Fluid mechanics and hydraulics miscellaneous Easy Questions and Answers | Page - 1

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

1. 0.95 m³ /s
1. 0.100 m³ / s
1. 0.105 m³ / s
1. 10.5 m³ / s

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Apply Froude’s law.
Density of fluid does not affect model study.
Q_r = Q_m = L_r^{5 / 2}
Q_p

Q_p = 10000 m³ / s
L_r = 1
100

⇒ Q_m = 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.
Q_r = Q_m = L_r^{5 / 2}
Q_p

Q_p = 10000 m³ / s
L_r = 1
100

⇒ Q_m = 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)

1. 24.7 kPa
1. 26.5 kPa
1. 26.7 kPa
1. 28.9 kPa

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Initial case,

P_A .P_o .gh_o = P_B + ρ_w g.h_w + ρ_m g.h_m
⇒ P_A + 900 × 9.81 × 3 = P_B × 1000 × 9.81 × 1.5 + 13550 × 9.81 × 0.2
⇒ P_A – P_B = 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

P_A ' + 900 × 9.81 × 2.9 = P_B ' + 1000 × 9.81 × 1.6
∴ P_B ' – P_A ' = 9.908 kPa
∴ P_A ' – P_B ' = –9.908 kPa
∴ Pressure to be increased = (P_A. P_B) – (P_A ' – P_B ')
= 14.81 – (– 9.908) = 24.73 kPa

Correct Option: A

Initial case,

P_A .P_o .gh_o = P_B + ρ_w g.h_w + ρ_m g.h_m
⇒ P_A + 900 × 9.81 × 3 = P_B × 1000 × 9.81 × 1.5 + 13550 × 9.81 × 0.2
⇒ P_A – P_B = 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

P_A ' + 900 × 9.81 × 2.9 = P_B ' + 1000 × 9.81 × 1.6
∴ P_B ' – P_A ' = 9.908 kPa
∴ P_A ' – P_B ' = –9.908 kPa
∴ Pressure to be increased = (P_A. P_B) – (P_A ' – P_B ')
= 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)

1. 0.57 m
1. 0.91 m
1. 1.02 m
1. 1.57 m

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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 m

Correct 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 ?

1. 15°
1. 30°
1. 45°
1. 60°

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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 (σ_c) 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

1. 9.80 m
1. 6.20 m
1. 5.50 m
1. none of the above

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σ_c = NPSH
H

⇒ 0.144 = NPSH
25

∴ NPSH = 3.6 m
NPSH = H_atm - h_v - h_s - h₂
P_atm = H_atm = 9.8 m of water
P_v = H_v = 0.3 m of water
h₂ = 0.4
⇒ 3.6 = 9.8 – 0.3 – h_s – 0.4
⇒ h_s = 5.5 m

Correct Option: C

σ_c = NPSH
H

⇒ 0.144 = NPSH
25

∴ NPSH = 3.6 m
NPSH = H_atm - h_v - h_s - h₂
P_atm = H_atm = 9.8 m of water
P_v = H_v = 0.3 m of water
h₂ = 0.4
⇒ 3.6 = 9.8 – 0.3 – h_s – 0.4
⇒ h_s = 5.5 m