BE Civil Engineering (IOE, TU) Fluid Mechanics (IOE, CE 553) Question Paper 2077 Nepal

(a) Derivation

Consider an elementary horizontal strip of area $dA = b\,dy$ at depth $y$ below the free surface. The gauge pressure on the strip is $p = \rho g y$.

Force on the strip: $dF = \rho g y \, dA$.

Total force:

$$F = \int \rho g y\, dA = \rho g \int y\, dA = \rho g\,\bar{y}A$$

where $\bar{y} = \frac{1}{A}\int y\,dA$ is the depth of the centroid. Thus $\boxed{F = \rho g \bar{y} A}$.

For the centre of pressure, take moments of $dF$ about the free-surface line:

$$F\,y_p = \int y \, dF = \rho g \int y^2 dA = \rho g\, I_o$$

where $I_o = \int y^2 dA = I_c + A\bar{y}^2$ (parallel-axis theorem). Hence

$$y_p = \frac{\rho g I_o}{\rho g \bar y A} = \bar y + \frac{I_c}{A\bar y}$$

so $\boxed{y_p = \bar y + \dfrac{I_c}{A\bar y}}$, always below the centroid.

(b) Numerical evaluation

Centroid depth: $\bar y = 1 + \dfrac{3}{2} = 2.5\,\text{m}$. Area: $A = b\,h = 2 \times 3 = 6\,\text{m}^2$.

$$F = \rho g \bar y A = 1000 \times 9.81 \times 2.5 \times 6 = 147\,150\,\text{N} = \mathbf{147.15\,kN}$$

Second moment about centroidal axis: $I_c = \dfrac{b h^3}{12} = \dfrac{2 \times 3^3}{12} = 4.5\,\text{m}^4$.

$$y_p = \bar y + \frac{I_c}{A \bar y} = 2.5 + \frac{4.5}{6 \times 2.5} = 2.5 + 0.3 = \mathbf{2.8\,m}$$

(c) Force to keep gate closed

The hinge is at the bottom edge, which lies at depth $1 + 3 = 4\,\text{m}$. The line of action of $F$ acts at $y_p = 2.8\,\text{m}$, i.e. a vertical distance $4 - 2.8 = 1.2\,\text{m}$ above the hinge. The applied force $P$ acts at the top edge, $3\,\text{m}$ above the hinge.

Taking moments about the bottom hinge ($\sum M = 0$):

$$P \times 3 = F \times 1.2$$ $$P = \frac{147.15 \times 1.2}{3} = \mathbf{58.86\,kN}$$

A horizontal force of 58.86 kN applied at the top edge keeps the gate closed.

(a) Assumptions and derivation

Assumptions: (i) steady flow; (ii) incompressible fluid; (iii) inviscid (frictionless) flow; (iv) flow along a single streamline; (v) only gravity and pressure forces act.

Euler's equation along a streamline (coordinate $s$):

$$\frac{1}{\rho}\frac{\partial p}{\partial s} + V\frac{\partial V}{\partial s} + g\frac{\partial z}{\partial s} = 0$$

For steady flow, multiplying through by $ds$ and integrating:

$$\int \frac{dp}{\rho} + \int V\,dV + \int g\,dz = \text{const}$$

For incompressible flow ($\rho$ constant):

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

Dividing by $g$ gives Bernoulli's equation in head form:

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

The three terms are the pressure head, velocity head, and elevation head; their sum (total head) is constant along a streamline for ideal flow.

(b) Numerical solution

Areas:

$$A_1 = \frac{\pi}{4}(0.2)^2 = 0.031416\,\text{m}^2, \quad A_2 = \frac{\pi}{4}(0.1)^2 = 0.0078540\,\text{m}^2$$

Velocities from continuity $Q = AV$:

$$V_1 = \frac{0.03}{0.031416} = \mathbf{0.955\,m/s}, \quad V_2 = \frac{0.03}{0.0078540} = \mathbf{3.820\,m/s}$$

Apply Bernoulli between 1 and 2 (horizontal, $z_1 = z_2$):

$$p_2 = p_1 + \tfrac{1}{2}\rho(V_1^2 - V_2^2)$$ $$p_2 = 200\,000 + \tfrac{1}{2}(1000)(0.955^2 - 3.820^2)$$ $$p_2 = 200\,000 + 500(0.912 - 14.592) = 200\,000 - 6\,840 = 193\,160\,\text{Pa}$$ $$\boxed{p_2 \approx 193.16\,\text{kPa (gauge)}}$$

The pressure drops as the pipe contracts and the velocity (and velocity head) rises, consistent with Bernoulli's principle.

(a) Derivation

Apply Bernoulli between inlet (1) and throat (2) for a horizontal meter ($z_1 = z_2$) and continuity $A_1V_1 = A_2V_2$:

$$\frac{p_1 - p_2}{\rho g} = \frac{V_2^2 - V_1^2}{2g} = h$$

where $h$ is the pressure-head difference. With $V_1 = (A_2/A_1)V_2$:

$$h = \frac{V_2^2}{2g}\left(1 - \frac{A_2^2}{A_1^2}\right) \Rightarrow V_2 = \sqrt{\frac{2gh}{1 - (A_2/A_1)^2}}$$

Theoretical discharge $Q_{th} = A_2 V_2$. Introducing $C_d$:

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

(b) Numerical evaluation

Areas:

$$A_1 = \frac{\pi}{4}(0.3)^2 = 0.070686\,\text{m}^2,\quad A_2 = \frac{\pi}{4}(0.15)^2 = 0.017671\,\text{m}^2$$

Equivalent water head from the mercury manometer:

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

Denominator:

$$\sqrt{A_1^2 - A_2^2} = \sqrt{0.070686^2 - 0.017671^2} = \sqrt{0.0049965 - 0.0003123} = \sqrt{0.0046842} = 0.068441$$ $$Q = 0.98 \times \frac{0.070686 \times 0.017671}{0.068441}\sqrt{2 \times 9.81 \times 3.15}$$ $$= 0.98 \times \frac{0.0012490}{0.068441}\times \sqrt{61.80} = 0.98 \times 0.018249 \times 7.861$$ $$Q = 0.14061\,\text{m}^3/\text{s}$$ $$\boxed{Q \approx 0.1406\,\text{m}^3/\text{s} = 140.6\,\text{L/s}}$$

(a) Buckingham $\pi$-theorem

If a physical phenomenon involves $n$ dimensional variables and these are expressed in terms of $m$ fundamental dimensions (here $M$, $L$, $T$), then the relationship can be reduced to $(n - m)$ independent dimensionless groups, called $\pi$-terms. The functional relationship $f(X_1, X_2, \dots, X_n)=0$ becomes $\phi(\pi_1, \pi_2, \dots, \pi_{n-m}) = 0$.

(b) Application

Variables: $F, D, V, \rho, \mu$, so $n = 5$. Fundamental dimensions $M, L, T$, so $m = 3$. Number of $\pi$-terms $= n - m = 2$.

Dimensions:

$$[F] = MLT^{-2},\ [D] = L,\ [V] = LT^{-1},\ [\rho] = ML^{-3},\ [\mu] = ML^{-1}T^{-1}$$

Repeating variables: $\rho, V, D$ (they contain all three dimensions and are independent).

$\pi_1$ (containing $F$): $\pi_1 = \rho^a V^b D^c F$

$$M^0 L^0 T^0 = (ML^{-3})^a (LT^{-1})^b (L)^c (MLT^{-2})$$

• $M$: $a + 1 = 0 \Rightarrow a = -1$
• $T$: $-b - 2 = 0 \Rightarrow b = -2$
• $L$: $-3a + b + c + 1 = 0 \Rightarrow 3 - 2 + c + 1 = 0 \Rightarrow c = -2$

$$\pi_1 = \frac{F}{\rho V^2 D^2}$$

$\pi_2$ (containing $\mu$): $\pi_2 = \rho^a V^b D^c \mu$

$$M^0 L^0 T^0 = (ML^{-3})^a (LT^{-1})^b (L)^c (ML^{-1}T^{-1})$$

• $M$: $a + 1 = 0 \Rightarrow a = -1$
• $T$: $-b - 1 = 0 \Rightarrow b = -1$
• $L$: $-3a + b + c - 1 = 0 \Rightarrow 3 - 1 + c - 1 = 0 \Rightarrow c = -1$

$$\pi_2 = \frac{\mu}{\rho V D} = \frac{1}{Re}$$

Since $\pi_2$ and its reciprocal are equally valid dimensionless groups, replace $\pi_2$ by $Re = \rho V D/\mu$. The relationship $\phi(\pi_1, \pi_2)=0$ rearranges to:

$$\boxed{\frac{F}{\rho V^2 D^2} = \phi(Re)}$$

as required.

(a) Momentum equation

For steady flow through a control volume, the resultant external force equals the net rate of change of momentum of the fluid (Newton's second law applied to a control volume):

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

The force the fluid exerts on a boundary is equal and opposite to the force the boundary exerts on the fluid. This lets us compute thrusts on plates, vanes, bends and nozzles without knowing internal pressure details.

(b) Numerical solution

Jet area:

$$A = \frac{\pi}{4}(0.05)^2 = 0.0019635\,\text{m}^2$$

(i) Stationary plate. Discharge striking the plate $Q = AV = 0.0019635 \times 20 = 0.039270\,\text{m}^3/\text{s}$. The jet is brought to rest in the normal direction, so the force is:

$$F = \rho Q V = 1000 \times 0.039270 \times 20 = 785.4\,\text{N}$$ $$\boxed{F \approx 785.4\,\text{N}}$$

(ii) Plate moving at $u = 5\,\text{m/s}$ in the jet direction. The relative velocity is $(V - u) = 20 - 5 = 15\,\text{m/s}$. The mass rate actually intercepted is based on the relative velocity:

$$F = \rho A (V - u)^2 = 1000 \times 0.0019635 \times 15^2 = 1000 \times 0.0019635 \times 225 = 441.8\,\text{N}$$ $$\boxed{F \approx 441.8\,\text{N}}$$

(iii) Work done per second (power).

$$P = F \times u = 441.8 \times 5 = 2209\,\text{W} \approx \mathbf{2.21\,kW}$$

The moving plate delivers about 2.21 kW.

(a) Dynamic vs kinematic viscosity

• Dynamic (absolute) viscosity $\mu$: the constant of proportionality between shear stress and velocity gradient, $\tau = \mu\,\dfrac{du}{dy}$. SI unit: $\text{Pa·s}$ (= $\text{N·s/m}^2$ = $\text{kg/(m·s)}$).
• Kinematic viscosity $\nu$: the ratio of dynamic viscosity to mass density, $\nu = \mu/\rho$. It represents the diffusivity of momentum. SI unit: $\text{m}^2/\text{s}$.

(b) Capillary rise

For a tube, balancing the surface-tension pull against the weight of the raised column:

$$h = \frac{4\sigma\cos\theta}{\rho g d}$$

Substituting $\theta = 0^\circ$ ($\cos\theta = 1$), $d = 0.001\,\text{m}$:

$$h = \frac{4 \times 0.0728 \times 1}{1000 \times 9.81 \times 0.001} = \frac{0.2912}{9.81} = 0.02968\,\text{m}$$ $$\boxed{h \approx 29.7\,\text{mm}}$$

(a) Definitions

• Streamline: an imaginary line drawn in the flow such that its tangent at every point gives the direction of the velocity at that instant. No flow crosses a streamline.
• Pathline: the actual trajectory traced by a single fluid particle over a period of time.
• Streakline: the locus, at a given instant, of all particles that have previously passed through a fixed point (e.g. a dye filament).

For steady flow, streamlines, pathlines and streaklines all coincide.

(b) Continuity for 2-D incompressible flow

The continuity equation is:

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$$

Given $u = 3x + 2y$, so $\dfrac{\partial u}{\partial x} = 3$.

Therefore:

$$\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x} = -3$$

Integrating with respect to $y$:

$$v = -3y + f(x)$$

Applying the condition $v = 0$ at $y = 0$ gives $f(x) = 0$. Hence:

$$\boxed{v = -3y \ \text{m/s}}$$

(a) Reynolds number

$$Re = \frac{\rho V D}{\mu} = \frac{850 \times 2 \times 0.05}{0.05} = \frac{85}{0.05} = 1700$$

Since $Re = 1700 < 2000$, the flow is laminar.

(b) Head loss due to friction

For laminar flow the Darcy friction factor is:

$$f = \frac{64}{Re} = \frac{64}{1700} = 0.03765$$

Darcy–Weisbach head loss:

$$h_f = f\,\frac{L}{D}\,\frac{V^2}{2g} = 0.03765 \times \frac{10}{0.05} \times \frac{2^2}{2 \times 9.81}$$ $$= 0.03765 \times 200 \times \frac{4}{19.62} = 0.03765 \times 200 \times 0.20387 = 1.535\,\text{m}$$ $$\boxed{h_f \approx 1.54\,\text{m of oil}}$$

(Check via Hagen–Poiseuille: $h_f = \dfrac{32 \mu L V}{\rho g D^2} = \dfrac{32 \times 0.05 \times 10 \times 2}{850 \times 9.81 \times 0.05^2} = \dfrac{32}{20.846} = 1.535\,\text{m}$, confirming the result.)

(a) Definitions

• Boundary layer thickness ($\delta$): the distance from the wall at which the local velocity reaches $99\%$ of the free-stream velocity ($u = 0.99 U$).
• Displacement thickness ($\delta^*$): the distance by which the boundary would have to be displaced outward to compensate for the reduction in mass flow due to the boundary layer, $\delta^* = \int_0^\delta \left(1 - \dfrac{u}{U}\right)dy$.
• Momentum thickness ($\theta$): the loss of momentum flux due to the boundary layer expressed as an equivalent free-stream thickness, $\theta = \int_0^\delta \dfrac{u}{U}\left(1 - \dfrac{u}{U}\right)dy$.

(b) Boundary-layer thickness at $x = 2\,\text{m}$

Local Reynolds number:

$$Re_x = \frac{U x}{\nu} = \frac{3 \times 2}{1.5\times 10^{-5}} = \frac{6}{1.5\times 10^{-5}} = 4\times 10^{5}$$

Since $Re_x = 4\times 10^5 < 5\times 10^5$, the layer is laminar; Blasius applies:

$$\delta = \frac{5x}{\sqrt{Re_x}} = \frac{5 \times 2}{\sqrt{4\times 10^5}} = \frac{10}{632.46} = 0.01581\,\text{m}$$ $$\boxed{\delta \approx 15.8\,\text{mm}}$$

(a) Mean velocity

$$A = \frac{\pi}{4}(0.3)^2 = 0.070686\,\text{m}^2$$ $$V = \frac{Q}{A} = \frac{0.05}{0.070686} = 0.7074\,\text{m/s}$$ $$\boxed{V \approx 0.707\,\text{m/s}}$$

(b) Friction head loss (Darcy–Weisbach)

$$h_f = f\,\frac{L}{D}\,\frac{V^2}{2g} = 0.02 \times \frac{500}{0.3} \times \frac{0.7074^2}{2 \times 9.81}$$ $$= 0.02 \times 1666.7 \times \frac{0.5004}{19.62} = 0.02 \times 1666.7 \times 0.025505 = 0.8501\,\text{m}$$ $$\boxed{h_f \approx 0.85\,\text{m of water}}$$

(c) Power to overcome friction

$$P = \rho g Q h_f = 1000 \times 9.81 \times 0.05 \times 0.8501 = 417.0\,\text{W}$$ $$\boxed{P \approx 0.417\,\text{kW}}$$

(a) Orifice coefficients

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

Relation: $\boxed{C_d = C_v \times C_c}$.

(b) Actual discharge

Orifice area:

$$a = \frac{\pi}{4}(0.1)^2 = 0.0078540\,\text{m}^2$$

Theoretical velocity (Torricelli): $V_{th} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 4} = \sqrt{78.48} = 8.859\,\text{m/s}$.

Actual discharge:

$$Q = C_d\,a\sqrt{2gH} = 0.6 \times 0.0078540 \times 8.859 = 0.04175\,\text{m}^3/\text{s}$$ $$\boxed{Q \approx 0.0417\,\text{m}^3/\text{s} = 41.7\,\text{L/s}}$$