BE Civil Engineering (IOE, TU) Fluid Mechanics (IOE, CE 553) Question Paper 2076 Nepal

(a) Definitions

Dynamic (absolute) viscosity $\mu$ is the property of a fluid that quantifies its resistance to shear deformation. By Newton's law of viscosity, $\tau = \mu\,\dfrac{du}{dy}$. SI unit: $\text{Pa·s} = \text{N·s/m}^2$.

Kinematic viscosity $\nu = \dfrac{\mu}{\rho}$ is the ratio of dynamic viscosity to mass density. SI unit: $\text{m}^2/\text{s}$.

Temperature dependence: For liquids, viscosity decreases with increasing temperature because cohesive intermolecular forces (the dominant resistance) weaken as molecules gain thermal energy. For gases, viscosity increases with temperature because molecular momentum exchange between layers (the dominant mechanism) intensifies as molecular speed rises.

(b) Terminal velocity of plate

At terminal velocity the net force is zero: the gravity component down the slope equals the viscous resistance.

Driving force along incline:

$$F = W\sin\theta = 300\times\sin 30^{\circ} = 300\times0.5 = 150\,\text{N}$$

Shear stress from oil film (linear profile):

$$\tau = \mu\frac{V}{t}$$

Resisting force $=\tau\,A$, with $A = 0.8\times0.8 = 0.64\,\text{m}^2$ and $t = 1.2\,\text{mm}=0.0012\,\text{m}$.

Equate:

$$W\sin\theta = \mu\frac{V}{t}A$$ $$150 = 0.9\times\frac{V}{0.0012}\times0.64$$ $$150 = 480\,V$$ $$V = \frac{150}{480} = 0.3125\,\text{m/s}$$

Terminal velocity $V = 0.3125\,\text{m/s} \approx 0.313\,\text{m/s}$.

(c) Capillary rise

$$h = \frac{4\sigma\cos\theta}{\rho g d}$$ $$h = \frac{4\times0.0728\times\cos 0^{\circ}}{1000\times9.81\times0.002}$$ $$h = \frac{0.2912}{19.62} = 0.01484\,\text{m}$$

Capillary rise $h \approx 14.8\,\text{mm}$.

(a) Derivation

Consider a vertical plane surface submerged in a liquid of specific weight $\gamma=\rho g$. The centroid lies at depth $\bar{h}$. For an elemental strip of area $dA$ at depth $h$, pressure $p=\gamma h$, so $dF=\gamma h\,dA$.

$$F=\int \gamma h\,dA = \gamma\int h\,dA = \gamma\,\bar{h}\,A$$

because $\int h\,dA = \bar{h}A$ (first moment of area). Thus total force $F=\gamma\bar{h}A$.

The centre of pressure $h^{*}$ is found by taking moments about the free surface:

$$F\,h^{*}=\int \gamma h^{2}\,dA = \gamma I_{0}$$

where $I_{0}=\int h^{2}dA = I_{G}+A\bar{h}^{2}$ (parallel-axis theorem). Hence

$$h^{*}=\frac{I_{G}+A\bar{h}^{2}}{A\bar{h}}=\bar{h}+\frac{I_{G}}{A\bar{h}}$$

(b) Numerical force and centre of pressure

Area $A = 2\times3 = 6\,\text{m}^2$.

Depth of centroid: $\bar{h}=1.5+\dfrac{3}{2}=3.0\,\text{m}$.

$$F=\gamma\bar{h}A = 9810\times3.0\times6 = 176{,}580\,\text{N}=176.58\,\text{kN}$$

$I_{G}=\dfrac{b d^{3}}{12}=\dfrac{2\times3^{3}}{12}=\dfrac{54}{12}=4.5\,\text{m}^4$.

$$h^{*}=\bar{h}+\frac{I_{G}}{A\bar{h}}=3.0+\frac{4.5}{6\times3.0}=3.0+0.25=3.25\,\text{m}$$

$F=176.58\,\text{kN}$, centre of pressure $=3.25\,\text{m}$ below the free surface.

(c) Force at top edge

Take moments about the bottom hinge. The line of action of $F$ is at $h^{*}=3.25\,\text{m}$ below the surface. The bottom edge is at $1.5+3=4.5\,\text{m}$ below the surface, so the force acts at a height above the hinge:

$$y_{F}=4.5-3.25=1.25\,\text{m}$$

The applied force $P$ at the top edge acts $3\,\text{m}$ above the hinge. Moment balance about hinge:

$$P\times3 = F\times y_{F}=176.58\times1.25=220.725\,\text{kN·m}$$ $$P=\frac{220.725}{3}=73.575\,\text{kN}$$

Required horizontal force at top edge $P\approx73.6\,\text{kN}$.

(a) Assumptions of Bernoulli's equation

1. The fluid is ideal (inviscid, non-viscous) — no friction losses.
2. The flow is steady.
3. The fluid is incompressible (constant density).
4. The flow is irrotational / along a single streamline.
5. Only gravity and pressure forces act (no shaft work or heat addition).

For a real fluid, between sections 1 and 2 (energy equation with loss):

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

where $h_L$ is the head loss between the two sections.

(b) Discharge through venturimeter

Areas:

$$A_1=\frac{\pi}{4}(0.20)^2=0.031416\,\text{m}^2,\qquad A_2=\frac{\pi}{4}(0.10)^2=0.0078540\,\text{m}^2$$

Pressure head difference from manometer:

$$h=x\left(\frac{S_{m}}{S_{w}}-1\right)=0.250\,(13.6-1)=0.250\times12.6=3.15\,\text{m of water}$$

Venturimeter discharge:

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

Compute the geometric term:

$$A_1 A_2 = 0.031416\times0.0078540 = 2.4674\times10^{-4}\,\text{m}^4$$ $$A_1^{2}-A_2^{2}=(0.031416)^2-(0.0078540)^2 = 9.8696\times10^{-4}-6.1685\times10^{-5}=9.2528\times10^{-4}$$ $$\sqrt{A_1^{2}-A_2^{2}}=0.030418\,\text{m}^2$$ $$\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}=\frac{2.4674\times10^{-4}}{0.030418}=8.1116\times10^{-3}\,\text{m}^2$$

Velocity head term:

$$\sqrt{2gh}=\sqrt{2\times9.81\times3.15}=\sqrt{61.803}=7.8615\,\text{m/s}$$

Therefore:

$$Q=0.98\times8.1116\times10^{-3}\times7.8615 = 0.06250\,\text{m}^3/\text{s}$$

Discharge $Q\approx0.0625\,\text{m}^3/\text{s}=62.5\,\text{L/s}$.

(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 a relation among $(n-m)$ independent dimensionless groups, called $\pi$-terms.

(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: $\rho, V, D$.

$\pi_1$ term (with $F$):

$$\pi_1=\rho^{a}V^{b}D^{c}F$$ $$M^{0}L^{0}T^{0}=(ML^{-3})^{a}(LT^{-1})^{b}(L)^{c}(MLT^{-2})$$

Equate exponents:

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

So:

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

$\pi_2$ term (with $\mu$):

$$\pi_2=\rho^{a}V^{b}D^{c}\mu$$ $$M^{0}L^{0}T^{0}=(ML^{-3})^{a}(LT^{-1})^{b}(L)^{c}(ML^{-1}T^{-1})$$

Equate exponents:

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

So:

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

The functional relation $\pi_1=\phi_1(\pi_2)$:

$$\frac{F}{\rho V^{2}D^{2}}=\phi_1\!\left(\frac{\mu}{\rho V D}\right)$$

Since any function of $\dfrac{\mu}{\rho VD}$ is equivalently a function of its reciprocal $\dfrac{\rho VD}{\mu}$ (the Reynolds number), we may write

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

as required.

(a) Discharge including all losses

Applying the energy equation between the two reservoir surfaces (both at atmospheric pressure, negligible velocities), the total available head equals the sum of all losses:

$$H = h_{entry}+h_{friction}+h_{exit}$$ $$12 = 0.5\frac{V^{2}}{2g}+\frac{fL}{D}\frac{V^{2}}{2g}+1.0\frac{V^{2}}{2g}$$

Compute the friction multiplier:

$$\frac{fL}{D}=\frac{0.02\times300}{0.2}=\frac{6}{0.2}=30$$

So:

$$12 = \frac{V^{2}}{2g}\,(0.5+30+1.0)=\frac{V^{2}}{2g}\times31.5$$ $$\frac{V^{2}}{2g}=\frac{12}{31.5}=0.38095\,\text{m}$$ $$V^{2}=0.38095\times2\times9.81=7.4743$$ $$V=\sqrt{7.4743}=2.7340\,\text{m/s}$$

Area:

$$A=\frac{\pi}{4}(0.2)^2=0.031416\,\text{m}^2$$

Discharge:

$$Q=AV=0.031416\times2.7340=0.08589\,\text{m}^3/\text{s}$$

Discharge $Q\approx0.0859\,\text{m}^3/\text{s}=85.9\,\text{L/s}$.

(b) Percentage of head lost to friction

Friction head loss:

$$h_{friction}=30\times\frac{V^{2}}{2g}=30\times0.38095=11.4286\,\text{m}$$

Fraction of total head:

$$\frac{h_{friction}}{H}=\frac{11.4286}{12}=0.9524$$

About $95.2\%$ of the available head is dissipated as pipe friction loss (the remaining $\approx4.8\%$ being entry and exit minor losses).

(a) Continuity check

For 2-D incompressible flow: $\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}=0$.

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

Continuity is satisfied, so the flow is physically possible.

(b) Rotation check

The $z$-component of rotation: $\omega_z=\dfrac{1}{2}\left(\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial y}\right)$.

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

The flow is irrotational.

(c) Stream function

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

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 $-v$:

$$\frac{\partial\psi}{\partial x}=6xy+f'(x)=-v=6xy$$ $$\Rightarrow f'(x)=0 \Rightarrow f(x)=C\;(\text{constant})$$

Taking the constant as zero:

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

Velocities

$$A_1=\frac{\pi}{4}(0.4)^2=0.125664\,\text{m}^2,\quad V_1=\frac{Q}{A_1}=\frac{0.4}{0.125664}=3.1831\,\text{m/s}$$ $$A_2=\frac{\pi}{4}(0.2)^2=0.031416\,\text{m}^2,\quad V_2=\frac{Q}{A_2}=\frac{0.4}{0.031416}=12.732\,\text{m/s}$$

Outlet pressure (Bernoulli, horizontal, no loss)

$$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}$$ $$p_2=p_1+\frac{\rho}{2}(V_1^2-V_2^2)=120000+\frac{1000}{2}(3.1831^2-12.732^2)$$ $$=120000+500(10.132-162.103)=120000+500(-151.971)=120000-75985=44015\,\text{Pa}$$

Pressure forces

$$p_1A_1=120000\times0.125664=15079.6\,\text{N}$$ $$p_2A_2=44015\times0.031416=1382.8\,\text{N}$$

Momentum flux $\dot m = \rho Q = 1000\times0.4 = 400\,\text{kg/s}$.

x-direction (inlet along $+x$, outlet has no $x$-velocity):

$$F_x = p_1A_1 - p_2A_2\cos 90^{\circ} + \dot m(V_{1x}-V_{2x})$$

Let $R_x$ be the force from bend on water. Momentum equation:

$$p_1A_1 - 0 + R_x = \dot m (V_{2x}-V_{1x}) = 400(0-3.1831)=-1273.2\,\text{N}$$ $$R_x = -1273.2 - 15079.6 = -16352.8\,\text{N}$$

y-direction (outlet along $+y$, inlet has no $y$-velocity):

$$0 + R_y - p_2A_2 = \dot m(V_{2y}-V_{1y})=400(12.732-0)=5092.8\,\text{N}$$ $$R_y = 5092.8 + 1382.8 = 6475.6\,\text{N}$$

$R_x, R_y$ are the components of force from the bend on the water. The force from water on the bend is equal and opposite:

$$F_x = 16352.8\,\text{N},\qquad F_y = -6475.6\,\text{N}$$

Resultant magnitude

$$F=\sqrt{16352.8^{2}+6475.6^{2}}=\sqrt{2.6741\times10^{8}+4.1933\times10^{7}}=\sqrt{3.0934\times10^{8}}=17588\,\text{N}$$

Direction (below $+x$ axis):

$$\theta=\tan^{-1}\!\left(\frac{6475.6}{16352.8}\right)=\tan^{-1}(0.3960)=21.6^{\circ}$$

Resultant force on bend $\approx17.59\,\text{kN}$, acting at $21.6^{\circ}$ below the inlet axis (in the $+x,\,-y$ quadrant).

(a) Discharge (Hagen–Poiseuille)

For laminar flow in a circular pipe:

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

Data: $\Delta p=80000\,\text{Pa}$, $D=0.04\,\text{m}$, $\mu=0.10\,\text{Pa·s}$, $L=50\,\text{m}$.

$$D^{4}=(0.04)^4=2.56\times10^{-6}\,\text{m}^4$$ $$Q=\frac{\pi\times80000\times2.56\times10^{-6}}{128\times0.10\times50}$$

Numerator: $\pi\times80000\times2.56\times10^{-6}=\pi\times0.2048=0.643398$.

Denominator: $128\times0.10\times50=640$.

$$Q=\frac{0.643398}{640}=1.00531\times10^{-3}\,\text{m}^3/\text{s}$$

Discharge $Q\approx1.005\times10^{-3}\,\text{m}^3/\text{s}=1.005\,\text{L/s}$.

(b) Reynolds number check

Mean velocity:

$$A=\frac{\pi}{4}(0.04)^2=1.25664\times10^{-3}\,\text{m}^2$$ $$V=\frac{Q}{A}=\frac{1.00531\times10^{-3}}{1.25664\times10^{-3}}=0.80000\,\text{m/s}$$

Density $\rho=0.90\times1000=900\,\text{kg/m}^3$.

$$Re=\frac{\rho V D}{\mu}=\frac{900\times0.80\times0.04}{0.10}=\frac{28.8}{0.10}=288$$

Since $Re=288 < 2000$, the flow is laminar, confirming the validity of the Hagen–Poiseuille assumption.

(a) Definitions

• Boundary layer thickness $\delta$: the distance from the solid boundary to the point where the local velocity reaches $99\%$ of the free-stream velocity ($u=0.99U$).
• Displacement thickness $\delta^{*}$: the distance by which the boundary would have to be displaced outward so that the ideal (frictionless) flow carries the same mass flux as the actual viscous flow; $\delta^{*}=\int_0^{\delta}\left(1-\dfrac{u}{U}\right)dy$.
• Momentum thickness $\theta$: the distance equivalent to the loss of momentum flux due to the boundary layer; $\theta=\int_0^{\delta}\dfrac{u}{U}\left(1-\dfrac{u}{U}\right)dy$.

(b) Displacement thickness

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

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

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

Integrate term by term:

$$\int_0^1 1\,d\eta = 1$$ $$\int_0^1 \frac{3}{2}\eta\,d\eta = \frac{3}{2}\cdot\frac{1}{2}=\frac{3}{4}$$ $$\int_0^1 \frac{1}{2}\eta^{3}\,d\eta = \frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$$

Therefore:

$$\delta^{*}=\delta\left(1-\frac{3}{4}+\frac{1}{8}\right)=\delta\left(\frac{8-6+1}{8}\right)=\delta\cdot\frac{3}{8}$$

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

(a) The three coefficients

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

Relation: $C_d = C_v\times C_c$.

(b) Numerical determination

Coefficient of velocity from the jet trajectory. With horizontal $x=2.6\,\text{m}$, vertical drop $y=0.50\,\text{m}$, head $H=4\,\text{m}$:

$$C_v=\frac{x}{\sqrt{4yH}}=\frac{2.6}{\sqrt{4\times0.50\times4}}=\frac{2.6}{\sqrt{8}}=\frac{2.6}{2.8284}=0.9192$$

$C_v\approx0.919$.

Coefficient of discharge. Orifice area:

$$a=\frac{\pi}{4}(0.05)^2=1.96350\times10^{-3}\,\text{m}^2$$

Theoretical velocity:

$$V_{th}=\sqrt{2gH}=\sqrt{2\times9.81\times4}=\sqrt{78.48}=8.8589\,\text{m/s}$$

Theoretical discharge:

$$Q_{th}=a\,V_{th}=1.96350\times10^{-3}\times8.8589=1.73952\times10^{-2}\,\text{m}^3/\text{s}=17.395\,\text{L/s}$$

Actual discharge $Q=7.0\,\text{L/s}=7.0\times10^{-3}\,\text{m}^3/\text{s}$:

$$C_d=\frac{Q}{Q_{th}}=\frac{7.0\times10^{-3}}{1.73952\times10^{-2}}=0.40241$$

$C_d\approx0.402$.

Coefficient of contraction:

$$C_c=\frac{C_d}{C_v}=\frac{0.40241}{0.9192}=0.4378$$

$C_c\approx0.438$.

(a) Pressure-measuring devices

• Simple piezometer: a transparent vertical tube open at the top, inserted into the pipe; liquid rises to a height proportional to the gauge pressure ($p=\rho g h$). Suitable only for moderate, positive (above-atmospheric) pressures of liquids; cannot measure gas pressure, vacuum, or high pressures.
• U-tube manometer: a U-shaped tube containing a heavier manometric fluid (e.g. mercury). One limb connects to the point of measurement, the other is open to atmosphere. It can measure high pressures, gas pressures and vacuum by balancing the manometric-fluid column.
• Differential manometer: a U-tube whose two limbs connect to two different points; it directly gives the difference of pressures between those points, not the absolute/gauge value at one point.

(b) Pressure difference

We write a manometric balance from A to B through the fluid columns. Let the mercury level on side A be the datum. Given: pipe A is water, pipe B is oil (SG 0.85), manometric fluid mercury (SG 13.6), deflection $x=0.30\,\text{m}$ with the A-side mercury level lower.

Define heights:

• From centre of A down to the higher mercury level (on the B-limb side): the A-limb mercury is $0.30\,\text{m}$ lower, and A is $0.20\,\text{m}$ above the higher mercury level. So water column on side A above its (lower) mercury surface $= 0.20+0.30 = 0.50\,\text{m}$.
• Pipe B is $0.40\,\text{m}$ below pipe A. Its higher mercury level is $0.20\,\text{m}$ below A, i.e. B is $0.40-0.20=0.20\,\text{m}$ below that higher mercury level. So the oil column on side B above its mercury surface $=0.20\,\text{m}$.

Balance of pressures at the lower mercury surface (A-limb level), going A $\to$ down water $\to$ across mercury $\to$ up oil $\to$ B:

$$p_A + \rho_w g(0.50) - \rho_{Hg} g(0.30) - \rho_{oil} g(0.20) = p_B$$

Using $\gamma_w=9810\,\text{N/m}^3$:

$$p_A - p_B = -\,\gamma_w(0.50) + \gamma_{Hg}(0.30) + \gamma_{oil}(0.20)$$ $$= -9810(0.50) + 13.6\times9810\,(0.30) + 0.85\times9810\,(0.20)$$

Term by term:

$$-9810\times0.50 = -4905\,\text{Pa}$$ $$13.6\times9810\times0.30 = 133416\times0.30 = 40024.8\,\text{Pa}$$ $$0.85\times9810\times0.20 = 8338.5\times0.20 = 1667.7\,\text{Pa}$$

Sum:

$$p_A - p_B = -4905 + 40024.8 + 1667.7 = 36787.5\,\text{Pa}$$

$p_A - p_B \approx 36.79\,\text{kPa}$ (pressure at A is higher than at B).