(a) if the pressure in the tank is 1.5 bar and atmospheric pressure is 1 bar, determine L, in m, for water with density of 997 kg/m^3 as the manometer liquid.
Let g = 9.81 m/s^2.
(b) determine L , in cm, if the manometer liquiud is mercury with a density of 13.59 g/cm^3 and the gas pressure is 1.3 bar. A barometer indicates the local atmospheric pressure is 750 mmHg. Let g = 9.81 m/s^2.

a) L= 511 cm

b) L= 22 cm

Explanation:

since when the manometer is in equilibrium , the pressure at the bottom of an U-tube manometer should be the same taking into account the arm that is in contact with the tank and the one open to the atmosphere. Thus

Pressure at the bottom on the side of the tank = Ptank + ρ*g*y₁

Pressure at the bottom on the open side = Patm + ρ*g*y₂

where P= pressure , ρ= density and y denotes height with respect to the bottom. Denoting L as the difference between the liquid heights

L=y₂-y₁

and

Ptank + ρ*g*y₁= Patm + ρ*g*y₂

Ptank - Patm = ρ*g*(y₂- y₁) = ρ*g*L

L= (Ptank - Patm)/(ρ*g)

thus

for a)

L= (Ptank - Patm)/(ρ*g) = (1.5 bar - 1 bar)/(997 kg/m³* 9.81 m/s²) * 10⁵ Pa/Bar = 5.11 m = 511 cm

L= 511 cm

for b)

L= (Ptank - Patm)/(ρ*g) = (1.3 bar * 10⁵ Pa/Bar - 750 mm Hg* 133.32 Pa/mm Hg)/(13.59*1000 kg/m³* 9.81 m/s²) = 0.22 m = 22 cm

L= 22 cm

answer:

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explanation: