Question

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

Answer 1

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

Answer 2

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.