BE Civil Engineering (IOE, TU) Fluid Mechanics (IOE, CE 553) Question Paper 2080 Nepal

(a) Definitions

• Dynamic (absolute) viscosity $\mu$ is the constant of proportionality between shear stress and velocity gradient in Newton's law of viscosity, $\tau = \mu\,\dfrac{du}{dy}$. SI unit: $\text{Pa·s}$ (= $\text{N·s/m}^2$ = $\text{kg/(m·s)}$).
• Kinematic viscosity $\nu = \mu/\rho$. SI unit: $\text{m}^2/\text{s}$.

Viscous resistance

Velocity gradient (linear profile across thin film):

$$\frac{du}{dy} = \frac{0.4}{1.2\times10^{-3}} = 333.33\ \text{s}^{-1}$$

Shear stress:

$$\tau = \mu\frac{du}{dy} = 0.85\times 333.33 = 283.33\ \text{Pa}$$

Viscous (resisting) force:

$$F_\mu = \tau A = 283.33\times 0.6 = \mathbf{170.0\ N}$$

Net driving force. The component of weight along the incline:

$$W\sin30^\circ = 250\times0.5 = 125\ \text{N}$$

Net force $= W\sin30^\circ - F_\mu = 125 - 170 = -45\ \text{N}$.

Comment. The viscous resistance ($170\ \text{N}$) exceeds the gravity component ($125\ \text{N}$), so the assumed steady downward velocity of $0.4\ \text{m/s}$ cannot be sustained by gravity alone; an external pull of $\mathbf{45\ N}$ down the slope is required to keep the plate moving at $0.4\ \text{m/s}$. Without it the plate would settle to a lower terminal speed where $W\sin30^\circ = F_\mu$, i.e. $u = \dfrac{125\times1.2\times10^{-3}}{0.85\times0.6} = 0.294\ \text{m/s}$.

(b) Capillary rise

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

Water:

$$h = \frac{4\times0.0728\times\cos0^\circ}{1000\times9.81\times1.5\times10^{-3}} = \frac{0.2912}{14.715} = 0.01979\ \text{m} \approx \mathbf{19.8\ mm\ (rise)}$$

Mercury: $\cos130^\circ = -0.6428$

$$h = \frac{4\times0.485\times(-0.6428)}{13600\times9.81\times1.5\times10^{-3}} = \frac{-1.2470}{200.12} = -0.006231\ \text{m} \approx \mathbf{-6.23\ mm}$$

Interpretation. Water wets glass ($\theta<90^\circ$) so $\cos\theta>0$ giving a positive rise of $19.8\ \text{mm}$. Mercury does not wet glass ($\theta>90^\circ$) so $\cos\theta<0$ giving a negative value, i.e. a capillary depression of $6.23\ \text{mm}$ below the free surface.

(a) Derivation

Consider a vertical surface; depth measured from the free surface. An elemental strip of area $dA$ at depth $h$ experiences pressure $p=\rho g h$, so

$$F = \int \rho g h\, dA = \rho g \int h\, dA = \rho g\, \bar h\, A$$

where $\bar h$ is the depth of the centroid (since $\int h\,dA=\bar h A$).

The centre of pressure $h_{cp}$ is found by taking moments about the free surface:

$$F\,h_{cp} = \int \rho g h^2\, dA = \rho g\, I_o$$

Using the parallel-axis theorem $I_o = I_G + A\bar h^2$,

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

(b) Numerical values

• Centroid depth: $\bar h = 1.5 + 3/2 = 3.0\ \text{m}$
• Area: $A = 2\times3 = 6\ \text{m}^2$
• $I_G = \dfrac{b d^3}{12} = \dfrac{2\times3^3}{12} = 4.5\ \text{m}^4$

Total force:

$$F = \rho g \bar h A = 1000\times9.81\times3.0\times6 = 176580\ \text{N} \approx \mathbf{176.6\ kN}$$

Centre of pressure:

$$h_{cp} = 3.0 + \frac{4.5}{6\times3.0} = 3.0 + 0.25 = \mathbf{3.25\ m\ below\ surface}$$

(c) Hinge / holding force

Measured from the top edge (the hinge), the line of action of $F$ acts at

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

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

$$P\times 3.0 = F\times y_{cp} = 176.58\times1.75 = 309.02\ \text{kN·m}$$ $$P = \frac{309.02}{3.0} = \mathbf{103.0\ kN}$$

A horizontal force of about $103\ \text{kN}$ at the bottom edge is required to hold the gate closed.

(a) Continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 2 + (-2) = 0$$

Continuity is satisfied, so the flow is a possible incompressible flow.

(b) Rotationality The $z$-component of vorticity:

$$\omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 3 - 3 = 0$$

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

(c) Stream function ($u = \partial\psi/\partial y,\ v = -\partial\psi/\partial x$):

$$\psi = \int u\, dy = \int (2x+3y)\,dy = 2xy + \tfrac{3}{2}y^2 + f(x)$$

Differentiate w.r.t. $x$: $\partial\psi/\partial x = 2y + f'(x) = -v = -(3x-2y) = -3x + 2y$, so $f'(x) = -3x$, $f(x) = -\tfrac{3}{2}x^2$.

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

Velocity potential ($u = \partial\phi/\partial x,\ v = \partial\phi/\partial y$):

$$\phi = \int u\, dx = \int(2x+3y)\,dx = x^2 + 3xy + g(y)$$

Differentiate w.r.t. $y$: $3x + g'(y) = v = 3x - 2y \Rightarrow g'(y) = -2y \Rightarrow g(y) = -y^2$.

$$\boxed{\phi = x^2 + 3xy - y^2 + C}$$

(d) Velocity at (2, 1)

$$u = 2(2)+3(1) = 7\ \text{m/s}, \qquad v = 3(2)-2(1) = 4\ \text{m/s}$$ $$V = \sqrt{7^2+4^2} = \sqrt{65} = \mathbf{8.06\ m/s}$$

Direction: $\theta = \tan^{-1}(v/u) = \tan^{-1}(4/7) = \mathbf{29.7^\circ}$ above the $x$-axis.

Velocities ($A=\frac{\pi}{4}D^2$):

$$A_1 = \tfrac{\pi}{4}(0.3)^2 = 0.070686\ \text{m}^2,\quad V_1 = \frac{0.25}{0.070686} = 3.537\ \text{m/s}$$ $$A_2 = \tfrac{\pi}{4}(0.2)^2 = 0.031416\ \text{m}^2,\quad V_2 = \frac{0.25}{0.031416} = 7.958\ \text{m/s}$$

Outlet pressure (Bernoulli, horizontal, no loss):

$$p_2 = p_1 + \tfrac{1}{2}\rho(V_1^2 - V_2^2) = 200000 + \tfrac{1}{2}(1000)(3.537^2 - 7.958^2)$$ $$= 200000 + 500(12.51 - 63.33) = 200000 - 25410 = 174590\ \text{Pa}$$

Pressure forces:

$$p_1A_1 = 200000\times0.070686 = 14137\ \text{N}$$ $$p_2A_2 = 174590\times0.031416 = 5485\ \text{N}$$

Mass flow: $\dot m = \rho Q = 1000\times0.25 = 250\ \text{kg/s}$.

Take $x$ along the inlet direction, $y$ transverse. Inlet flow is along $+x$; outlet flow makes $45^\circ$ with $+x$.

x-momentum (force on water $R_x'$):

$$p_1A_1 - p_2A_2\cos45^\circ + R_x' = \dot m(V_2\cos45^\circ - V_1)$$ $$14137 - 5485(0.7071) + R_x' = 250(7.958\times0.7071 - 3.537)$$ $$14137 - 3879 + R_x' = 250(5.627 - 3.537) = 250(2.090) = 522.5$$ $$R_x' = 522.5 - 10258 = -9735\ \text{N}$$

y-momentum:

$$0 - p_2A_2\sin45^\circ + R_y' = \dot m(V_2\sin45^\circ - 0)$$ $$-3879 + R_y' = 250(5.627) = 1406.8$$ $$R_y' = 1406.8 + 3879 = 5286\ \text{N}$$

$R_x', R_y'$ are forces on the water; the force on the bend is equal and opposite:

$$F_x = 9735\ \text{N},\qquad F_y = -5286\ \text{N}$$

Resultant:

$$F = \sqrt{9735^2 + 5286^2} = \sqrt{9.476\times10^7 + 2.794\times10^7} = \sqrt{1.227\times10^8} = \mathbf{11.08\ kN}$$ $$\theta = \tan^{-1}\!\left(\frac{5286}{9735}\right) = \mathbf{28.5^\circ}\ \text{below the inlet axis.}$$

(a) Energy equation between reservoir surfaces

The available head equals the sum of all losses:

$$H = \left(k_e + \frac{fL}{D} + k_x\right)\frac{V^2}{2g}$$

Friction term:

$$\frac{fL}{D} = \frac{0.02\times600}{0.25} = 48.0$$

Total loss coefficient:

$$K = 0.5 + 48.0 + 1.0 = 49.5$$

Solve for $V$:

$$20 = 49.5\times\frac{V^2}{2\times9.81} \Rightarrow V^2 = \frac{20\times19.62}{49.5} = 7.927 \Rightarrow V = 2.815\ \text{m/s}$$

Area: $A = \frac{\pi}{4}(0.25)^2 = 0.049087\ \text{m}^2$.

$$Q = AV = 0.049087\times2.815 = 0.1382\ \text{m}^3/\text{s} \approx \mathbf{138.2\ L/s}$$

(b) Distribution of head

Velocity head: $\dfrac{V^2}{2g} = \dfrac{7.927}{19.62} = 0.4040\ \text{m}$.

Friction accounts for 97 % of the total head; the combined minor losses are only about 3 %.

Comment. For this long pipe ($L/D = 2400$) the minor losses are small relative to friction and could be neglected for a quick estimate (doing so gives $V=\sqrt{20\times19.62/48}=2.859\ \text{m/s}$, $Q=0.1404\ \text{m}^3/\text{s}$, an error of only $\approx1.6\%$). For short pipes, however, minor losses dominate and must be retained.

Set-up. For a differential manometer in which both pipe points carry the same fluid (oil), the net manometer relation is

$$p_A - p_B = \Delta h\,(\gamma_{Hg} - \gamma_{oil}) + \gamma_{oil}\,(z_B - z_A)$$

where $\Delta h = 0.25\ \text{m}$ is the mercury deflection and $z_B - z_A = 0.4\ \text{m}$ (B is higher). The mercury being higher on the $B$-limb is consistent with $p_A > p_B$.

Specific weights (using $\gamma_w = 9810\ \text{N/m}^3$):

$$\gamma_{oil} = 0.85\times9810 = 8338.5\ \text{N/m}^3$$ $$\gamma_{Hg} = 13.6\times9810 = 133416\ \text{N/m}^3$$

Compute:

$$p_A - p_B = 0.25(133416 - 8338.5) + 8338.5(0.4)$$ $$= 0.25(125077.5) + 3335.4 = 31269.4 + 3335.4 = 34604.8\ \text{Pa}$$ $$\boxed{p_A - p_B \approx \mathbf{34.6\ kPa}}$$

The positive value confirms $p_A > p_B$, consistent with the mercury being pushed lower on the $A$-side limb.

Areas:

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

Equivalent head of water from the mercury manometer:

$$h = x\left(\frac{S_{Hg}}{S_w} - 1\right) = 0.180\left(\frac{13.6}{1} - 1\right) = 0.180\times12.6 = 2.268\ \text{m of water}$$

Discharge formula:

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

Denominator:

$$A_1^2 - A_2^2 = (0.031416)^2 - (0.0078540)^2 = 9.8696\times10^{-4} - 6.1685\times10^{-5} = 9.2528\times10^{-4}$$ $$\sqrt{A_1^2 - A_2^2} = 0.030418\ \text{m}^2$$

Numerator factor:

$$\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}} = \frac{0.031416\times0.0078540}{0.030418} = \frac{2.4674\times10^{-4}}{0.030418} = 8.112\times10^{-3}$$ $$\sqrt{2gh} = \sqrt{2\times9.81\times2.268} = \sqrt{44.50} = 6.671\ \text{m/s}$$

Discharge:

$$Q = 0.98\times 8.112\times10^{-3}\times 6.671 = \mathbf{0.05304\ m^3/s} \approx \mathbf{53.0\ L/s}$$

(a) Buckingham $\pi$-theorem. If a physical problem involves $n$ dimensional variables that can be expressed using $m$ fundamental dimensions, then the relationship can be reduced to a relation among $(n-m)$ independent dimensionless groups ($\pi$-terms).

(b) Variables: $F, V, L, \rho, \mu, g$ → $n = 6$. Fundamental dimensions $M, L, T$ → $m = 3$. Number of $\pi$-terms $= 6 - 3 = 3$.

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

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

$\pi_1$ (with $F$): $\pi_1 = \rho^a V^b L^c F$. $M:\ a+1=0\Rightarrow a=-1$. $T:\ -b-2=0\Rightarrow b=-2$. $L:\ -3a+b+c+1=0\Rightarrow 3 -2 + c +1=0\Rightarrow c=-2$.

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

$\pi_2$ (with $\mu$): $\pi_2 = \rho^a V^b L^c \mu$. $M:\ a+1=0\Rightarrow a=-1$. $T:\ -b-1=0\Rightarrow b=-1$. $L:\ -3a+b+c-1=0\Rightarrow 3-1+c-1=0\Rightarrow c=-1$.

$$\pi_2 = \frac{\mu}{\rho V L} = \frac{1}{Re}\ \Rightarrow\ \text{use its inverse } \frac{\rho V L}{\mu}=Re$$

$\pi_3$ (with $g$): $\pi_3 = \rho^a V^b L^c g$. $M:\ a=0$. $T:\ -b-2=0\Rightarrow b=-2$. $L:\ -3a+b+c+1=0\Rightarrow -2+c+1=0\Rightarrow c=1$.

$$\pi_3 = \frac{gL}{V^2} = \frac{1}{Fr^2}\ \Rightarrow\ \text{use } \frac{V}{\sqrt{gL}}=Fr$$

Therefore

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

The two groups are the Reynolds number $Re = \rho VL/\mu$ (viscous similarity) and the Froude number $Fr = V/\sqrt{gL}$ (gravity/wave-making similarity). For ship-model testing Froude similarity governs wave resistance.

(a) Reynolds number

$$Re = \frac{\rho V D}{\mu} = \frac{900\times1.2\times0.05}{0.1} = \frac{54}{0.1} = 540$$

Since $Re = 540 < 2000$, the flow is laminar.

(b) Hagen–Poiseuille pressure drop

$$\Delta p = \frac{32\,\mu\,V\,L}{D^2} = \frac{32\times0.1\times1.2\times80}{(0.05)^2} = \frac{307.2}{0.0025} = 122880\ \text{Pa}$$ $$\boxed{\Delta p \approx \mathbf{122.9\ kPa}}$$

(Equivalent head loss $h_f = \Delta p/\rho g = 122880/(900\times9.81) = 13.92\ \text{m}$ of oil.)

(c) Maximum velocity. For laminar pipe flow the centre-line velocity is twice the mean:

$$u_{max} = 2V = 2\times1.2 = \mathbf{2.4\ m/s}$$

(a) Definitions

• Boundary-layer thickness $\delta$: the distance from the wall at which the local velocity reaches $99\%$ of the free-stream velocity ($u = 0.99U$).
• Displacement thickness $\delta^*$: the distance by which the boundary is conceptually displaced outward to account for the reduction in mass flow due to the velocity deficit,

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

• Momentum thickness $\theta$: the thickness representing the loss of momentum flux in the boundary layer,

$$\theta = \int_0^\delta\frac{u}{U}\left(1 - \frac{u}{U}\right)dy.$$

(b) Boundary-layer thickness at $x = 0.5\ \text{m}$

Local Reynolds number:

$$Re_x = \frac{U x}{\nu} = \frac{3\times0.5}{1.5\times10^{-5}} = \frac{1.5}{1.5\times10^{-5}} = 1.0\times10^{5}$$

Since $Re_x = 1.0\times10^5 < 5\times10^5$ (critical), the flow is laminar at this location.

Blasius thickness:

$$\delta = \frac{5.0\,x}{\sqrt{Re_x}} = \frac{5.0\times0.5}{\sqrt{1.0\times10^5}} = \frac{2.5}{316.23} = 0.007906\ \text{m}$$ $$\boxed{\delta \approx \mathbf{7.91\ mm}}$$

(For reference, displacement thickness $\delta^* = 1.721x/\sqrt{Re_x} = 1.721\times0.5/316.23 = 2.72\ \text{mm}$.)

(a) Bernoulli's equation

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

The three terms are the pressure head $p/\rho g$, the velocity (kinetic) head $V^2/2g$, and the elevation (potential) head $z$, all in metres of fluid.

Assumptions: (i) steady flow, (ii) incompressible fluid, (iii) inviscid/frictionless (no energy loss), (iv) flow along a streamline, (v) no energy added/removed by machines.

(b) Pressure at section 2

Areas and velocities:

$$A_1 = \tfrac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2,\quad V_1 = \frac{0.04}{0.017671} = 2.264\ \text{m/s}$$ $$A_2 = \tfrac{\pi}{4}(0.10)^2 = 0.0078540\ \text{m}^2,\quad V_2 = \frac{0.04}{0.0078540} = 5.093\ \text{m/s}$$

Apply Bernoulli (1 → 2), solving for $p_2$:

$$p_2 = p_1 + \tfrac{1}{2}\rho(V_1^2 - V_2^2) + \rho g(z_1 - z_2)$$

Compute each term (Pa):

• $p_1 = 250000$
• $\tfrac{1}{2}\rho(V_1^2 - V_2^2) = 0.5\times1000\times(2.264^2 - 5.093^2) = 500\times(5.126 - 25.94) = 500\times(-20.81) = -10406$
• $\rho g(z_1 - z_2) = 1000\times9.81\times(0 - 4) = -39240$

$$p_2 = 250000 - 10406 - 39240 = 200354\ \text{Pa}$$ $$\boxed{p_2 \approx \mathbf{200.4\ kPa}}$$

The pressure falls because the fluid both rises (gaining elevation head) and accelerates (gaining velocity head) at the expense of pressure head.