Fluid Mechanics and Machines & Hydrology PYQ – TN TRB Lecturer
21. A smooth stationary flat plate is held at θ° to the direction of the jet. After striking the plate, the jet gets divided into two streams Q1 upwards and Q2 downwards. The ratio of Q1/Q2 is _____ .
[TN TRB 2017: 1 Mark]
- 1 – cos θ 1 + cos θ
- 1 + cos θ 1 – cos θ
- 1 – sin θ 1 + sin θ
- 1 + sin θ 1 – sin θ
Explanation:
From the derivation of jet striking a flat smooth stationary plate at an angle θ,
Q2
=
1 – cos θ
22. The highest water saving method of irrigation is _____ .
[TN TRB 2017: 1 Mark]
- Sprinkler
- Drip
- Sub-surface
- Basin
Explanation:
The wastage of water is minimum,
- In Sprinkler irrigation, the losses due to surface runoff and percolation are eliminated. Efficiency is 80%.
- In drip irrigation, the losses due to evaporation and seepage are eliminated. Efficiency is 90%.
23.  Pressure inside water droplet is _____ if the surface tension is ‘σ’ and diameter is ‘d’.
[TN TRB 2017: 1 Mark]
- 3σ/d
- 4σ/d
- 8σ/d
- 16σ/d
Explanation:
Let,
σ – Surface tension of water droplet
p – pressure inside the water droplet in addition to outside pressure
Tensile force due to surface tension = σ x πd
Pressure force inside the water droplet = p x
4
For equilibrium, σ x πd = p x
4
⇒ p = 4σ/d
24.  The dimensions of pressure is _____ .
[TN TRB 2017: 1 Mark]
- ML-1T-1
- ML-1T-2
- MLT-1
- L2T-1
Explanation:
As per Newton’s II law, Force = Mass x Acceleration
Pressure =
Area
=
Area
⇒ Pressure =
L2
⇒ Pressure = ML-1T-2
25.  For what value of the constant ‘a’, the vector field Vˆ = (axy – z3) iˆ + (a-2)x2 jˆ + (1-a)xz2 kˆ is irrotational ?
[TN TRB 2017: 1 Mark]
- 4
- 5
- -2
- 7
Explanation:
Let, u = axy – z3, v = (a – 2)x2, w = (1 – a) xz2
For an irrotational flow, ωx = ωy = ωz = 0
ωx = ½{∂w/∂y – ∂v/∂z}, ωy = ½{∂w/∂x – ∂u/∂z}, ωz = ½{∂v/∂x – ∂u/∂y}
Considering ωz = 0
½(∂v/∂x – ∂u/∂y) = 0
⇒
∂x
–
∂y
= 0
⇒
∂x
–
∂y
= 0
⇒ 2x(a – 2) – ax = 0
⇒ 2a – 4 – a = 0
⇒ a = 4
26.  The diameter of a pipe is 200 mm. The velocity of flow through the pipe is 4 m/s. The discharge through the pipe is _____ .
[TN TRB 2017: 1 Mark]
- 127.60 lps
- 126.60 lps
- 130.60 lps
- 125.60 lps
Explanation:
Discharge, Q = Area x Velocity
⇒ Q =
4
x V =
4
x 4
⇒ Q = 0.1256 m3/s (or) 125.6 lt/s
27.  A river has an average surface width of 20 m. If the evaporation measured in the velocity of the river is 0.5 mm/day, the volume of water evaporated in 60 km stretch of the river in a month of 30 days in m3 is _____ .
[TN TRB 2017: 1 Mark]
- 1800
- 180
- 18000
- 18
Explanation:
Evaporation measured per day = 0.5 mm
Evaporation for 30 days = 0.5 x 30 = 15 mm
Volume of water evaporated in 30 days = (15 x 10-3) x 20 x (60 x 103)
⇒ Volume of water evaporated in 30 days = 18000 m3
28.  If a two-dimensional velocity potential function is given by ɸ = 3xy, the discharge between the streamlines passing through (2,6) and (6,6) will be _____ .
[TN TRB 2017: 2 Marks]
- 14.68 units
- 18.97units
- 25.45 units
- 48 units
Explanation:
From the given velocity potential function, φ = 3xy,
u = –
∂x
= –
∂x
= – 3y
v = –
∂y
= –
∂y
= – 3x
The relation between the velocity components (u, v) and stream function (ψ) is
u = –
∂y
and v =
∂x
Consider, v =
∂x
⇒ – 3x =
∂x
Integrate with respect to ‘y’ on both sides,
–
2
+ C = ψ
At (2,6), ψ1 = – 1.5x2 + C = -(1.5 x 22) + C = -6 + C
At (6,6), ψ1 = -1.5x2 + C = -(1.5 x 62) + C = -54 + C
Discharge, Q = |ψ1 – ψ2| = |-54 + 6|
⇒ Q = 48 units
Note: Avoid proceeding u = –
∂y
because the points (2,6) and (6,6) varies along the x-coordinates.
29.  The equation for computing the velocity of flow by Manning’s formula is _____ .
[TN TRB 2017: 2 Marks]
- 1 nR1/2 S2/3
- 1 nR3/4 S1/2
- 1 nR2/3 S1/2
- 1 nR1/2 S3/4