Both a gage and a manometer are attached to a gas tank to measure its pressure (

Question

Both a gage and a manometer are attached to a gas tank to measure its pressure (see figure). If the reading of the pressure gage is 80 kPa, determine the distance between the two fluid levels of the manometer if the fluid is: (a) mercury (rho = 13, 600 kg/m^3), and (b) water (rho = 1,000 kg/m^3). Assume g = 9.81m/s^2. The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer as shown in the figure. Determine the gage pressure of air in the tank if h_1 = 0.2 m, h_2 = 0.3 m, and h_3 = 0.4 m. The densities of water, oil, and mercury are 1,000 kg/m^3, 850 kg/m^3, and 13, 600 kg/m^3, respectively. Assume g = 9.81m/s^2.

Explanation / Answer

3. Pressure gage Pg= 80Kpa

mercury density=13600kg/m^3

water density=1000kg/m^3

now we consider P= Patmosphere then we can use expression Pg-density*g*height= P

now Pg-p =Pgage

density*g*height = 80Kpa

then for mercury height= 80*10^3/13600*9.81 = .59m a)

for water height =80*10^3/1000*9.81 = 8.15m b)

4. we know that gage pressure is the pressure difference between absolute aand atmospheric pressure

combining the pressure equation for multipoints

we get Pgage=P1-P2 = densityhg(g/gc)h3 -densityw(g/gc)h1- densityoil(g/gc)h2

therefore using values

Pgage= (13600*.4 -1000*.2- 850* .3)9.81Kpa [gc =1kg-m/NS^2 ]

or Pgage =48902Pa =48.9Kpa

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