Fluid mechanics and hydraulics miscellaneous
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For a two dimensional flow field, the stream function y is given as ψ = 3 (y² − x²). 2
The magnitude of discharge occurring between the stream lines passing through points (0, 3) and (3, 4) is
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Q = ψ1 – ψ2
= 3 (9 − 0) − 3 (16 − 9) 2 2 = 27 − 24 + 7 = 3 units 2 2 Correct Option: B
Q = ψ1 – ψ2
= 3 (9 − 0) − 3 (16 − 9) 2 2 = 27 − 24 + 7 = 3 units 2 2
- A 2 km long pipe of 0.2 m diameter connects two reservoirs. The difference between water levels in the reservoirs is 8 m. The Darcy-Weisbach friction factor of the pipe is 0.04. Accounting for frictional, entry and exit losses, the velocity in the pipe (in m/s) is
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Frictional loss
hf = fLV² d × 2g
Loss of head at entrance,hL = 0.5V² 2g
Loss at exit of pipehc = V² 2g ∴ 8 = fLV² + 0.5V² + V² d × 2g 2g 2g ⇒ 8 = 0.04 × 2000 × V² + V² + 0.5V² 2 × g × 0.2 2g 2g
⇒ 8 = 20.3873 V² + 0.07645 V²
⇒ V = 0.625 m/s ≌ 0.63 m/sCorrect Option: A
Frictional loss
hf = fLV² d × 2g
Loss of head at entrance,hL = 0.5V² 2g
Loss at exit of pipehc = V² 2g ∴ 8 = fLV² + 0.5V² + V² d × 2g 2g 2g ⇒ 8 = 0.04 × 2000 × V² + V² + 0.5V² 2 × g × 0.2 2g 2g
⇒ 8 = 20.3873 V² + 0.07645 V²
⇒ V = 0.625 m/s ≌ 0.63 m/s
- The normal depth in a wide rectangular channel is increased by 10%. The percentage increase in the discharge in the channel is
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The discharge in the channel is given by
Q = 1 AR2/3S1/2 n ⇒ Q = 1 (By) 
By 
2/3 S1/2 n B + 2y
For a wide rectangular channel
B >> y∴ Q = 1 (By)(y)2/3S1/2 n
∴ Q ∝ y5/3
Let Q1 = k(y)5/3
∴ Q2 = k(1.1y)5/3∴ Q2 − Q1 × 100 Q1 = (1.1y)5/3 − (y)5/3 × 100 = 17.21% (y)5/3 Correct Option: D
The discharge in the channel is given by
Q = 1 AR2/3S1/2 n ⇒ Q = 1 (By) By 2/3 S1/2 n B + 2y
For a wide rectangular channel
B >> y∴ Q = 1 (By)(y)2/3S1/2 n
∴ Q ∝ y5/3
Let Q1 = k(y)5/3
∴ Q2 = k(1.1y)5/3∴ Q2 − Q1 × 100 Q1 = (1.1y)5/3 − (y)5/3 × 100 = 17.21% (y)5/3
- The velocity components of a two dimensional plane motion of a fluid are:
u = y³ + 2x − x²y and v = xy² − 2y² − x³ 3 3
The correct statement is:
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For incompressible flow,
∂u + ∂v = 0 ∂x ∂y u = y³ + 2x − x²y 3 v = xy² − 2y − x³ 3 ⇒ ∂ y³ + 2x − x²y + ∂ xy² − 2y − x³ ∂x 3 ∂x 3
= 2 − 2xy + 2xy – 2 = 0
∴ Flow is incompressible
For irrotational flow,1 ∂v − ∂u = 0 2 ∂x ∂y 1 
∂ xy² − 2y − x³ − ∂ y³ + 2x − x²y 
2 ∂x 3 ∂y 3 = 1 [y² − x² − y² + x²] = 0 2
∴ Flow is irrotationalCorrect Option: A
For incompressible flow,
∂u + ∂v = 0 ∂x ∂y u = y³ + 2x − x²y 3 v = xy² − 2y − x³ 3 ⇒ ∂ y³ + 2x − x²y + ∂ xy² − 2y − x³ ∂x 3 ∂x 3
= 2 − 2xy + 2xy – 2 = 0
∴ Flow is incompressible
For irrotational flow,1 ∂v − ∂u = 0 2 ∂x ∂y 1 ∂ xy² − 2y − x³ − ∂ y³ + 2x − x²y 2 ∂x 3 ∂y 3 = 1 [y² − x² − y² + x²] = 0 2
∴ Flow is irrotational
- A rectangular channel having a bed slope 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m³ /s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)
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The width of section should be such that specific energy should be critical at contracted section.
Ec = 3 q2 1/3 2 g 
0.7893 = 3 (1)2/3 2 g1/3(Bmin)2/3
Bmin = 0.836 mCorrect Option: B
The width of section should be such that specific energy should be critical at contracted section.
Ec = 3 q2 1/3 2 g 0.7893 = 3 (1)2/3 2 g1/3(Bmin)2/3
Bmin = 0.836 m
