BE Civil Engineering (IOE, TU) Fluid Mechanics (IOE, CE 553) Question Paper 2078 Nepal

(a) Derivation

Consider a plane surface of area $A$ inclined at angle $\theta$ to the free surface. Take an elemental strip of area $dA$ at depth $h$ below the surface.

Pressure on the strip: $p = \rho g h$. Force on strip: $dF = \rho g h\, dA$.

Total force:

$$F = \int \rho g h\, dA = \rho g \sin\theta \int y\, dA = \rho g \sin\theta\, (\bar{y} A) = \rho g \bar{h} A$$

where $\bar{h}$ is the depth of the centroid.

Centre of pressure (taking moments of $dF$ about the surface line and equating to the moment of $F$):

$$F\, y_{cp} = \int y\, dF = \rho g \sin\theta \int y^2 dA = \rho g \sin\theta\, I_o$$

Using $I_o = I_G + A\bar{y}^2$ (parallel-axis theorem), for a vertical surface ($\theta = 90^\circ$) this reduces to:

$$\boxed{\,h_{cp} = \bar{h} + \dfrac{I_G}{A\,\bar{h}}\,}$$

where $I_G = \dfrac{b d^3}{12}$ for a rectangle.

(b) Numerical computation

Geometry: width $b = 2\ \text{m}$, gate height $d = 3 - 1 = 2\ \text{m}$.

Area: $A = b\,d = 2 \times 2 = 4\ \text{m}^2$.

Depth of centroid: $\bar{h} = \dfrac{1 + 3}{2} = 2\ \text{m}$.

Total force:

$$F = \rho g \bar{h} A = 1000 \times 9.81 \times 2 \times 4 = 78\,480\ \text{N}$$ $$\boxed{F = 78.48\ \text{kN}}$$

Centre of pressure:

$$I_G = \frac{b d^3}{12} = \frac{2 \times 2^3}{12} = 1.333\ \text{m}^4$$ $$h_{cp} = \bar{h} + \frac{I_G}{A\bar{h}} = 2 + \frac{1.333}{4 \times 2} = 2 + 0.1667 = 2.167\ \text{m}$$ $$\boxed{h_{cp} = 2.17\ \text{m below the free surface}}$$

The centre of pressure lies $0.167\ \text{m}$ below the centroid, as expected for a submerged plane surface.

(a) Bernoulli's equation

Assumptions: (i) steady flow, (ii) incompressible fluid, (iii) inviscid (frictionless) flow, (iv) flow along a streamline, (v) no energy added or extracted.

Real-flow energy equation between two sections:

$$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_L$$

For ideal flow $h_L = 0$.

(b) Numerical computation

Areas:

$$A_1 = \frac{\pi}{4}(0.20)^2 = 0.03142\ \text{m}^2, \qquad A_2 = \frac{\pi}{4}(0.10)^2 = 0.007854\ \text{m}^2$$

Velocities (continuity $Q = AV$):

$$V_1 = \frac{0.05}{0.03142} = 1.592\ \text{m/s}$$ $$V_2 = \frac{0.05}{0.007854} = 6.366\ \text{m/s}$$

Pressure at section 2 (horizontal, $z_1 = z_2$, $h_L = 0$):

$$p_2 = p_1 + \tfrac{1}{2}\rho\,(V_1^2 - V_2^2)$$ $$p_2 = 200\,000 + \tfrac{1}{2}(1000)\,(1.592^2 - 6.366^2)$$ $$p_2 = 200\,000 + 500\,(2.534 - 40.53) = 200\,000 - 18\,998 = 181\,002\ \text{Pa}$$ $$\boxed{V_1 = 1.59\ \text{m/s}, \quad V_2 = 6.37\ \text{m/s}, \quad p_2 = 181.0\ \text{kPa}}$$

The pressure drops as the pipe contracts and kinetic energy rises, consistent with energy conservation.

(a) Momentum equation

For steady flow through a control volume, the net external force equals the net rate of efflux of momentum:

$$\sum \vec{F} = \rho Q\,(\vec{V}_{out} - \vec{V}_{in})$$

External forces include the pressure forces on the inlet and outlet faces and the reaction $\vec{R}$ from the bend on the fluid. The force on the bend is equal and opposite to $\vec{R}$.

(b) Numerical computation

Areas and velocities:

$$A_1 = \frac{\pi}{4}(0.30)^2 = 0.07069\ \text{m}^2,\quad A_2 = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2$$ $$V_1 = \frac{0.10}{0.07069} = 1.415\ \text{m/s},\quad V_2 = \frac{0.10}{0.01767} = 5.659\ \text{m/s}$$

Pressure at outlet (Bernoulli, horizontal, frictionless):

$$p_2 = p_1 + \tfrac{1}{2}\rho(V_1^2 - V_2^2) = 150\,000 + 500(1.415^2 - 5.659^2) = 134\,989\ \text{Pa}$$

Coordinate system: inlet flow along $+x$, outlet flow along $+y$.

$x$-direction (force on fluid):

$$F_x = p_1 A_1 + \rho Q V_1 = 150\,000(0.07069) + 1000(0.10)(1.415)$$ $$F_x = 10\,602.9 + 141.5 = 10\,744.4\ \text{N}$$

$y$-direction:

$$F_y = p_2 A_2 + \rho Q V_2 = 134\,989(0.01767) + 1000(0.10)(5.659)$$ $$F_y = 2385.5 + 565.9 = 2951.3\ \text{N}$$

Resultant force on the bend:

$$R = \sqrt{F_x^2 + F_y^2} = \sqrt{10\,744.4^2 + 2951.3^2} = 11\,142\ \text{N}$$ $$\theta = \tan^{-1}\!\left(\frac{F_y}{F_x}\right) = \tan^{-1}\!\left(\frac{2951.3}{10\,744.4}\right) = 15.4^\circ$$ $$\boxed{R = 11.14\ \text{kN at } 15.4^\circ \text{ from the inlet axis}}$$

The force the water exerts on the bend is equal in magnitude and opposite in sense to this reaction.

(a) Buckingham $\pi$-analysis

Variables: $\Delta p,\ V,\ D,\ \rho,\ \mu,\ \varepsilon$ — total $n = 6$. Fundamental dimensions: $M, L, T$ — so $m = 3$. Number of $\pi$ groups: $n - m = 6 - 3 = 3$.

Choose repeating variables $\rho, V, D$ (dimensionally independent, spanning $M, L, T$).

$$\pi_1 = \frac{\Delta p}{\rho V^2} \quad(\text{Euler / pressure coefficient})$$ $$\pi_2 = \frac{\mu}{\rho V D} = \frac{1}{Re}\quad\Rightarrow\quad Re = \frac{\rho V D}{\mu}$$ $$\pi_3 = \frac{\varepsilon}{D}\quad(\text{relative roughness})$$

Hence:

$$\boxed{\frac{\Delta p}{\rho V^2} = \phi\!\left(Re,\ \frac{\varepsilon}{D}\right)}$$

which is the dimensionless form underlying the Darcy friction-factor (Moody) correlation.

(b) Froude similitude

For free-surface (gravity-dominated) flow the Froude number is matched: $Fr_m = Fr_p$. Length scale ratio: $L_r = L_m/L_p = 1/36$.

Under Froude scaling (with density ratio $\rho_r = 1$):

• Velocity ratio: $V_r = L_r^{1/2} = (1/36)^{1/2} = 1/6$
• Discharge ratio: $Q_r = L_r^{5/2} = (1/36)^{2.5} = 1/7776$
• Force ratio: $F_r = \rho_r L_r^3 = (1/36)^3 = 1/46\,656$

Prototype discharge:

$$Q_p = \frac{Q_m}{Q_r} = 0.045 \times 7776 = 349.9\ \text{m}^3/\text{s}$$ $$\boxed{Q_p \approx 350\ \text{m}^3/\text{s}}$$

Prototype force:

$$F_p = \frac{F_m}{F_r} = 12 \times 46\,656 = 559\,872\ \text{N}$$ $$\boxed{F_p \approx 559.9\ \text{kN}}$$

(a) Darcy–Weisbach derivation

Consider steady flow in a pipe of diameter $D$ and length $L$. The wall shear stress $\tau_0$ acts over the wetted surface. Force balance on the fluid cylinder:

$$(p_1 - p_2)\frac{\pi D^2}{4} = \tau_0\,(\pi D L)$$ $$\Rightarrow\ h_f = \frac{p_1 - p_2}{\rho g} = \frac{4 \tau_0 L}{\rho g D}$$

Expressing the wall shear via the friction factor, $\tau_0 = \dfrac{f \rho V^2}{8}$, gives:

$$\boxed{h_f = \frac{f L V^2}{2 g D}}$$

(b) Numerical computation

Mean velocity:

$$A = \frac{\pi}{4}(0.25)^2 = 0.04909\ \text{m}^2$$ $$V = \frac{Q}{A} = \frac{0.06}{0.04909} = 1.222\ \text{m/s}$$

Friction head loss:

$$h_f = \frac{f L V^2}{2 g D} = \frac{0.02 \times 500 \times 1.222^2}{2 \times 9.81 \times 0.25}$$ $$h_f = \frac{0.02 \times 500 \times 1.4937}{4.905} = \frac{14.937}{4.905} = 3.046\ \text{m}$$ $$\boxed{h_f = 3.05\ \text{m of water}}$$

Power to overcome friction:

$$P = \rho g Q h_f = 1000 \times 9.81 \times 0.06 \times 3.046 = 1793\ \text{W}$$ $$\boxed{P = 1.79\ \text{kW}}$$

Solution

Manometer (equivalent head of water):

$$h = x\left(\frac{S_{Hg}}{S_w} - 1\right) = 0.25\,(13.6 - 1) = 0.25 \times 12.6 = 3.15\ \text{m of water}$$

Areas:

$$A_1 = \frac{\pi}{4}(0.30)^2 = 0.07069\ \text{m}^2,\qquad A_2 = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2$$

Venturimeter discharge equation:

$$Q = C_d\,\frac{A_1 A_2}{\sqrt{A_1^2 - A_2^2}}\,\sqrt{2 g h}$$ $$\frac{A_1 A_2}{\sqrt{A_1^2 - A_2^2}} = \frac{0.07069 \times 0.01767}{\sqrt{0.07069^2 - 0.01767^2}} = \frac{1.2491\times10^{-3}}{\sqrt{4.9970\times10^{-3} - 3.123\times10^{-4}}}$$ $$= \frac{1.2491\times10^{-3}}{0.06846} = 0.018245\ \text{m}^2$$ $$\sqrt{2 g h} = \sqrt{2 \times 9.81 \times 3.15} = \sqrt{61.80} = 7.861\ \text{m/s}$$ $$Q = 0.98 \times 0.018245 \times 7.861 = 0.1406\ \text{m}^3/\text{s}$$ $$\boxed{Q = 0.141\ \text{m}^3/\text{s} = 140.6\ \text{L/s}}$$

(a) Flow regime

Discharge: $Q = 0.1\ \text{L/s} = 1\times10^{-4}\ \text{m}^3/\text{s}$.

$$A = \frac{\pi}{4}(0.04)^2 = 1.2566\times10^{-3}\ \text{m}^2$$ $$V = \frac{Q}{A} = \frac{1\times10^{-4}}{1.2566\times10^{-3}} = 0.0796\ \text{m/s}$$ $$Re = \frac{\rho V D}{\mu} = \frac{900 \times 0.0796 \times 0.04}{0.10} = 28.6$$

Since $Re = 28.6 \ll 2000$, the flow is laminar.

(b) Hagen–Poiseuille pressure drop

$$\Delta p = \frac{128\,\mu L\,Q}{\pi D^4} = \frac{128 \times 0.10 \times 10 \times 1\times10^{-4}}{\pi \times (0.04)^4}$$

Denominator: $\pi \times (0.04)^4 = \pi \times 2.56\times10^{-6} = 8.042\times10^{-6}$. Numerator: $128 \times 0.10 \times 10 \times 1\times10^{-4} = 1.28\times10^{-2}$.

$$\Delta p = \frac{0.0128}{8.042\times10^{-6}} = 1591.5\ \text{Pa}$$ $$\boxed{\Delta p = 1.59\ \text{kPa}}$$

Head loss:

$$h_f = \frac{\Delta p}{\rho g} = \frac{1591.5}{900 \times 9.81} = 0.180\ \text{m}$$ $$\boxed{h_f = 0.18\ \text{m of oil}}$$

(a) Boundary layer

The boundary layer is the thin region adjacent to a solid surface where viscous effects are significant and the velocity rises from zero at the wall (no-slip) to about $99\%$ of the free-stream value.

• Laminar boundary layer: smooth, ordered fluid layers; thinner; lower wall shear; occurs at low $Re_x$ (below $\approx 5\times10^5$ on a flat plate).
• Turbulent boundary layer: chaotic mixing of eddies; thicker; higher wall shear and steeper near-wall velocity gradient; occurs at higher $Re_x$.

(b) Blasius (laminar) computation

Reynolds number:

$$Re_x = \frac{U x}{\nu} = \frac{3 \times 1.5}{1.5\times10^{-5}} = 3.0\times10^{5}$$

Since $Re_x = 3\times10^5 < 5\times10^5$, the laminar Blasius relations apply.

Boundary-layer thickness:

$$\delta = \frac{5x}{\sqrt{Re_x}} = \frac{5 \times 1.5}{\sqrt{3.0\times10^5}} = \frac{7.5}{547.7} = 0.01369\ \text{m}$$ $$\boxed{\delta = 13.7\ \text{mm}}$$

Local skin-friction coefficient:

$$C_{f,x} = \frac{0.664}{\sqrt{Re_x}} = \frac{0.664}{547.7} = 1.212\times10^{-3}$$ $$\boxed{C_{f,x} = 1.21\times10^{-3}}$$

(a) Continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 3 + (-3) = 0$$

Continuity is satisfied — a valid incompressible flow.

(b) Rotationality and velocity potential

Vorticity ($z$-component):

$$\omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 2 - 2 = 0$$

Since $\omega_z = 0$, the flow is irrotational, so a velocity potential exists.

Using $u = \dfrac{\partial \phi}{\partial x}$ and $v = \dfrac{\partial \phi}{\partial y}$:

$$\frac{\partial \phi}{\partial x} = u = 3x + 2y \Rightarrow \phi = \frac{3x^2}{2} + 2xy + f(y)$$

Differentiate w.r.t. $y$ and equate to $v$:

$$\frac{\partial \phi}{\partial y} = 2x + f'(y) = v = 2x - 3y \Rightarrow f'(y) = -3y \Rightarrow f(y) = -\frac{3y^2}{2}$$ $$\boxed{\phi = \frac{3x^2}{2} + 2xy - \frac{3y^2}{2} + C}$$

(c) Stream function

Using $u = \dfrac{\partial \psi}{\partial y}$ and $v = -\dfrac{\partial \psi}{\partial x}$:

$$\frac{\partial \psi}{\partial y} = u = 3x + 2y \Rightarrow \psi = 3xy + y^2 + g(x)$$

Differentiate w.r.t. $x$ and equate to $-v$:

$$\frac{\partial \psi}{\partial x} = 3y + g'(x) = -v = -2x + 3y \Rightarrow g'(x) = -2x \Rightarrow g(x) = -x^2$$ $$\boxed{\psi = 3xy + y^2 - x^2 + C}$$

(a) Surface tension, capillarity and rise

Surface tension $\sigma$ is the tensile force per unit length acting in the plane of a liquid's free surface, arising from unbalanced cohesive forces at the surface. Capillarity is the rise or fall of a liquid in a small-bore tube due to the combined effect of surface tension and adhesion/cohesion.

Capillary rise:

$$h = \frac{4\sigma\cos\theta}{\rho g d} = \frac{4 \times 0.0728 \times \cos 0^\circ}{1000 \times 9.81 \times 0.001}$$ $$h = \frac{0.2912}{9.81} = 0.02968\ \text{m}$$ $$\boxed{h = 29.7\ \text{mm}}$$

(b) Bulk modulus

Bulk modulus of elasticity $K$ is the ratio of an applied compressive stress (pressure increase) to the resulting volumetric strain:

$$K = -\frac{dp}{dV/V}$$

Required pressure increase for $dV/V = 0.5\% = 0.005$:

$$dp = K \times \frac{dV}{V} = 2.2\times10^9 \times 0.005 = 1.1\times10^7\ \text{Pa}$$ $$\boxed{dp = 11.0\ \text{MPa}}$$

(a) Orifice coefficients

• Coefficient of velocity $C_v$ = actual velocity at the vena contracta divided by the theoretical velocity $\sqrt{2gH}$ (typically $\approx 0.97$–$0.99$).
• Coefficient of contraction $C_c$ = area of the jet at the vena contracta divided by the area of the orifice (typically $\approx 0.61$–$0.64$).
• Coefficient of discharge $C_d$ = actual discharge divided by the theoretical discharge (typically $\approx 0.60$–$0.62$).

Relation:

$$\boxed{C_d = C_v \times C_c}$$

(b) Actual discharge

Orifice area:

$$A = \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2$$

Theoretical velocity:

$$V_{th} = \sqrt{2 g H} = \sqrt{2 \times 9.81 \times 2.0} = \sqrt{39.24} = 6.264\ \text{m/s}$$

Actual discharge:

$$Q = C_d\, A\, \sqrt{2 g H} = 0.60 \times 1.9635\times10^{-3} \times 6.264$$ $$Q = 7.380\times10^{-3}\ \text{m}^3/\text{s}$$ $$\boxed{Q = 7.38\ \text{L/s}}$$