The pressure gauges $G_{1} a n d G_{2}$ installed on the system show pressure of $P_{G 1}$ =5.00 bar and $P_{G 2}$ =1.00 bar. The value of unknown pressure P is [Atmospheric pressure = 1.01 bar]

1.01 bar

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2.01 bar

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5.00 bar

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7.01 bar

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Solution

The correct option is D 7.01 bar

$P_{G 1} = 5 b a r$
$P_{G 2} = 1 b a r$
$P_{a t m} = 1.01 b a r$
$F o r p r e s s u r e g a u g e G_{2},$
Absolute pressure
=Atmospheric pr+Gauge pr
$p_{a b s} = p_{a t m} + p_{G 2} = 1.01 + 1 = 2.01 b a r$
$= L o c a l a t m o s p h e r i c p r . f o r p r . g a u g e G_{1}$
$= (p_{a t m})_{1}$
$F o r p r e s s u r e g a u g e G_{1}$
Absolute pressure = Local atm. pressure+Gauge pressure
$P = (P_{a t m})_{1} + P_{G 1}$
= 2.01 + 5 = 7.01 bar

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