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attashe74 [19]
3 years ago
5

The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer as shown. determine the gage

pressure of air in the tank if h1 = 0.2 m, h2 = 0.3 m, and h3 = 0.46 m. take the densities of water, oil, and mercury to be 1000 kg/m3 , 850 kg/m3 , and 13,600 kg/m3 , respectively

Physics
2 answers:
Sphinxa [80]3 years ago
8 0

Answer : 56.9 kPa

Solution : Given :-

ρ (H_{2} O) = 1000 kg/m^{3} ; ρ (oil) = 850 kg/m^{3}; ρ (Hg) = 13,600 kg/m^{3}

h_{1} = 0.2 m; h_{2} = 0.3 m; h_{3} = 0.46 m

Formula : P_{1} + ρ (H_{2} O) g h_{1} + ρ (oil)g h_{2} - ρ (Hg) g h_{3} = P_{atm}

On rearranging we get,

P_{1} = P_{atm} - ρ (H_{2} O) g h_{1} - ρ (oil)g h_{2} + ρ (Hg) g h_{3}

Now,

P_{1} - P_{atm} = g (ρ (Hg) h_{3} - ρ (H_{2} O) h_{1} - ρ (oil)h_{2})

We know that, P_{1} - P_{atm} = P_{1}_gage

On substitution we get,

= (9.81 m/s^2) X [13,600 kg/m^3) X (0.46m) - (1000kg/m^3) X (0.2 m)- (850kg/m^3) X (0.3m) X X\frac{1N} {1kg . m/s^{2}} X \frac {1kPa} {1000. N/m^{2}}

= 56.9 kPa.

Therefore, P_{1}_gage = 56.9 kPa.

GenaCL600 [577]3 years ago
6 0
There is something wrong with the given, the second should be 3, 850 kg/m3. So here is the answer for this question:
The formula is P1 + Pwatergh1 + Poilgh2 - Pmercurygh3 = Patm
In order to solve for P1 we need to rearrange the formula, so it will look like:P1 = P atm - Pwatergh1 - Poilgh2 + Pmercurygh3
Noting that P1gage would be equal to P1 - Patm and we can substitute the given above:P1,gage = (9.81 m/s^2)[13,600 kg/m^3)(0.46m) - (1000kg/m^3)(0.2 m)
= (850kg/m^3)(0.3m) (1N / 1kg * m/s^2) (1kPa/ 1000 N/m^2)
= 56.9 kPa
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