GATE Mechanical: Fluid Mechanics, previous-year questions solved

1. The whole rod is under water, so its buoyancy (ρ_wgV_r) beats its weight (0.6ρ_wgV_r). The net upward force 0.4ρ_wgV_r acts at the rod's centre, L/2 from A.
2. The ball's net downward force is (11.4 − 1)ρ_wgV_b = 10.4ρ_wgV_b, acting at L from A (ball radius ≪ L).
3. Moments about the hinge A (every force is vertical, so the angle cancels): 0.4V_r(L/2) = 10.4V_bL, so V_b = 0.2V_r/10.4.
4. V_r = 3.14 × 0.02² × 8 = 0.010048 m³, so V_b = 1.932 × 10⁻⁴ m³.
5. r = (3V_b/4π)^1/3 = 3.59 cm.

1. The oil pushes on the piston only along the cylinder axis. Resolve the 5000 N weight: the component along the axis is W sin 60° = 4330 N. The perpendicular component, W cos 60°, is taken as a side load by the rod and cylinder.
2. Piston area A = π × 0.15² = 0.07069 m².
3. p = 4330/0.07069 = 61.26 kPa (gauge).

Using the full weight gives 70.7 kPa, close to option (D), which is the trap.

Bernoulli from the free surface (atmospheric, nearly at rest) to the jet (atmospheric): ρgh = ½ρV². Density cancels, so V = √(2gh) for any liquid. This is Torricelli's result.

With losses neglected, viscosity plays no part either, so V₁ = V₂.

1. The pitot tip brings the air to rest, so it senses stagnation pressure p₀ = p + ½ρ_airV². The wall tap senses static pressure p.
2. Manometer: p₀ − p = (ρ_Hg − ρ_air)gh = (13600 − 1.23) × 10 × 0.030 = 4079.6 Pa.
3. V = √(2 × 4079.6/1.23) = 81.4 m/s.

Dropping ρ_air from the manometer term changes the answer by less than 0.01 m/s.

1. Take a streamline from far upstream at the inlet depth (pressure p_atm + ρgh₁, speed V) to the tube exit (pressure p_atm, height h₁ + h₂ above the inlet, speed u).
2. Bernoulli: p_atm + ρgh₁ + ½ρV² = p_atm + ½ρu² + ρg(h₁ + h₂).
3. The h₁ terms cancel: u = √(V² − 2gh₂), so Q = (π/4)D²√(V² − 2gh₂).

The inlet depth drops out because the hydrostatic pressure at depth h₁ exactly pays for lifting the liquid back up to the free surface. If V² < 2gh₂, nothing flows and the liquid stands in the tube: that is a pitot tube.

Integrating Euler's equation along a streamline gives p/ρ + V²/2 + gz = constant on that streamline, for steady, inviscid, incompressible flow, rotational or not. That covers (C) and (D).

The constant is the same on every streamline only if the flow is irrotational, which covers (B). In a rotational flow the Bernoulli constant changes from one streamline to the next, so it cannot be applied between two arbitrary points: option (A).

Fluid going round the bend follows a curved path, so it needs a net force towards the centre of curvature. Only a pressure gradient can supply it: ∂p/∂r = ρV²/r, with r measured outward. Pressure therefore rises from the inner wall to the outer wall: P_o > P_i.

The difference scales roughly as ρV²(D/R): it grows with the square of velocity and falls as the bend radius R grows, so (C) and (D) are both wrong.

1. A probe that moves with the fluid sees the material (substantial) derivative: DT/Dt = ∂T/∂t + u ∂T/∂x + v ∂T/∂y.
2. At (1, 1, 1): u = 2x = 2, v = y + 3t = 4; ∂T/∂t = 4, ∂T/∂x = 4x + y = 5, ∂T/∂y = x = 1.
3. DT/Dt = 4 + 2 × 5 + 4 × 1 = 18.

Option (A), 4, is only the local term ∂T/∂t: what a probe fixed in space would record.

1. u = ∂ψ/∂y = kx³, v = −∂ψ/∂x = −3kx²y. At (1, 1): u = 1, v = -3 m/s.
2. Steady flow, so only convective terms: a_x = u ∂u/∂x + v ∂u/∂y = x³(3x²) + 0 = 3.
3. a_y = u ∂v/∂x + v ∂v/∂y = x³(−6xy) + (−3x²y)(−3x²) = 3x⁴y = 3.
4. |a| = √(3² + 3²) = √18 = 4.24 m/s².

• (A) False: both u and v are non-zero and depend on x and y, so the field is two-dimensional.
• (B) True: ∂u/∂x + ∂v/∂y = 4y − 4y = 0.
• (C) True: ∂v/∂x − ∂u/∂y = 4x − 4x = 0.
• (D) True: the flow is steady and u = v = 0 at the origin, so the convective acceleration u∂V/∂x + v∂V/∂y is zero there. The origin is a stagnation point.

1. Euler's equation without body forces: ∇p = −ρ(V·∇)V.
2. At (1, 1): u = y² − x² = 0, v = 2xy = 2.
3. a_x = u(−2x) + v(2y) = 4; a_y = u(2y) + v(2x) = 4.
4. ∇p = −1.5 × (4î + 4ĵ) = −6î − 6ĵ.

(D) is the acceleration with the sign flipped but without the density.

1. Continuity: ∂v/∂y = −∂u/∂x = −2 cosh x.
2. Integrate in y: v = −2y cosh x + f(x). The condition v(x, 0) = cosh x gives f(x) = cosh x, so v = cosh x (1 − 2y).
3. v(0, −1) = 1 × (1 + 2) = 3.

1. Irrotational means ∇ × V = 0. With u = 3z, v = 0, w = Cx, the x and z components of the curl are zero automatically.
2. The y component: ∂u/∂z − ∂w/∂x = 3 − C = 0.
3. C = 3.0.

1. A = π/4 m²; ṁ = ρVA = 1 × 100 × π/4 = 25π kg/s (the same at both sections).
2. Momentum on the gas between the sections: (p₁ − p₂)A + F_{pipe on gas} = ṁ(V₂ − V₁).
3. F_{pipe on gas} = 25π × 70 − 10 000 × π/4 = 1750π − 2500π = −750π N.
4. The gas pushes on the pipe with the opposite force: 750π N in the flow direction (wall friction drags the pipe along).

1. Q = 4.5/60 = 0.075 m³/s. Exit radius r = 0.15 + 0.4 × 0.5² = 0.25 m.
2. A₁ = 0.07069 m², A₂ = 0.19635 m²; V₁ = 1.0610 m/s, V₂ = 0.3820 m/s.
3. Bernoulli with p₂ = 0 (gauge): p₁ = ½ × 1000 × (V₂² − V₁²) = -489.9 Pa. The inlet is below atmospheric.
4. Momentum on the water in the diffuser: p₁A₁ + F = ρQ(V₂ − V₁), so F = 75 × (0.3820 − 1.0610) − (-489.9)(0.07069) = -16.30 N.
5. The magnitude carried through the flange is 16.30 N.

Only the radius at the two ends matters; the x² profile in between does not change the answer for frictionless flow.

1. With c_D = 24μ/(ρVd), the drag simplifies to D = 3πμVd (Stokes' law). The liquid density cancels.
2. Constant velocity: weight = drag, ρ_sg(πd³/6) = 3πμVd.
3. V = ρ_sgd²/(18μ) = 11000 × 10 × (10⁻⁴)²/(18 × 1.1 × 10⁻³) = 1/18 m/s ≈ 0.0556 m/s.

1. II is true: N, E and S are at the same pressure and all three pipes start at J, so p_J − p_N = p_J − p_E = p_J − p_S.
2. Darcy-Weisbach with the same f and D: Δp = f(L/D)(ρV²/2) ∝ LQ². Equal Δp gives L Q_N² = 10L Q_E², so Q_E = Q_N/√10.
3. Continuity at J: Q = 2Q_N + Q_E = Q_E(2√10 + 1), so Q_E = Q/7.32, not Q/21. I is false.

Q/21 is what you would get if Δp were proportional to Q (laminar, Hagen-Poiseuille flow, where f itself depends on Q). With a fixed friction factor the drop goes as Q², and that is the trap.

• (A) True: the laminar parabola has V_avg = 0.5V_max. The turbulent profile is much flatter; the 1/7th power law gives V_avg ≈ 0.82V_max.
• (B) False: it's the other way round. Below Mach 0.3, density changes by under about 5%, and the flow is treated as incompressible.
• (C) True: laminar f = 64/Re, with no roughness term. The roughness bumps sit inside the viscous flow and don't trip it.
• (D) False: f = 64/Re increases as Re falls.

1. Dimensions: γ = M T⁻², R = L, μ = M L⁻¹ T⁻¹, √t = T^1/2.
2. Match powers on both sides. M: a + c = 0. L: b − c = 1. T: −2a − c + ½ = 0.
3. Substituting c = −a into the T equation: −a + ½ = 0, so a = 0.5, with c = -0.5 and b = 0.5.

That is the Lucas-Washburn law, L ∝ √(γRt/μ).

1. Froude similarity: V/√(gL) is the same for model and prototype, with g the same in both.
2. V_p/V_m = √(L_p/L_m) = √100 = 10.
3. V_p = 10 m/s.

• A purely convergent nozzle can accelerate a gas only up to Mach 1, reached at the exit (the throat). Raising P₀ can't push the exit Mach number past 1, so (B) is true and (D) is false.
• Choked mass flow is ṁ = A*P₀√(γ/RT₀)(2/(γ+1))^{(γ+1)/(2(γ−1))}. It is fixed only for fixed inlet conditions and grows in proportion to P₀, so (C) is true and (A) is false.

"Choked" means lowering the back pressure no longer increases the flow. Raising the inlet pressure still does: the gas at the throat is simply denser.

In isentropic supersonic flow the area-velocity relation dA/A = (M² − 1) dV/V says a growing area speeds the flow up, so pressure should fall from A to B. A pressure rise means the flow left the isentropic path: a normal shock between A and B jumps the flow to subsonic, raising the pressure, and the subsonic flow then decelerates further in the diverging section.

(A) treats the gas as incompressible: in a compressible flow the mass flow rate is constant, not the volume flow rate. The official key accepts only (B).

Across a normal shock the flow goes from supersonic (M_E > 1) to subsonic (M_F < 1) in a very thin, irreversible layer.

• Static pressure, density and temperature all jump up: (A) and (B) true.
• Mach number drops: (C) false.
• The shock is adiabatic but irreversible, so entropy rises (and stagnation pressure falls): (D) true.

1. The flow is pressure driven, so pressure falls along the pipe.
2. Isothermal ideal gas: ρ = p/RT, so density falls in step with pressure.
3. Steady flow in a uniform pipe: ρV is constant, so V = (ρV)/ρ increases along the flow.

1. With no friction in the bucket (equal relative velocities), the power per unit mass flow is proportional to u(V − u)(1 − cos β), where u is the bucket speed and V the jet speed.
2. u(V − u) is largest at u = V/2 = 62.83 m/s.
3. N = 60u/(πD) = 60 × 62.83/(π × 1) = 1200.0 rpm.

1. Blade speed at inlet: u₁ = πDN/60 = π × 1 × 600/60 = 31.416 m/s.
2. The guide blade angle α = 15° sets the direction of the absolute velocity: V_w1 = V_f1/tan α = 10/tan 15° = 37.321 m/s.
3. The inlet velocity triangle gives the runner vane angle: tan β = V_f1/(V_w1 − u₁) = 10/5.905.
4. β = 59.4°.