express atmospheric pressure in meter of water column in cm of mercury column. I

Question

express atmospheric pressure in meter of water column in cm of mercury column.

I need answers for questions from 1 to 7.

Express atmospheric pressure in meter of water column and in cm of mercury column. Determine the intensity of pressure in water at a depth of 5m. below the free surface of pgh = 1000 times 9.81times 5=49050 nim2 Specific gravity of sea water is 1.01. Determine the intensity of pressure in Pa and force in kN on a horizontal lamina held at a depth of 1 km from the free surface of sea water. The surface area of the lamina is 3m Times 2m.Convert intensity of pressure of 2MPa into equivalent pressure head of oil of specific gravity 0.8.The specific gravity of oil is 0.75. Determine the depth of a point within the oil at which the intensity of pressure is 0.1 N/mm2.What are the gauge pressure and absolute pressure at a point 3m below the free surface of a liquid having a density of 1500 kg/m3 if the atmospheric pressure is equivalent to 750 mm of mercury ? Density of water is 1000 kg /m3.The following figure shows a pipe line containing water. Determine the pressure head at I in metr

Explanation / Answer

1) atmospheric pressure = 76 cm of Hg = 10m of water

2) P = density*g*(h) + 101.325 kPa = 1000*9.8*(5) + 101.325kPa = 150.325 kPa

3) P = density*g*h = 1010*9.8*(1000) + 101.325kPa = 10 MPa

F = P(A) = 10 MPa * 3*2 = 60 MPa

4) 800*9.8*h = 2 MPa => h = 255.1 m

5) 750*9.8*h = 10^5 Pa => h = 13.6 m

6) Pg = P - Patm

P = 1500*9.8*3 + 750mm Hg = 144.06 kPa

Pg = 1500*9.8*3 = 44.1 kPa

7) h = 90 + 30/sqrt(2) + 20 = 131.21 m

P = 1000*9.8*131.21 + 0.101325 MPa = 1.386 MPa

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