refrigeration effect and COP are Increased Reduced

Explanation / Answer

1.1

From pump affinity laws, flow rate Q is proportional to N*D where N is rotational speed and D is impeller dia.

Q1 / Q2 = (N1 / N2)*(D1 / D2)

We have D2 = D1 and N2 = 2*N1

Thus, Q1 / Q2 = 1/2 * 1 = 1/2

or Q2 = 2*Q1

Hence, option b is correct.

1.2

Specific speed of pump Ns = N*Q1/2 / H3/4

where N = rotational speed, Q = volume flow rate, H = Head rise

We have N = 2000 rpm, Q = 15000 gal/min, H = 400 ft

Thus, Ns = 2738.6 rpm...Option b is correct

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