TGPSC HMWSSB Manager Fluid Mechanics & Hydraulics previous year questions with answers
Every Fluid Mechanics & Hydraulics question asked in the TGPSC HMWSSB Manager (Engineering) papers on this site, grouped by topic. Answers come from TSPSC's final key; each has a short explanation.
Practise Fluid Mechanics & Hydraulics onlineFluid Mechanics
The dimensions for Kinematic Viscosity is
- FL⁻²T
- ML⁻¹T⁻¹
- L²T⁻²
- L²T⁻¹
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Answer: (4) L²T⁻¹
Kinematic viscosity ν = μ/ρ, with units m²/s, so its dimensions are L²T⁻¹. FL⁻²T and ML⁻¹T⁻¹ are the dimensions of dynamic viscosity μ.
The stream function for a potential flow field is given by ψ=x²-y², the corresponding potential function (Φ), assuming zero potential at the origin is
- x²+y²
- 2xy
- x²-y²
- x-y
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Answer: (2) 2xy
From ψ = x² − y², u = ∂ψ/∂y = −2y and v = −∂ψ/∂x = −2x. With u = −∂Φ/∂x and v = −∂Φ/∂y this gives Φ = 2xy (zero at the origin); Φ and ψ are harmonic conjugates, and the opposite sign convention only changes the sign to −2xy.
Water flows through a large size pipe. The stagnation pressure and static pressures measured by a pitot tube are 0.3m and 0.24m of water. The velocity of flow in, m/minute is
- 1.08
- 65.00
- 10.8
- 0.65
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Answer: (2) 65.00
Dynamic head = 0.30 − 0.24 = 0.06 m, so V = √(2gh) = √(2 × 9.81 × 0.06) = 1.085 m/s. In m/minute, V = 1.085 × 60 ≈ 65 m/minute.
Poise is the CGS unit of
- Kinematic viscosity
- Dynamic viscosity
- Mass Density
- Weight Density
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Answer: (2) Dynamic viscosity
1 poise = 1 dyne-s/cm² = 0.1 N-s/m², a unit of dynamic viscosity; the CGS unit of kinematic viscosity is the stokes (cm²/s).
The velocity gradient is 1000/s. The viscosity is 1.2×10⁻⁴ N-s/m². The shear stress is
- 0.12N/m²
- 1.2×10⁻⁷ N/m²
- 12N/m²
- 12×10⁻⁵ N/m²
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Answer: (1) 0.12N/m²
By Newton's law of viscosity τ = μ(du/dy) = 1.2×10⁻⁴ × 1000 = 0.12 N/m².
If ψ = 3x²y - y³, the values of u and v are
- 6xy, 3x² - 3y²
- 3x² - 3y², 6xy
- (3x² - 3y²), -6xy
- 3y² - 3x², 6xy
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Answer: (3) (3x² - 3y²), -6xy
With the convention u = ∂ψ/∂y and v = −∂ψ/∂x: u = 3x² − 3y² and v = −6xy. Some Indian textbooks use the opposite sign convention (u = −∂ψ/∂y, v = ∂ψ/∂x), which gives option (4), but the official key follows the first.
In a three dimensional motion of a fluid, the component of rotation about the x-axis, ω_x is
- (1/2)(∂w/∂y - ∂v/∂z)
- (1/2)(∂u/∂z - ∂w/∂y)
- (1/2)(∂v/∂x - ∂u/∂y)
- (1/2)(∂v/∂x - ∂w/∂y)
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Answer: (1) (1/2)(∂w/∂y - ∂v/∂z)
Rotation about the x-axis involves only the velocities in the y-z plane: ω_x = ½(∂w/∂y − ∂v/∂z); option (3) is ω_z.
Existence of velocity potential implies that
- Fluid is in continuum
- Fluid is irrotational
- Fluid is ideal
- Fluid is compressible
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Answer: (2) Fluid is irrotational
If u = −∂φ/∂x, v = −∂φ/∂y, w = −∂φ/∂z, every rotation component such as ½(∂v/∂x − ∂u/∂y) becomes zero, so a velocity potential can exist only for irrotational flow.
Piezometric head is the summation of
- Velocity head and elevation head
- Velocity head and pressure head
- Pressure head and elevation head
- Total head
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Answer: (3) Pressure head and elevation head
Piezometric head = p/γ + z (pressure head + elevation head), the level to which water rises in a piezometer; adding the velocity head gives the total head.
The Bernoulli's equation is written with usual notation as p/γ + v²/2g + z = Constant. In this equation each of the terms represents
- Energy in kg.m/kg mass of fluid
- Energy in Nm/kg mass of fluid
- Energy in Nm/N weight of fluid
- Energy in KW/kg mass of fluid
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Answer: (3) Energy in Nm/N weight of fluid
In this form every term has the unit of length (m), i.e. energy per unit weight (N·m/N), which is why the terms are called heads.
Reynolds number of a flow is the ratio of
- Gravity forces to viscous forces
- Gravity forces to pressure forces
- Inertia forces to viscous forces
- Viscous forces to pressure forces
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Answer: (3) Inertia forces to viscous forces
Re = ρVD/μ = inertia force / viscous force; the ratio involving gravity (inertia to gravity) is the Froude number.
Ratio of the average velocity to maximum velocity for steady laminar flow in circular pipe is
- 1/2
- 2/3
- 3/2
- 2
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Answer: (1) 1/2
For laminar (Hagen-Poiseuille) flow the velocity profile is parabolic, u = u_max(1 − r²/R²), whose mean over the circular section is u_max/2.
Difference in elevation between TEL and HGL of pipe flow at a point is equal to
- Datum head
- Velocity head
- Pressure head
- Piezometric head
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Answer: (2) Velocity head
The HGL is at p/γ + z and the TEL at p/γ + z + V²/2g, so the TEL lies above the HGL by the velocity head V²/2g.
Boundary layer separation takes place when
- ∂p/∂x>0
- ∂v/∂y≤0
- ∂p/∂x>0 & ∂v/∂y≤0
- ∂p/∂x<0
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Answer: (3) ∂p/∂x>0 & ∂v/∂y≤0
Separation needs an adverse pressure gradient (∂p/∂x > 0) that slows the flow near the wall until the velocity gradient at the wall becomes zero or negative; a favourable gradient (∂p/∂x < 0) prevents separation, and an adverse gradient alone does not guarantee it.
Hydraulics & Hydraulic Machines
A rectangular channel has a width of 1.8m and carries a discharge of 1.8m³/sec at a depth of 0.20m. The specific energy is
- 1.03m
- 1.47m
- 1.87m
- 2.0m
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Answer: (2) 1.47m
Velocity V = Q/(by) = 1.8/(1.8 × 0.20) = 5 m/s, so velocity head = 5²/(2 × 9.81) = 1.274 m. Specific energy E = y + V²/2g = 0.20 + 1.274 = 1.47 m.
A 50mm diameter jet having a velocity of 25m/s strikes a flat stationary plate, the normal of which is inclined at 60° to the axis of the jet. The normal force exerted on the plate in Newtons is
- 460N
- 360 N
- 640 N
- 630 N
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Answer: (1) 460N
With the plate normal at 60° to the jet, F_n = ρaV² cos 60° = 1000 × (π/4)(0.05)² × 25² × 0.5 ≈ 614 N (reading 60° as the angle of the plate itself gives ρaV² sin 60° ≈ 1063 N instead). The official key gives 460 N, which neither standard working reproduces, so treat this key with caution. (Official final key; some sources differ.)
For a uniform flow with a depth of 0.6m and Froude number of 2.0 in a rectangular channel, the specific energy will be
- 2.4m
- 0.8m
- 2.6m
- 1.8m
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Answer: (4) 1.8m
Fr² = V²/(gy) = 4, so V²/2g = 2y = 1.2 m and E = y + V²/2g = 0.6 + 1.2 = 1.8 m.
A water turbine of expected efficiency 85% operates under 36m at a flow rate of 10m³/s. The unit weight of water is 10kN/m³. The nearest shaft power in MW is
- 3.06
- 3060
- 3600
- 850
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Answer: (1) 3.06
Shaft power = ηγQH = 0.85 × 10 kN/m³ × 10 m³/s × 36 m = 3060 kW = 3.06 MW (3060 is the answer in kW, not MW).