Fluid mechanics and hydraulics miscellaneous
- A laboratory model of a river is built to a geometric scale of 1 : 00. The fluid used in the model is oil of mass density 900 kg/m3. The heighest flood in the river is 10,000 m3 / s. The corresponding discharge in the model shall be
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Apply Froude’s law.
Density of fluid does not affect model study.Qr = Qm = Lr5 / 2 Qp
Qp = 10000 m3 / sLr = 1 100 ⇒ Qm = 10000 × 
1 
5 / 2 100 = 10000 × 1 5 = 0.1 m3 / s 10 Correct Option: B
Apply Froude’s law.
Density of fluid does not affect model study.Qr = Qm = Lr5 / 2 Qp
Qp = 10000 m3 / sLr = 1 100 ⇒ Qm = 10000 × 1 5 / 2 100 = 10000 × 1 5 = 0.1 m3 / s 10
- Two pipelines, one carrying oil (mass density = 900 kg/m3) and the other water, are connected to a manometer as shown in the figure, By what amount the pressure in the water pipe should be increased so that the mercury levels in both the limbs of the manometer become equal? (Mass density of mercury = 13,550 kg/m3 and g = 9.81 m/s2)
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Initial case,
PA .Po .gho = PB + ρw g.hw + ρm g.hm
⇒ PA + 900 × 9.81 × 3 = PB × 1000 × 9.81 × 1.5 + 13550 × 9.81 × 0.2
⇒ PA – PB = 14.81 × 103 N/m2 = 14.81 kPa
To make mercury level equal, lower mercury by 0.1 m at side B
So that at side A it rises by 0.1 m and becomes equal.
Resultant figure
PA ' + 900 × 9.81 × 2.9 = PB ' + 1000 × 9.81 × 1.6
∴ PB ' – PA ' = 9.908 kPa
∴ PA ' – PB ' = –9.908 kPa
∴ Pressure to be increased = (PA. PB) – (PA ' – PB ')
= 14.81 – (– 9.908) = 24.73 kPaCorrect Option: A
Initial case,
PA .Po .gho = PB + ρw g.hw + ρm g.hm
⇒ PA + 900 × 9.81 × 3 = PB × 1000 × 9.81 × 1.5 + 13550 × 9.81 × 0.2
⇒ PA – PB = 14.81 × 103 N/m2 = 14.81 kPa
To make mercury level equal, lower mercury by 0.1 m at side B
So that at side A it rises by 0.1 m and becomes equal.
Resultant figure
PA ' + 900 × 9.81 × 2.9 = PB ' + 1000 × 9.81 × 1.6
∴ PB ' – PA ' = 9.908 kPa
∴ PA ' – PB ' = –9.908 kPa
∴ Pressure to be increased = (PA. PB) – (PA ' – PB ')
= 14.81 – (– 9.908) = 24.73 kPa
- A hydraulic jump takes place in a triangular channel of vertex angle 90°, as shown in figure. The discharge is 1m3 /s and the pre-jump depth is 0.5m. What will be the post-jump? (Take g = 9.81 m/s2)
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Taking specific force at 2 sections:
z = y 3 ⇒ 12 + y12 × y1 = 12 + y22 × y2 9.81 × y12 3 9.81 × y12 3
y1 = 0.5 m⇒ 0.449 = 1 + y22 9.81y23 3
⇒ y2 = 1.02 mCorrect Option: C
Taking specific force at 2 sections:
z = y 3 ⇒ 12 + y12 × y1 = 12 + y22 × y2 9.81 × y12 3 9.81 × y12 3
y1 = 0.5 m⇒ 0.449 = 1 + y22 9.81y23 3
⇒ y2 = 1.02 m
- A horizontal jet strikes a frictionless vertical plate (the plan view is shown in the figure). It is then divided into two parts, as shown in the figure. If the impact loss can be neglected, what is the value of θ ?
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Q2 = 1 + cosα Q1 1 - cosα 
α-angle between plate and jet⇒ 0.75 = 1 + cosα 0.25 1 - cosα ⇒ 3 = 1 + cosα 1 - cosα
3 - 3cosα = 1 + cosα
2 = 4cosα⇒ cosα = 1 2
∴ α = 60°
φ = 90 - α = 30°Correct Option: B
Q2 = 1 + cosα Q1 1 - cosα
α-angle between plate and jet⇒ 0.75 = 1 + cosα 0.25 1 - cosα ⇒ 3 = 1 + cosα 1 - cosα
3 - 3cosα = 1 + cosα
2 = 4cosα⇒ cosα = 1 2
∴ α = 60°
φ = 90 - α = 30°
- A pump can lift water at a discharge of 0.15 m3 /s to a head of 25 m. The critical cavitation number (σc) for the pump is found to be 0.144. The pump is to be installed at a location where the barometric pressure is 9.8 m of water and the vapour pressure of water is 0.30 of water. The intake pipe friction loss is 0.40 m. Using the minimum value of NPSH (Net Positive Suction Head), the maximum allowable elevation above the sump water surface at which the pump can be located is
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σc = NPSH H ⇒ 0.144 = NPSH 25
∴ NPSH = 3.6 m
NPSH = Hatm - hv - hs - h2
Patm = Hatm = 9.8 m of water
Pv = Hv = 0.3 m of water
h2 = 0.4
⇒ 3.6 = 9.8 – 0.3 – hs – 0.4
⇒ hs = 5.5 mCorrect Option: C
σc = NPSH H ⇒ 0.144 = NPSH 25
∴ NPSH = 3.6 m
NPSH = Hatm - hv - hs - h2
Patm = Hatm = 9.8 m of water
Pv = Hv = 0.3 m of water
h2 = 0.4
⇒ 3.6 = 9.8 – 0.3 – hs – 0.4
⇒ hs = 5.5 m
