23) the right limb of a simple u-tube manometer containing mercury is open to the atmosphere while the left limb is connected to a pipe in which a fluid of sp.Gr 0.9 is flowing .The center of the pipe is 12cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe if the difference of mercury level in the two limbs is 20cm.
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Given the right limb of a simple u-tube manometer containing mercury is open to the atmosphere while the left limb is connected to a pipe in which a fluid of sp.Gr 0.9 is flowing .The center of the pipe is 12cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe if the difference of mercury level in the two limbs is 20cm.
- According to question
- specific gravity S1 = 0.9
- Therefore density ρ1 = S1 x 1000
- = 0.9 x 1000
- = 900 kg / m^3
- Also specific gravity S2 = 13.6
- Therefore density ρ2 = 13.6 x 1000
- = 13600 kg/m^3
- Now h2 = 20 cm = 0.2 m
- Now difference of mercury level will be h1 = 20 – 12
- = 8 cm
- = 0.08 m
- Let P be the pressure of fluid in pipe.
- So we have P + ρ1gh1 = p2 + ρ2gh2
- Atmospheric pressure p2 will be 0
- So P + 900 x 9.8 x 0.08 = 13,600 x 9.8 x 0.2
- P + 705.6 = 26,656
- P = 26,656 – 705.6
- P = 25,950 N/m^2
P = 2.595 N / cm^2
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Equating the pressure above A – A:
Ppipe + Pleft limb = PRight limb
Ppipe + ρ1gh1 = ρ2gh2
Ppipe + 900 × 9.81 × 0.08 = 13600 × 9.81 × 0.20
Ppipe = 25976.88 N/m2 = 2.597 N/cm2
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