Oil of specific gravity 0.93 flows steadily through the system shown. The oil-water manometer shows a deflection of 6 cm. Calculate the exit velocity neglecting all losses through the system.
Answer: V_exit= 0.718 m/s

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Answer 1

A steady flow of oil with a specific gravity of 0.93 is present. The exit velocity neglecting all losses through the system V=0.7179 m/s.

P1-P2/sg=h=-6*[ρm/ρoil-1]=-6[1/0.93-1]

h=-0.4516cm=-0.4516*10⁻²cm

V₂²-V₁²/2g=h=-0.4516*10⁻²

V₂-(A₂/A₁)².V²₂/2g=-0.4516*10⁻²

V₂²=0.0132=0.11486 m/s

A₂V₂=Aexit.Vexit

Vexit=A₂/Aexit*V₂

=π/4.5²/ π/4 2²*0.1148

Velocityexit=0.7179 m/s.

Exit losses are calculated as a coefficient multiplied by the change in velocity head from the downstream end of the culvert, just inside it, to the cross section immediately downstream of it. A pipe's head loss is also inversely correlated with its length. So it makes sense to keep the pipe work as straight and compact as possible. Friction loss must be taken into account during system design since it has an impact on the flow rate and fluid pressure inside the pipe system.

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