Water is pumped from a storage tank A to a storage tank B by a centrifugal pump. The levels of the surfaces of the liquid in both tanks are kept constant. The elevation of the level of the surface of the liquid in the tank B is 10 m higher than that in the tank A. The vacuum pressure of the space above the level of the surface of the liquid in the tank A is 50 kPa, and the gauge pressure the space above the level of the surface of the liquid in the tank of B is 50 kPa. The dimension of the suction pipe of the pump is phi 108 m m times 4 mm, and the length of the suction pipe is 50 m; the dimension of the discharge pipe of the pump is phi 80 mm times 4.5 mm, and the length of the discharge pipe is 100m. The length values of the pipes mentioned above include all the equivalent lengths of local friction losses except for local friction loss of inlet and outlet of pipelines The friction factors of all the pipes are 0,02. The reading R of the orifice the discharge pipe is 100 mm, and the indicator liquid is the mercury. The pump efficiency 75%. The density of the water at the operation temperature 1000 kg/m^3 The density of the mercury is 13.6 times 10^3 kg/m^3. The orifice meter coefficient C_o = 0.6. The orifice diameter is 45 mm. The local atmospheric pressure is 100 kPa. The acceleration of gravity g = 10 m/s^2. The velocity. through orifice V_o = C_o Squareroot 2gR (p_A - p)/p. What is the power supplied to the pump N (kw); When the opening of the outlet valve of the pump is reduced in certain degree, please point out the variation trend of the vacuum pressure in the inlet of the pump and that of the gauge pressure in the outlet of the pump by concrete analyses.
part (a): the velocity thru the meter is calculated:
V = Co*sqrt( 2gR(Drho)/rhow)) = 0.6*sqrt( 2*10* (13.6-1.2)/1) = 3 m/s approx
This is the velocity of the flow thru the pipe.
Head lsses: friction factor for the two sections of pipe: inlet, outlet to be calculated
discharge K1 =f(L/D)= .02*(100/.0845) = 23.67
suction K2= f(L/D) = .02 *(50/.104) = 9.62
Total fricition factor 33.3
Head loss over total pipe: K V^2/2g= 33.3 *9/20=15m approx
Total head to be considered given 10m + loss =25 m
Net Power of pump: rho*Q*g*H= 1000* ( velocity*area of pipe) *10* 25
Area of pipe: take average dia, .095 m
Net power = 3000* 250*pi/4*(.095)^2 = 5.3 KW
Actual power , use efficieny of .75 is 5.3/.75 = 7KW approx
b) if opening is reduced, f(L/D) increases, so friction head will increase. There wil be more load on the pump. Same calculation to be done. Only formula to use is the meter formula, where V increases, so R decreases.For this to see please differentiate the meter formula:
DV= constant*( R ^-.5), so R will go down if V increase due to smaller dia
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