SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 115 of 200

Marks: +1, -0

A U-tube manometer containing a liquid of specific gravity

{S}

measures gauge pressure of water flowing in a pipe.

{h}_{2}=

difference in level of manometric liquid in the two limbs

{h}_{1}=

height of water between point of connection of pressure tap and level of manometric liquid in the limb to which it is connected by a tube bending down.
The measured gauge pressure is:

${h}_{2} {m}$ of manometric liquid
${h}_{2} . {Sm}$ water
$({h}_{2} \.{S}-{h}_{1}) {m}$ of water
${h}_{1} {m}$ water

Validate

Solution:

Let

P_{A}

be the pressure in the pipe and

P_{0}

be the atmospheric pressure.

\ρ_{m}=

Density of mercury,

\ρ_{w}=

Density of water

{S}=

Specific gravity

=\ρ_{{m}} \/ \ρ_{{w}} \⇒\ρ_{{m}}={S} \ρ_{{w}}

Balancing the pressure about datum (dotted line is the datum),

{P}_{{A}}+\ρ_{{w}} {gh}_{1}={P}_{0}+\ρ_{{m}} {gh}_{2}

P_{A}-P_{0}=\ρ_{m} g h_{2}-\ρ_{w} g h_{1}

\Δ{P}

(Gauge pressure)

={g}(\ρ_{{m}} {h}_{2} \.-\ρ_{{u}} {h}_{1})

\Δ{P}={g} \ρ_{{W}}({Sh}_{2}-{h}_{1})

\Δ{P} \/(\ρ_{{w}} {g})=({h}_{2} \.{S}-{h}_{1}) {m}

of water

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