BE Civil Engineering (IOE, TU) Fluid Mechanics (IOE, CE 553) Question Paper 2079 Nepal

Given: width $b = 2.5\ \text{m}$, height $h = 3.0\ \text{m}$, depth of top edge $= 1.5\ \text{m}$, vertical gate.

Area $A = b \times h = 2.5 \times 3.0 = 7.5\ \text{m}^2$.

Depth of centroid below free surface:

$$\bar{h} = 1.5 + \frac{3.0}{2} = 3.0\ \text{m}.$$

(a) Total hydrostatic force

$$F = \rho g \bar{h} A = 1000 \times 9.81 \times 3.0 \times 7.5 = 220725\ \text{N} \approx \mathbf{220.73\ kN}.$$

(b) Centre of pressure

Second moment of area about the centroidal horizontal axis:

$$I_G = \frac{b h^3}{12} = \frac{2.5 \times 3.0^3}{12} = \frac{2.5 \times 27}{12} = 5.625\ \text{m}^4.$$

For a vertical plane the depth of centre of pressure is:

$$h_{cp} = \bar{h} + \frac{I_G}{\bar{h} A} = 3.0 + \frac{5.625}{3.0 \times 7.5} = 3.0 + \frac{5.625}{22.5} = 3.0 + 0.25 = \mathbf{3.25\ m\ below\ the\ free\ surface}.$$

(c) Force at bottom edge to hold gate closed

Take moments about the top hinge. The hydrostatic resultant acts at the centre of pressure, which is at a distance below the hinge of:

$$y_{cp} = h_{cp} - 1.5 = 3.25 - 1.5 = 1.75\ \text{m}.$$

The applied force $P$ acts at the bottom edge, $3.0\ \text{m}$ from the hinge. Moment balance about hinge:

$$P \times 3.0 = F \times 1.75$$ $$P = \frac{220.725 \times 1.75}{3.0} = \frac{386.27}{3.0} = \mathbf{128.76\ kN}.$$

Free surface --------------------
      | 1.5 m
  ====O==== hinge (top edge)
      |     <- F acts at 1.75 m below hinge
      |  -> P at bottom (3.0 m below hinge)
  =========  bottom edge

(a) Derivation of Bernoulli's equation

Consider a fluid element of length $ds$ and cross-sectional area $dA$ along a streamline. Euler's equation of motion along the streamline (per unit mass) for steady flow is:

$$\frac{1}{\rho}\frac{dp}{ds} + g\frac{dz}{ds} + V\frac{dV}{ds} = 0.$$

Multiplying by $ds$:

$$\frac{dp}{\rho} + g\,dz + V\,dV = 0.$$

Integrating along the streamline for an incompressible fluid ($\rho$ = constant):

$$\frac{p}{\rho} + gz + \frac{V^2}{2} = \text{constant}.$$

Dividing by $g$ gives the head form:

$$\boxed{\frac{p}{\rho g} + z + \frac{V^2}{2g} = \text{constant (total head)}}$$

Assumptions: (1) flow is steady, (2) fluid is incompressible, (3) fluid is inviscid/frictionless (no energy loss), (4) flow is along a streamline, (5) no energy added/removed by machines.

(b) Venturimeter discharge

Inlet area $A_1 = \frac{\pi}{4}(0.200)^2 = \frac{\pi}{4}(0.04) = 0.0314159\ \text{m}^2$.

Throat area $A_2 = \frac{\pi}{4}(0.100)^2 = \frac{\pi}{4}(0.01) = 0.0078540\ \text{m}^2$.

Pressure head difference (horizontal, so $z_1 = z_2$):

$$h = \frac{\Delta p}{\rho g} = \frac{35000}{1000 \times 9.81} = 3.5678\ \text{m of water}.$$

Venturimeter discharge formula:

$$Q = C_d \frac{A_1 A_2}{\sqrt{A_1^2 - A_2^2}}\sqrt{2gh}.$$

Compute terms:

$$A_1 A_2 = 0.0314159 \times 0.0078540 = 2.46740\times10^{-4}\ \text{m}^4.$$ $$A_1^2 - A_2^2 = (0.0314159)^2 - (0.0078540)^2 = 9.8696\times10^{-4} - 6.1685\times10^{-5} = 9.2528\times10^{-4}.$$ $$\sqrt{A_1^2 - A_2^2} = 0.030418\ \text{m}^2.$$ $$\sqrt{2gh} = \sqrt{2 \times 9.81 \times 3.5678} = \sqrt{69.99} = 8.366\ \text{m/s}.$$

Therefore:

$$Q = 0.98 \times \frac{2.46740\times10^{-4}}{0.030418} \times 8.366 = 0.98 \times 8.1116\times10^{-3} \times 8.366.$$ $$Q = 0.98 \times 0.067862 = 0.06650\ \text{m}^3/\text{s}.$$

Discharge $Q \approx 0.0665\ \text{m}^3/\text{s} = 66.5\ \text{L/s}$.

Given: $Q = 0.30\ \text{m}^3/\text{s}$, $D_1 = 0.300\ \text{m}$, $D_2 = 0.200\ \text{m}$, $\theta = 60^\circ$, $p_1 = 200\ \text{kPa}$, horizontal bend.

Areas and velocities

$$A_1 = \frac{\pi}{4}(0.3)^2 = 0.070686\ \text{m}^2,\quad A_2 = \frac{\pi}{4}(0.2)^2 = 0.031416\ \text{m}^2.$$ $$V_1 = \frac{Q}{A_1} = \frac{0.30}{0.070686} = 4.244\ \text{m/s}.$$ $$V_2 = \frac{Q}{A_2} = \frac{0.30}{0.031416} = 9.549\ \text{m/s}.$$

Outlet pressure (Bernoulli, horizontal, frictionless)

$$p_1 + \tfrac{1}{2}\rho V_1^2 = p_2 + \tfrac{1}{2}\rho V_2^2.$$ $$p_2 = 200000 + \tfrac{1}{2}(1000)(4.244^2 - 9.549^2).$$ $$V_1^2 = 18.01,\ V_2^2 = 91.18,\ \Rightarrow V_1^2 - V_2^2 = -73.17.$$ $$p_2 = 200000 + 500\times(-73.17) = 200000 - 36585 = 163415\ \text{Pa} = 163.42\ \text{kPa}.$$

Momentum equation

Let the inlet flow be along the +x direction; the outlet makes $60^\circ$ with the inlet.

Mass flow rate $\dot{m} = \rho Q = 1000 \times 0.30 = 300\ \text{kg/s}.$

Pressure forces: $p_1 A_1 = 200000 \times 0.070686 = 14137.2\ \text{N}$; $p_2 A_2 = 163415 \times 0.031416 = 5133.9\ \text{N}$.

x-direction (let $F_x$ = force of bend on water):

$$p_1 A_1 - p_2 A_2\cos\theta + F_x = \dot{m}(V_2\cos\theta - V_1).$$ $$14137.2 - 5133.9\cos60^\circ + F_x = 300(9.549\cos60^\circ - 4.244).$$ $$14137.2 - 2567.0 + F_x = 300(4.7745 - 4.244) = 300(0.5305) = 159.2.$$ $$11570.2 + F_x = 159.2 \Rightarrow F_x = -11411.0\ \text{N}.$$

y-direction:

$$0 - p_2 A_2\sin\theta + F_y = \dot{m}(V_2\sin\theta - 0).$$ $$-5133.9\sin60^\circ + F_y = 300(9.549\sin60^\circ).$$ $$-4446.0 + F_y = 300(8.2697) = 2480.9.$$ $$F_y = 2480.9 + 4446.0 = 6926.9\ \text{N}.$$

Force of bend on water $= (-11411.0,\ 6926.9)\ \text{N}$. By Newton's third law, force of water on bend is:

$$R_x = 11411.0\ \text{N},\quad R_y = -6926.9\ \text{N}.$$

Resultant magnitude:

$$R = \sqrt{11411.0^2 + 6926.9^2} = \sqrt{1.3021\times10^8 + 4.7982\times10^7} = \sqrt{1.7819\times10^8} = 13349\ \text{N}.$$

Direction:

$$\alpha = \tan^{-1}\left(\frac{6926.9}{11411.0}\right) = \tan^{-1}(0.6070) = 31.3^\circ\ \text{below the inlet (+x) axis}.$$

Resultant force on the bend $R \approx 13.35\ kN$, acting at $31.3^\circ$ below the inlet flow direction.

Given: $H = 25\ \text{m}$, $L = 1200\ \text{m}$, $D = 0.300\ \text{m}$, $f = 0.020$, $k_{entry}=0.5$, $k_{exit}=1.0$.

(a) Discharge

Applying the energy equation between the two reservoir surfaces, the total available head equals the sum of all losses:

$$H = \left(k_{entry} + \frac{fL}{D} + k_{exit}\right)\frac{V^2}{2g}.$$

Friction coefficient term:

$$\frac{fL}{D} = \frac{0.020 \times 1200}{0.300} = \frac{24}{0.300} = 80.$$

Sum of coefficients:

$$K = 0.5 + 80 + 1.0 = 81.5.$$

So:

$$25 = 81.5 \times \frac{V^2}{2 \times 9.81} = 81.5 \times \frac{V^2}{19.62}.$$ $$V^2 = \frac{25 \times 19.62}{81.5} = \frac{490.5}{81.5} = 6.0184\ \text{m}^2/\text{s}^2.$$ $$V = 2.4533\ \text{m/s}.$$

Area $A = \frac{\pi}{4}(0.3)^2 = 0.070686\ \text{m}^2.$

$$Q = A V = 0.070686 \times 2.4533 = 0.17342\ \text{m}^3/\text{s}.$$

Discharge $Q \approx 0.1734\ \text{m}^3/\text{s} = 173.4\ \text{L/s}$.

(b) Percentage of head consumed by minor losses

Velocity head: $\frac{V^2}{2g} = \frac{6.0184}{19.62} = 0.30675\ \text{m}.$

Minor (entry + exit) losses:

$$h_{minor} = (0.5 + 1.0)\times 0.30675 = 1.5 \times 0.30675 = 0.4601\ \text{m}.$$

Percentage of total head:

$$\%_{minor} = \frac{0.4601}{25}\times 100 = \mathbf{1.84\%}.$$

(For verification, friction loss $h_f = 80 \times 0.30675 = 24.54\ \text{m}$, and $24.54 + 0.46 = 25.0\ \text{m}$, which equals $H$.)

(a) Buckingham $\pi$-theorem

The drag force depends on: $F = f(\rho, V, L, \mu)$. Total variables $n = 5$ ($F, \rho, V, L, \mu$).

Number of fundamental dimensions $m = 3$ (M, L, T). Number of $\pi$-groups $= n - m = 5 - 3 = 2$.

Choose repeating variables $\rho, V, L$ (they include all three dimensions and are independent).

Group $\pi_1$ (contains $F$): $\pi_1 = \rho^a V^b L^c F$.

Dimensions: $[\rho]=ML^{-3}$, $[V]=LT^{-1}$, $[L]=L$, $[F]=MLT^{-2}$.

$$M:\ a + 1 = 0 \Rightarrow a = -1.$$ $$T:\ -b - 2 = 0 \Rightarrow b = -2.$$ $$L:\ -3a + b + c + 1 = 0 \Rightarrow 3 - 2 + c + 1 = 0 \Rightarrow c = -2.$$ $$\therefore \pi_1 = \frac{F}{\rho V^2 L^2}.$$

Group $\pi_2$ (contains $\mu$): $\pi_2 = \rho^a V^b L^c \mu$, with $[\mu]=ML^{-1}T^{-1}$.

$$M:\ a + 1 = 0 \Rightarrow a = -1.$$ $$T:\ -b - 1 = 0 \Rightarrow b = -1.$$ $$L:\ -3a + b + c - 1 = 0 \Rightarrow 3 - 1 + c - 1 = 0 \Rightarrow c = -1.$$ $$\therefore \pi_2 = \frac{\mu}{\rho V L} = \frac{1}{Re}.$$

Functional relation $\pi_1 = \phi(\pi_2)$:

$$\frac{F}{\rho V^2 L^2} = \phi\!\left(\frac{\mu}{\rho V L}\right) = \phi\!\left(\frac{\rho V L}{\mu}\right),$$

so

$$\boxed{F = \rho V^2 L^2\,\phi(Re)}.$$

(b) Model testing — Reynolds number similarity

For a fully submerged body, dynamic similarity requires equal Reynolds numbers:

$$\left(\frac{\rho V L}{\mu}\right)_m = \left(\frac{\rho V L}{\mu}\right)_p.$$

Solve for model velocity ($L_m/L_p = 1/20$, so $L_p/L_m = 20$):

$$V_m = V_p \cdot \frac{\rho_p}{\rho_m}\cdot\frac{\mu_m}{\mu_p}\cdot\frac{L_p}{L_m}.$$ $$V_m = 4 \times \frac{1025}{1000}\times\frac{1.00\times10^{-3}}{1.05\times10^{-3}}\times 20.$$ $$V_m = 4 \times 1.025 \times 0.95238 \times 20 = 4 \times 19.524 = 78.10\ \text{m/s}.$$

Required model speed $V_m \approx 78.1\ \text{m/s}$.

Drag-force ratio. With equal $Re$, $\phi(Re)$ is the same, so $\frac{F}{\rho V^2 L^2}$ is equal for model and prototype:

$$\frac{F_p}{F_m} = \frac{\rho_p V_p^2 L_p^2}{\rho_m V_m^2 L_m^2}.$$ $$= \frac{1025}{1000}\times\left(\frac{4}{78.10}\right)^2\times 20^2 = 1.025 \times (0.051216)^2 \times 400.$$ $$= 1.025 \times 0.0026231 \times 400 = 1.025 \times 1.04925 = 1.0755.$$

Drag-force ratio $F_p/F_m \approx 1.08$ (prototype drag is about $1.08$ times the model drag).

(a) Definitions

• Dynamic (absolute) viscosity $\mu$: the property of a fluid that quantifies its resistance to shear/angular deformation, defined by Newton's law of viscosity $\tau = \mu\frac{du}{dy}$. SI unit: $\text{Pa·s}$ (= $\text{N·s/m}^2$ = kg/(m·s)).
• Kinematic viscosity $\nu$: the ratio of dynamic viscosity to mass density, $\nu = \mu/\rho$. SI unit: $\text{m}^2/\text{s}$.

(b) Calculations

$\mu = 1.2\ \text{Pa·s}$, $A = 0.6\ \text{m}^2$, $t = 1.5\ \text{mm} = 0.0015\ \text{m}$, $u = 0.4\ \text{m/s}$.

Velocity gradient (linear profile):

$$\frac{du}{dy} = \frac{0.4}{0.0015} = 266.67\ \text{s}^{-1}.$$

Shear stress:

$$\tau = \mu\frac{du}{dy} = 1.2 \times 266.67 = 320\ \text{N/m}^2 = \mathbf{320\ Pa}.$$

Viscous force:

$$F = \tau A = 320 \times 0.6 = \mathbf{192\ N}.$$

Density: $\rho = 0.85 \times 1000 = 850\ \text{kg/m}^3.$

Kinematic viscosity:

$$\nu = \frac{\mu}{\rho} = \frac{1.2}{850} = 1.412\times10^{-3}\ \text{m}^2/\text{s} = \mathbf{1.41\times10^{-3}\ m^2/s}.$$

(a) Continuity check

For 2-D incompressible flow continuity requires $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$.

$$\frac{\partial u}{\partial x} = 6x,\qquad \frac{\partial v}{\partial y} = -6x.$$ $$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x - 6x = 0.\ \checkmark$$

Continuity is satisfied, so the flow is physically possible.

(b) Rotationality

The $z$-component of rotation:

$$\omega_z = \frac{1}{2}\left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right).$$ $$\frac{\partial v}{\partial x} = -6y,\qquad \frac{\partial u}{\partial y} = -6y.$$ $$\omega_z = \frac{1}{2}(-6y - (-6y)) = \frac{1}{2}(0) = 0.$$

Since $\omega_z = 0$, the flow is irrotational.

(c) Stream function $\psi$

By definition $u = \frac{\partial\psi}{\partial y}$ and $v = -\frac{\partial\psi}{\partial x}$.

Integrate $u$ with respect to $y$:

$$\psi = \int (3x^2 - 3y^2)\,dy = 3x^2 y - y^3 + f(x).$$

Differentiate with respect to $x$ and equate to $-v$:

$$\frac{\partial\psi}{\partial x} = 6xy + f'(x) = -v = 6xy.$$

So $f'(x) = 0 \Rightarrow f(x) = C$ (constant).

$$\boxed{\psi = 3x^2 y - y^3 + C}.$$

(Taking the constant as zero, $\psi = 3x^2 y - y^3$.)

Given: $\mu = 0.10\ \text{Pa·s}$, $S = 0.90 \Rightarrow \rho = 900\ \text{kg/m}^3$, $D = 0.040\ \text{m}$, $Q = 0.5\ \text{L/s} = 0.0005\ \text{m}^3/\text{s}$, $L = 50\ \text{m}$.

Mean velocity

$$A = \frac{\pi}{4}(0.040)^2 = \frac{\pi}{4}(0.0016) = 1.25664\times10^{-3}\ \text{m}^2.$$ $$V = \frac{Q}{A} = \frac{0.0005}{1.25664\times10^{-3}} = 0.39789\ \text{m/s}.$$

(a) Reynolds number

$$Re = \frac{\rho V D}{\mu} = \frac{900 \times 0.39789 \times 0.040}{0.10} = \frac{14.324}{0.10} = 143.2.$$

Since $Re = 143 < 2000$, the flow is laminar. $\checkmark$

(b) Pressure drop (Hagen-Poiseuille)

$$\Delta p = \frac{128\,\mu L Q}{\pi D^4}.$$

Compute $D^4 = (0.040)^4 = 2.56\times10^{-6}\ \text{m}^4.$

Numerator: $128 \times 0.10 \times 50 \times 0.0005 = 128 \times 0.10 \times 0.025 = 0.32.$

Denominator: $\pi \times 2.56\times10^{-6} = 8.04248\times10^{-6}.$

$$\Delta p = \frac{0.32}{8.04248\times10^{-6}} = 39789\ \text{Pa} \approx \mathbf{39.79\ kPa}.$$

(Equivalent head loss $h_f = \Delta p/(\rho g) = 39789/(900\times9.81) = 4.51\ \text{m of oil}$.)

(a) Definitions

• Boundary layer: the thin region of fluid adjacent to a solid surface in which viscous effects are significant and the velocity rises from zero at the wall (no-slip) to (essentially) the free-stream velocity $U$.
• Displacement thickness $\delta^*$: the distance by which the boundary surface would have to be displaced outward so that an ideal (inviscid) flow carries the same mass flow rate as the real (retarded) flow. $\delta^* = \int_0^\delta \left(1 - \frac{u}{U}\right)dy.$
• Momentum thickness $\theta$: the thickness representing the deficit of momentum flux caused by the boundary layer, $\theta = \int_0^\delta \frac{u}{U}\left(1 - \frac{u}{U}\right)dy.$

(b) Displacement thickness for the cubic profile

Let $\eta = y/\delta$, so $dy = \delta\,d\eta$, with $\frac{u}{U} = \frac{3}{2}\eta - \frac{1}{2}\eta^3.$

$$\delta^* = \int_0^\delta\left(1 - \frac{u}{U}\right)dy = \delta\int_0^1\left(1 - \frac{3}{2}\eta + \frac{1}{2}\eta^3\right)d\eta.$$

Integrate term by term:

$$\int_0^1 1\,d\eta = 1,\quad \int_0^1 \frac{3}{2}\eta\,d\eta = \frac{3}{2}\cdot\frac{1}{2} = \frac{3}{4},\quad \int_0^1 \frac{1}{2}\eta^3\,d\eta = \frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}.$$ $$\delta^* = \delta\left(1 - \frac{3}{4} + \frac{1}{8}\right) = \delta\left(\frac{8 - 6 + 1}{8}\right) = \frac{3}{8}\delta.$$

$\boxed{\delta^* = \dfrac{3}{8}\delta = 0.375\,\delta}$.

Given: gauge fluid mercury $S_{Hg} = 13.6$, working fluid water $S_w = 1.0$, manometer reading $x = 250\ \text{mm} = 0.25\ \text{m}$, points $A$ and $B$ at the same elevation (horizontal pipe).

For a differential U-tube manometer with both points carrying the same fluid (water) at the same level, the pressure difference is:

$$p_A - p_B = x\,g\,(\rho_{Hg} - \rho_w).$$

(The water columns of equal height on both limbs cancel because $A$ and $B$ are at the same elevation; only the net mercury-versus-water column over the reading $x$ contributes.)

Substitute:

$$\rho_{Hg} = 13.6\times1000 = 13600\ \text{kg/m}^3,\quad \rho_w = 1000\ \text{kg/m}^3.$$ $$p_A - p_B = 0.25 \times 9.81 \times (13600 - 1000).$$ $$p_A - p_B = 0.25 \times 9.81 \times 12600 = 0.25 \times 123606 = 30901.5\ \text{Pa}.$$

$p_A - p_B \approx 30.9\ \text{kPa}$.

Expressed as head of water:

$$h = \frac{p_A - p_B}{\rho_w g} = \frac{30901.5}{1000\times9.81} = 3.15\ \text{m of water}.$$

(a) Orifice coefficients

• Coefficient of contraction $C_c$: ratio of the cross-sectional area of the jet at the vena contracta to the area of the orifice, $C_c = a_c/a$.
• Coefficient of velocity $C_v$: ratio of the actual velocity of the jet at the vena contracta to the theoretical velocity $\sqrt{2gH}$, $C_v = V_{actual}/V_{theoretical}$.
• Coefficient of discharge $C_d$: ratio of actual discharge to theoretical discharge, $C_d = Q_{actual}/Q_{theoretical}$.

Relationship: $C_d = C_c \times C_v$.

(b) Calculations

Orifice area:

$$a = \frac{\pi}{4}(0.050)^2 = \frac{\pi}{4}(0.0025) = 1.96350\times10^{-3}\ \text{m}^2.$$

Theoretical velocity:

$$V_{th} = \sqrt{2gH} = \sqrt{2\times9.81\times3.0} = \sqrt{58.86} = 7.6720\ \text{m/s}.$$

Theoretical discharge:

$$Q_{th} = a\,V_{th} = 1.96350\times10^{-3}\times 7.6720 = 0.015064\ \text{m}^3/\text{s} = 15.064\ \text{L/s}.$$

Actual discharge $Q_{act} = 9.0\ \text{L/s} = 0.0090\ \text{m}^3/\text{s}.$

Coefficient of discharge:

$$C_d = \frac{Q_{act}}{Q_{th}} = \frac{9.0}{15.064} = 0.5974 \approx \mathbf{0.597}.$$

Coefficient of contraction (from $C_d = C_c C_v$):

$$C_c = \frac{C_d}{C_v} = \frac{0.5974}{0.97} = 0.6159 \approx \mathbf{0.616}.$$

These values ($C_d \approx 0.60$, $C_c \approx 0.62$) are typical for a sharp-edged orifice.