SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 160 of 200

Marks: +1, -0

Manometric, total static and friction-loss heads for a centrifugal pump are 30, 19 and 10.5 m respectively. The velocity of discharge of water is:

Solution:

{H}_{{M}}={h}_{{s}}+{h}_{{d}}+{h}_{{fs}}+{h}_{{fd}}+{{V}_{{d}}^{2}}/{2 {g}}

={H}+{H}_{{f}}+{{V}_{{d}}^{2}}/{2 {g}} \

⇒{{V}_{{d}}^{2}}/{2 {g}}={H}_{{m}}-{H}-{H}_{{f}}

{{V}_{{d}}^{2}}/{2 {g}}={H}_{{m}}-{H}-{H}_{{f}}

=30-19-10.5=0.5

{V}_{{d}}=(2 \× {g} \×0.5)^{0.5} {m} \/ {s}

{V}_{{d}}=({g})^{0.5} {m} \/ {s}

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