Water enters a turbo-machine at a velocity 12 ft/s. The flow occurs in the horiz
ID: 1823103 • Letter: W
Question
Water enters a turbo-machine at a velocity 12 ft/s. The flow occurs in the horizontal plane. The inside diameters of the inlet and exit pipes are 1 ft and 2 ft, and the gage pressures at the inlet and exit are 10 psi and 5 psi, respectively. The machine is anchored at the lower right corner, which is 18 in. to the right of the inlet pipe centerline and also 20 in. below the exit pipe centerline. Determine: The velocity at the exit and the flow rate. The x and y components of the reaction force at O. The reaction moment at O. The shaft power of the machine. Is it a pump or a turbine?Explanation / Answer
a) A1v1 = A2v2
/4*d12v1 = /4*d22v2
12*12 = 22*v2
Velocity at exit, v2 = 3 ft/s
Flowrate Q = A1v1 = /4*12*12 = 9.42 ft^3/s
b) X-component of reaction = [P2 + 1/2*(/g)V22]A2 = [5*12^2 + 1/2*(62.4/32.2)*3^2]*(/4*2^2) = 2288 lb
Y-component of reaction = [P1 + 1/2*(/g)V12]A1 = [10*12^2 + 1/2*(62.4/32.2)*12^2]*(/4*1^2) = 1240 lb
c) Reaction moment at O = 2288*(20/12) + 1240*(18/12) = 5673 lb-ft
d) [P2 + 1/2*(/g)V22] - [P1 + 1/2*(/g)V12] = [5*12^2 + 1/2*(62.4/32.2)*3^2] - [10*12^2 + 1/2*(62.4/32.2)*12^2]
= -851 lb/ft^2 (Negative sign indicates turbine.)
Since the outlet energy is lower than inlet, the device is a turbine.
Turbine shaft power = 851*9.42 = 8016 lb-ft/s = 8016/550 hp = 14.6 hp
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