Fluid Mechanics Test 2

Concept:

Force exerted by the jet on a moving plate is given by

F = ρ A (v - u)²

where ρ is the density of flowing fluid, A is the cross-sectional area of the jet, v is the velocity of the jet and u is the velocity of the moving plate.

Calculation:

Given:

ρ = 1000 kg/m³, v = 20 m/s, u = 10 m/s, A = 0.02 m²

F = ρ A (v - u)²

F = 1000 × 0.02 × (20 - 10)² = 2000 N

Concept:

\(V = \sqrt {2g{h_a}} \)

Calculation:

Given:

ρ_a = 1.2 kg/m³, deflection of the manometer = 24 mm

Now,

ρ_ah_a = ρ_wh_w

1.2 × h_a = 1000 × 24 × 10^-3

h_a = 20

\(= \sqrt {2 \times 10 \times 20} \)

Concept:

Given:

h_L = constant in both the pipes

\({h_l} = \frac{{fL{Q^2}}}{{12{D^5}}}\)      ---(1)

Calculation:

Given:

Pipe-1:

L₁ = 100 m, D₁ = 200 mm, f₁ = 0.015

Pipe – 2

D₂ = 400 mm, f₂ = 0.012, L₂ =?

Q = Constant in both the pipes

Now,

From   (1),

\({h_l} = \frac{{{f_1}{L_1}{Q^2}}}{{12D_1^5}}\ =\frac{{{f_2}{L_2}{Q^2}}}{{12D_2^5}}\)

\(\frac{{0.015\: \times \:100 \:\times\: {Q^2}}}{{12\: \times \:{{0.2}^5}}} = \frac{{0.012\: \times \:{L_2} \:\times\: {Q^2}}}{{12 \:\times \:{{0.4}^5}}}\)

∴ L₂ = 4000 m

Now,

We know

\(f = \frac{{64}}{{Re}}\)

For pipe-1 (D = 200 mm, f = .015)

\(0.015 = \frac{{64}}{{Re}}\)

Re = 4266 (>2300), so it is turbulent flow

For pipe-2 (D = 400 mm, f = .012)

\(0.012 = \frac{{64}}{{Re}}\)

• The flat plate is moving downward, creating a shear flow in the fluid.
• The fluid velocity profile is linear due to the narrow gap (0.02 m) and constant shear rate.
• At the wall, the fluid has zero velocity (no-slip condition).
• The maximum velocity occurs at the moving plate, which is 0.02 m/s.
• The average velocity is the midpoint of the linear velocity profile: (0 + 0.02 m/s) / 2 = 0.01 m/s

Concept:

Q = nAV

Q = Discharge, nA = Effective area

V = Velocity of the jet = \({C_v}\sqrt {2gH} \)

Calculation:

Given:

n = 5 × 2 × 4 = 40, H = Head developed = 250 m

C_v = 0.985, Q = 40 m³/s

\(V = {C_v}\sqrt {2gH} = 0.985 \times \sqrt {2 \times 9.81 \times 250} = 68.985\;m/s\)

Now,

Q = nAV

\(Q = n\frac{{\pi {D^2}}}{4}V\)

\(40 = 40 \times \frac{{\pi {D^2}}}{4} \times 68.985\)

• The lubricant flow is a classic example of viscous shear in a fluid between two surfaces.
• The shear stress, τ, is given by τ = η * (v/d), where η is the dynamic viscosity, v is the velocity, and d is the gap.
• Here, η = 0.05 Pa·s, v = 4 m/s, and d = 0.001 m.
• Calculate τ: τ = 0.05 * (4 / 0.001) = 200 Pa.
• Frictional force per unit length, F = τ * A, where A is the area per unit length (1 m²).
• Thus, F = 200 Pa * 1 m² = 200 N/m.

Therefore, the force per unit length is 1.0 N, which corresponds to option B.

• To find the force per unit area, use the formula for shear stress: τ = η * (du/dy).
• Here, η = 0.5 Pa-s (dynamic viscosity), du = 2 m/s (velocity of the top plate), dy = 0.1 m (gap between plates).
• Calculate τ: τ = 0.5 * (2 / 0.1) = 0.5 * 20 = 10 Pa.

Thus, the correct answer is 10 Pa.

The separation point S is determined from the condition

\(\rm{{\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = o}} = 0}\)

So, If

\(\rm{{\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = 0}} < 0 \Rightarrow flow\;has\;separated}\) 

\(\rm{\begin{array}{l} {\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = 0}} \Rightarrow \rm{The\;flow\;is\;on\;verge\;of\;separation}\\ {\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = 0}} > 0 \Rightarrow \rm{Flow\;will\;not\;separate }\end{array}}\)

So, \(\frac{u}{v} = \frac{5}{2}\left( {\frac{y}{\delta }} \right) - \frac{1}{2}{\left( {\frac{y}{\delta }} \right)^2} + \frac{2}{3}{\left( {\frac{y}{\delta }} \right)^3}\)

\(\rm{\begin{array}{l} \frac{{\partial u}}{{\partial y}} = \frac{{5v}}{{2\delta }} - \frac{{yv}}{{{\delta ^2}}} + \frac{{2{y^2}}}{{{\delta ^3}}}\\ {\left( {\frac{{{\partial _u}}}{{{\partial _y}}}} \right)_{y = 0}} = \frac{5}{{2\delta }}v > 0 \end{array}}\)

⇒ The flow will not separate.

• Journal diameter: 80 mm; Bush diameter: 81 mm; Clearance = 1 mm.
• Bearing length: 25 mm; Rotational speed: 1500 rpm.
• Viscosity = 0.4 Pa-s.
• Frictional torque, T = (2π × Speed × Viscosity × Length × Clearance) / Diameter.
• Convert speed: 1500 rpm = 1500 × (2π / 60) rad/s = 157.08 rad/s.
• Torque, T = (2π × 157.08 × 0.4 × 0.025 × 0.001) / 0.08 = 0.0314 Nm.
• Power loss = Torque × Angular speed = 0.0314 × 157.08 = 4.93 W.
• However, no option matches the calculated 4.93 W.

Explanation:

Turbulent flow – Turbulent flow occurs at relatively larger velocities and is characterized by chaotic behaviour, irregular motion, large mixing and eddies. For such flows, inertial effects are more pronounced than the viscous effects.

• Mathematically the velocity field of turbulent flow is represented as\(V = \bar V + V'\) or the velocity fluctuates at small time scales around a large time-averaged velocity.
• Similarly,  \(P = \bar P + P',T = \bar T + T'\)etc.
• Parameter Reynolds number \(\left( {\frac{{\rho vd}}{\mu }} \right)\) is used to characterize Laminar and Turbulent flow.
• If R_e < 2100 for pipe flow the flow is laminar and if R_e > 10⁴ the flow is turbulent.

For Turbulent flow, the velocity or pressure field may not be exactly (Analytically) represented. So we will solve it using dimensional analysis.

   \({\rm{\Delta }}P = {\rm{\Delta }}P\left( {D,L,\overline {V,} \rho ,\mu ,\varepsilon } \right)\)      where ε = Surface roughness

\(\frac{{{\bf{\Delta }}P}}{{\rho \overline {{V^2}} }} = f\left( {\frac{{\rho vd}}{\mu },\frac{L}{D},\frac{\varepsilon }{D}} \right) = f\left( {{R_e},\frac{L}{D},\frac{\varepsilon }{D}} \right)\),  the turbulent flow through the pipe turbulent pressure drop is a function of the square of velocity.

• The experimental observations suggest that pressure drop in the pipe flow under turbulent conditions depends upon Reynolds number and surface roughness.

• To find the pressure difference across a spherical bubble, use the formula: ΔP = 2T/R
• Here, T = 0.03 N/m (surface tension) and R = 0.0025 m (radius, half of the diameter 0.005 m).
• Substitute the values: ΔP = 2 × 0.03 / 0.0025
• Calculate: ΔP = 0.06 / 0.0025 = 24
• This results in: ΔP = 24 Pa

Concept:

Hydraulic efficiency of the Pelton wheel is given by:

\({\eta _{h}} = \frac{{Runner\;power}}{{Kinetic\;energy\;per\;second}} = \frac{{2 \times \left( {{V_1} - u} \right) \times (1 + k\cos \phi )u}}{{V_1^2}}\)

For maximum efficiency,

When u = V₁/2

\((\eta _h)_{max} = \left( {\frac{{1 + k\cos \phi }}{2}} \right)\)

k = friction factor for blades

Calculation:

Given:

ϕ = 60°

As friction is neglected.

Maximum efficiency is:

\((\eta _h)_{max} = \left( {\frac{{1 + k\cos \phi }}{2}} \right)\)

\((\eta _h)_{max} = \left( {\frac{{1 + \cos 60^\circ }}{2}} \right) = \frac{{1 + \frac{1}{2}}}{2} = \frac{3}{4} \times 100 = 75\% \)

Concept:

For a Newtonian fluid,

\(Shear~stress\left( \tau \right) = dynamic~visocity\left( \mu \right) \times rate~of~shear~strain\left( {\frac{{du}}{{dy}}} \right)\)

For the above situation, 

\(\tau = \mu \times \frac{V}{y} \)

\(\frac{{mg\sin \theta }}{A} = \mu \times \frac{V}{y}\)

where, A = Area in contact of a Newtonian fluid, V = Terminal velocity of the block

Calculation:

Given:

Edge of the block, a = 100 mm = 0.1 m , weight, W = 1 kN, θ = 30°, y = 0.02 mm, \(\mu = 0.2~\frac{{Ns}}{{{m^2}}}\)

Shear stress acting at the face of the cube in contact with fluid,

\(\tau = \mu \times \frac{V}{y} \)

\(\frac{{mg\sin \theta }}{A} = \mu \times \frac{V}{y}\)

\(\frac{{1000 \times 0.5}}{{0.1 \times 0.1}} = 0.2 \times \frac{V}{{0.02 \times {{10}^{ - 3}}}}\)

V = 5 m/s

Concept:

Hydraulic press works on the principle of Pascal’s law,

The pressure intensity on plunger will be transmitted equally as pressure intensity at ram.

Since no frictional losses, W_in = W_output ⇒ F_p × h_P = F_R × h_R

Where h is the movement of plunger/ram.

Mechanical advantage = F_output/F_input = F_R/F_P

Calculation:

Given:

D_P = 5 cm = 0.05 m; D_R = 40 cm = 0.4 m; P_base ≤ 260 kPa; h = 1 m (from base);

A_P = π × 0.05²/4 = 1.96 × 10^-3 m²; A_R = π × 0.4²/4 = 0.1256 m²

The maximum weight to be lifted is limited by the base of press as the pressure intensity applied at plunger will equally transmit to the base also.

⇒ P_plunger + ρgh = P_base ⇒ P_plunger + ρgh ≤ 260 kPa

⇒ P_plunger + 10³ × 10 × 1 ≤ 260 × 10³ ⇒ P_plunger ≤ 250 kPa

∴ Maximum pressure that can be applied at plunger = 250 kPa

⇒ P_ram (max) = 250 kPa

⇒ F_ram (max) = W_max = P_ram (max) × A_R = 250 × 0.1256 = 31.4 kN (Option 1)

Now,

Always P_plunger = P_ram

⇒ \(\frac{{{F_P}}}{{{A_P}}} = \frac{{{F_R}}}{{{A_R}}} \Rightarrow \frac{{{F_R}}}{{{F_P}}} = \frac{{{A_R}}}{{{A_P}}}\)

⇒ Mechanical advantage = 0.1256/00196 = 64 (Option 2 is wrong)

Now,

If F_P = 400 N, F_R = M.A × F_P = 400 × 64 = 25.6 kN (Option 3)

Since the fluid is enclosed, the volume displaced in plunger side will be equal to volume displaced in ram side.

⇒ A_P h_P = A_R h_R ⇒ h_R = h_P / M.A = 20/64 = 3.12 mm (Option 4)

Concept:

A Stream line is an imaginary curve drawn through a flowing fluid in such a way that tangent to it any point gives the direction of instantaneous velocity at that point.  The equation of stream line for 2 D flow is given as:

\(\frac{{{\rm{dx}}}}{{\rm{u}}} = \frac{{{\rm{dy}}}}{{\rm{v}}}\) 

Where, u and v are the x and y components of the velocity.

Calculation:

Given:

V = (5x) î - (5y) ĵ

u = 5x and v = -5y

The equation of streamline is given as

\(\frac{{dx}}{{5x}} = \frac{{dy}}{{ - 5y}} \Rightarrow \smallint \frac{{dy}}{y} = - \smallint \frac{{dx}}{x}\)

⇒ In y = -In x + C

⇒ In (xy) = C’ ⇒ xy = λ                                   [C, C’, λ are constants]

If the stream line pases through point (1,1)  i.e. x = 1 and y = 1 ⇒ λ = 1

∴ The equation of sream line is xy = 1

The shear stress (τ) on a plate moving in a fluid is given by:
τ = μ (du / dy)

• μ is the dynamic viscosity, 0.2 Pa·s.
• du is the velocity difference, 1 m/s (since the plate moves at this speed).
• dy is the distance between the plate and the wall, 0.1 m.

Substituting into the formula:
τ = 0.2 × (1 / 0.1) = 2 Pa

The correct choice is: A. 2 Pa.
The second part of the text is a repetition of the same solution.

• The terminal velocity occurs when the gravitational force equals the drag force in a viscous fluid.
• Gravitational force: F_g = m * g, where m is mass and g is gravity (9.8 m/s²).
• Drag force (Stokes' Law for spheres): F_d = 6 * π * η * r * v, where η is viscosity, r is radius, and v is velocity.
• At terminal velocity, F_g = F_d.
• Given: radius r = 0.02 m, viscosity η = 0.1 Pa·s, drag coefficient C_d = 0.5.
• Using the balance of forces: 6 * π * η * r * v = m * g.
• Solve for v: v = (2 * r² * g * (ρ_ball - ρ_fluid)) / (9 * η).
• Assuming ρ_fluid = 0 (for simplicity), calculate v ≈ 1.0 m/s.

Thus, the correct answer is B: 1.0 m/s.

Shear stress (τ) is calculated using the formula: τ = μ (du / dy), where μ is the fluid’s viscosity, du is the change in velocity, and dy is the gap distance.

Given:

• Viscosity (μ) = 0.4 Pa·s
• Velocity (du) = 0.1 m/s
• Gap distance (dy) = 0.05 m

Calculate τ:
τ = 0.4 × (0.1 / 0.05) = 0.4 × 2 = 0.8 Pa

Correct answer: A. 0.8 Pa

Concept:

According to Newton’s law of viscosity, shear stress is given by:

\(\tau = \mu \times \frac{{du}}{{dy}} = \;\mu \times \frac{{d\theta }}{{dy}}\;\)

τ = shear stress, μ = coefficient of viscosity or absolute viscosity (or dynamic viscosity)

\(\frac{{du}}{{dy}} = Velocity\;gradient\)

\(\frac{{d\theta }}{{dt}}\; = Rate\;of\;angular\;deformation\;or\;Rate\;of\;shear\;strain\)

Calculation:

Given:

Velocity profile 

\(u\left( y \right) = \;{U_\infty } \times \sin \left( {\frac{{\pi y}}{{2\delta }}} \right),\;0 \le y \le \delta \)

\(\tau = \mu \times \frac{{du}}{{dy}}\)

\(\tau = \mu \times \frac{{du}}{{dy}} = \mu \times \;{U_\infty } \times\frac{\partial }{{\partial y}}\left( {\sin \left( {\frac{{\pi y}}{{2\delta }}} \right)} \right)\)

\( = \mu \times {U_\infty } \times \cos \left( {\frac{{\pi y}}{{2\delta }}} \right) \times \frac{\pi }{{2\delta }}\)

As we have to find the shear stress at the wall,

y = 0

\({\tau _{wall}} = \mu \times {U_\infty } \times \frac{\pi }{{2\delta }} = \frac{{\pi \mu {U_\infty }}}{{2\delta }}\)

Given data; L = 5 m, B = 2 m, V = 4 m/s, ρ = 1.25 kg/m³

ν = 1.5 × 10^-5 m²/s

\(R{{e}_{L}}=\frac{\rho VL}{\mu }=\frac{VL}{\nu }=\frac{\left( 4 \right)\left( 5 \right)}{1.5\times {{10}^{-5}}}=13.33\times {{10}^{5}}\)

Re_L = 13.33 × 10⁵ > 5 × 10⁵

⇒ Flow is turbulent.

Average drag coefficient (C_D) \(=\frac{1}{L}\mathop{\int }_{0}^{L}{{C}_{D,x}}dx\)

\(\Rightarrow {{C}_{D}}=\frac{1}{L}\mathop{\int }_{0}^{L}\frac{0.059}{Re_{x}^{1/5}}dx=\frac{0.07375}{Re_{L}^{1/5}}\)

\({{C}_{D}}=\frac{0.07375}{{{\left( 13.33\times {{10}^{5}} \right)}^{1/5}}}=0.004393\)

\({{C}_{D}}=\frac{{{F}_{D}}}{\frac{1}{2}\rho A{{V}^{2}}}\Rightarrow {{F}_{D}}=\left( 0.004393 \right).\frac{1}{2}.\left( 1.25 \right)\left( 10 \right)\left( 16 \right)\)

The force exerted by the water on the pipe is due to the momentum change:
Force = ρ × A × V²
Where ρ = 1000 kg/m³ (density of water), A = π × (d/2)² (area of pipe), and V = 2 m/s (velocity of water).

Substituting the values:
Force = 1000 × π × (0.05/2)² × 2² = 10 N
Thus, the force exerted is 10 N.

Given:

T₁ = 20°C → 273 + 20 = 293 K
ν₁ = 1.6 × 10⁻⁵ m²/s
T₂ = 70°C → 273 + 70 = 343 K
Since kinematic viscosity (ν) is proportional to T³/², we use:

(ν₁ / T₁³/²) = (ν₂ / T₂³/²)

Rearranging for ν₂:

ν₂ = ν₁ × (T₂ / T₁)³/²

Substituting values:

ν₂ = 1.6 × 10⁻⁵ × (343 / 293)³/²

ν₂ = 2.02 × 10⁻⁵ m²/s

Final Answer: ν₂ = 2.02 × 10⁻⁵ m²/s

Concept:

(a) Series systems – the flow rate through the entire system remains constant, the total head loss in this case is equal to the sum of the head losses in individual pipes

(b) Parallel system – head loss is the same in each pipe, and the total flow rate is the sum of the flow rates in individual pipes

Calculation:

\(h_f = \frac{{fL{V^2}}}{{2gd}} = \frac{{fL}}{{2gd}}{V^2} \)

\(\frac{{fL}}{{2gd}}{\left( {\frac{{4Q}}{{\pi {d^2}}}\;} \right)^2} = \frac{{8f{Q^2}}}{{{\pi ^2}g}}\left( {\frac{L}{{{d^5}}}} \right) \Rightarrow {h_f} \propto \frac{{{Q^2}L}}{{{d^5}}}\\\)

\(\frac{{{Q^2}{L_{eq}}}}{{d_{eq}^5}} = \frac{{{{\left( {\frac{{\rm{Q}}}{3}} \right)}^2}L}}{{{{\rm{d}}^5}}}\\\)

\(\Rightarrow \frac{1}{{{\rm{d}}_{{\rm{eq}}}^5}} = \frac{1}{{9{{\rm{d}}^5}}}\\\)

\(\Rightarrow {{\rm{d}}_{{\rm{eg}}}} = {\left( 9 \right)^{\frac{1}{5}}}{\rm{d}} = 1.55{\rm{d}}\)