Question

Ethanol is in a closed tank and the level indicator has broken. Your boss requests that you determine the height (and therefore the volume) of the ethanol in the tank (the tank is partially full and there is a gas phase above the liquid). Connected to the very bottom of the tank is an open-ended manometer using mercury. On the tank side of the manometer tube, there is 500. mm of ethanol above the mercury. The height differential of the mercury is 300. mm. The pressure of the gas phase inside the tank is shown to be 0.800 psig from a gauge on top of the tank. Assume atmospheric pressure is 1.00 atm.

a) Draw the manometer and label all important terms.

b) Determine the depth of ethanol in the tank.

Answer 1

pressure as shown by pressure gauge= 0.8psig =14.7+0.8 psia =15.5 psia =(15.5/`14.7)*101325 N/m2 (1)

Pressure due to liquid head of 500mm =0.5m of ethanol (a density of 780 kg/m3 of ethanol is considereed= 780*500/1000*9.8 (2)

let the level of liquid in the tank be h meters

there fore static pressure =h*9.8*780 (3)

Hence total pressure on tank side = 15.6*101325+780*9.8(500/1000+h) ( Eq.1+Eq.2 + Eq.3) (4)

pressure on the other side of the tank= differential pressure + atmospheric pressure = (300/1000)*9.8*13600 +101325=39984 N/m2 +101325

under steady state Eq.4= Eq.5

1.05* 101325+780*9.8(h+0.5)= 39984+101325

780*9.8(h+0.5)= 39984-0.05*101325=34917.75

h+0.5= 4.57 and h= 4.07 m