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Centrifugal pump, Manometric efficiency, Velocity triangle, Vane angle, Discharge, Head, Impeller dimensions, Outlet velocity, Relative velocity
To find the vane angle at the outer periphery (outlet) of the impeller, we need to use the velocity triangle at the outlet. The manometric efficiency relates the actual head developed to the theoretical head imparted by the impeller. We'll use the given discharge, speed, head, impeller diameter, width, and efficiency to find the necessary velocities and then calculate the vane angle.
Given Data:
Calculate the velocity of flow at outlet (V f 2):
Q=π D 2B 2V f 2
So,
V f 2=π D 2B 2Q
Substitute values:
V f 2=π×0.25×0.05 0.118
V f 2=0.03927 0.118
V f 2≈3.006 m/s
Calculate the tangential velocity at outlet (u 2):
u 2=π D 2N/60
u 2=π×0.25×1450/60
u 2=3.1416×0.25×24.1667
u 2=0.7854×24.1667
u 2≈18.98 m/s
Manometric efficiency equation:
η man o=u 2V w 2g H m
Where V w 2 is the whirl velocity at outlet.
Rearrange to find V w 2:
V w 2=η man ou 2g H m
Substitute values:
g=9.81 m/s 2
H m=25 m
η man o=0.75
u 2=18.98 m/s
V w 2=0.75×18.98 9.81×25
V w 2=14.235 245.25
V w 2≈17.23 m/s
Draw velocity triangle at outlet and calculate vane angle (β 2):
At outlet:
The vane angle β 2 at outlet is given by:
tan β 2=u 2−V w 2V f 2
Substitute values:
tan β 2=18.98−17.23 3.006
tan β 2=1.75 3.006
tan β 2≈1.718
β 2=arctan(1.718)
β 2≈60.5∘
The vane angle at the outer periphery of the impeller is:
60.5∘
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