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centnifugal water pump discharges a flow rate of 1 6ms atrotational speed of 420pm. The impeler outer disete is 0 6 m and the inner diaeter is three to the tangent line with an absolute tangential velocity of t23ms and leaves it with the absolute sangrential velocity of 7 8s The radial now velacity is ensferem deoughous the punp c) Find the ideal head imparted to the fQuid (in m) 30 6 261 DOLL
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Answer #1

Given data

Velocity traingle →

eall 2. 久

Speed (N) 420 rpm

Outer diameter (D) = 0.6 m

0.6 Inside diameter (D ) 0.2 m

Inlet blade velocity (u) = 60

π* 0.2 * 420 Inlet blade velocity (ui 4.39 m/s

Outlet blade velocity (u2) 60

Outlet blade velocity (u2) = π * 0.6 * 4 13.19 m/s 60

extup{Absolute tangential Velocity at inlet}left ( V _{w1} ight )=2.3 extup{ m/s}

extup{Absolute tangential Velocity at Outlet}left ( V _{w2} ight )=7.8 extup{ m/s}

alpha _{1}=60^{o}

extup{a) angle of vane inlet}left ( eta _{1} ight )

extup{From inlet Velocity traingle} ightarrow

tanalpha _{1}=rac{V_{f1}}{V_{w1}}

tan60=rac{V_{f1}}{2.3}

Vf1 3,98 m/s

extup{angle of vane inlet}left ( eta _{1} ight )=taneta _{1}=rac{V_{f1}}{u_{1}-V_{w1}}

taneta _{1}=rac{3.983}{4.39-2.3}

eta _{1}=62.31^{0}

extup{b) angle of vane Outlet}left ( eta _{2} ight )

extup{From Outlet Velocity traingle} ightarrow

V_{f1}=V_{f2}=3.983 extup{ m/s Given}

extup{angle of vane Outlet}left ( eta _{2} ight )=taneta _{2}=rac{V_{f2}}{u_{2}-V_{w2}}

taneta _{2}=rac{3.983}{13.19-7.8}

eta _{2}=36.46^{0}

extup{c) Total Head(H)}

extup{Total Head(H)}=rac{V_{w1}*u_{1}}{g}+rac{V_{w2}*u_{2}}{g}

extup{Total Head(H)}=rac{2.3*4.39}{9.81}+rac{7.8*13.19}{9.81}

extup{Total Head(H)}=11.51 extup{ m}

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