The liquid in the open-tube manometer in the figure below is mercury, y1 = 2.40

Question

The liquid in the open-tube manometer in the figure below is mercury, y1 = 2.40 cm, and y2 = 6.75 cm. Atmospheric pressure is 989 millibars. Open-tube manometer Po= patm Pressure p FT Patm+P82 The pressure is the same at the bottoms of the two tubes (a) What is the absolute pressure at the bottom of the U-shaped tube? Pa (b) What is the absolute pressure in the open tube at a depth of 4.35 cm below the free surface? Pa (c) What is the absolute pressure of the gas in the tank? Pa (d) What is the gauge pressure of the gas in pascals? Pa

Explanation / Answer

a)

1 millibar =100 pascals

density of mercury p=13600 kg/m3

P=Po+pgh=(989*100)+13600*9.8*0.0675

P=1.08*105pa

b)

P=Po+pgh=(989*100)+13600*9.8*0.0435

P=1.05*105pa

c)

P=Po+pgh=(989*100)+13600*9.8*(0.0675-0.024)

P=1.05*105pa

d)

Gauge pressure

Pgauge =P-Po=1.05*105-989*100=5.8*103pa

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