Concept:

Young’s Modulus of elasticity (Y):

It is the ratio of Longitudinal (tensile or compressive) stress (σ) to the longitudinal strain (ϵ) is defined as Young’s modulus and is denoted by the symbol Y. (it is valid within the elastic limit)

$i.e.,Y=\frac{Langitudinalstress\left(\sigma \right)}{Langitudinalstrain\left(ϵ\right)}=\frac{\frac{F}{A}}{\frac{\mathrm{\Delta }L}{L}}$

The bulk modulus of elasticity (Β): It is the ratio of Hydraulic (compressive) stress (p) to the volumetric strain (ΔV/V) is defined as Bulk modulus and is denoted by the symbol K. (it is valid with in the elastic limit)

$i.e.,B=-\frac{p}{\frac{\mathrm{\Delta }V}{V}}$

And a unit of B is N/m2 or Pa (Pascal) since 1 N/m2 = 1 Pa

Modulus of rigidity or shear modulus of elasticity (η): It is the ratio of tangential stress to the shearing strain θ is defined as modulus of rigidity or shear modulus of elasticity and it is denoted by the symbol of η.

$i.e.,\eta =\frac{tangentialstress}{shearingstrain}=\frac{\frac{F}{A}}{\theta }or\frac{FL}{A\mathrm{\Delta }x}$

Explanation:

Given,

P₁ = 0.40 MPa, P₂ = 12.3 MPa, dV/V = 1.2%

dP = 12.3 – 0.40 = 11.9 MPa

$\frac{dV}{V}=1.2\mathrm{\%}=-0.012$

Bulk modulus of elasticity,

$k=-\frac{dP}{\left(\frac{dV}{V}\right)}=\frac{11.9}{0.012}=991.67MPa$

Hence the modulus of elasticity of the liquid is 991.7 MPa

Note most students get confused about the application of different modulus of elasticity, thus each modulus of elasticity can be summarized as 

Young's modulus ⇒ Used in linear expansion

Bulk Modulus/ Compressibility ⇒ Used for Volumetric expansion 

Rigidity modulus ⇒ Used to measure the change in the shape of an object due to deforming force 

Concept:

Force exerted by the jet on a moving plate is given by

F = ρ A (v - u)²

where ρ is the density of flowing fluid, A is the cross-sectional area of the jet, v is the velocity of the jet and u is the velocity of the moving plate.

Calculation:

Given:

ρ = 1000 kg/m³, v = 20 m/s, u = 10 m/s, A = 0.02 m²

F = ρ A (v - u)²

F = 1000 × 0.02 × (20 - 10)² = 2000 N

Concept:

$V=\sqrt{2g{h}_a}$

Calculation:

Given:

ρ_a = 1.2 kg/m³, deflection of the manometer = 24 mm

Now,

ρ_ah_a = ρ_wh_w

1.2 × h_a = 1000 × 24 × 10^-3

h_a = 20

$=\sqrt{2\times 10\times 20}$

Concept:

Given:

h_L = constant in both the pipes

$h_l=\frac{fL{Q}^2}{12D^5}$      ---(1)

Calculation:

Given:

Pipe-1:

L₁ = 100 m, D₁ = 200 mm, f₁ = 0.015

Pipe – 2

D₂ = 400 mm, f₂ = 0.012, L₂ =?

Q = Constant in both the pipes

Now,

From   (1),

$h_l=\frac{f_1{L}_1{Q}^2}{12D_1^5}=\frac{f_2{L}_2{Q}^2}{12D_2^5}$

$\frac{0.015\times 100\times Q^2}{12\times {0.2}^5}=\frac{0.012\times L_2\times Q^2}{12\times {0.4}^5}$

∴ L₂ = 4000 m

Now,

We know

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

For pipe-1 (D = 200 mm, f = .015)

$0.015=\frac{64}{Re}$

Re = 4266 (>2300), so it is turbulent flow

For pipe-2 (D = 400 mm, f = .012)

$0.012=\frac{64}{Re}$

Bernoulli’s Equation is known as the conservation of energy principle and states that in a steady, ideal flow of an incompressible fluid, the total energy at any point of the fluid is constant. The total energy consists of pressure energy, kinetic energy and potential energy or datum energy. And no external force acts on the liquid into the consideration except gravity. 

$\frac{P}{w}+\frac{v^2}{2g}+z=constant$

The following are the assumptions made in the derivation of Bernoulli's equation:

1. The fluid is ideal i.e. viscosity is zero
2. The flow is steady
3. The flow is incompressible
4. The flow is one dimensional

Concept:

Q = nAV

Q = Discharge, nA = Effective area

V = Velocity of the jet = $C_v\sqrt{2gH}$

Calculation:

Given:

n = 5 × 2 × 4 = 40, H = Head developed = 250 m

C_v = 0.985, Q = 40 m³/s

$V=C_v\sqrt{2gH}=0.985\times \sqrt{2\times 9.81\times 250}=68.985m/s$

Now,

Q = nAV

$Q=n\frac{\pi D^2}{4}V$

$40=40\times \frac{\pi D^2}{4}\times 68.985$

Concept:

When pipes are in Series:

1. The discharge will remain the same.

2. Total head loss = Sum of head loss of all the pipes connected in series

When pipes are in Parallel:

1.  Head loss in each pipe will remain the same

2. The discharge will be sum of the discharges in the individual pump.

Explanation:

Based on the above concept,

Pipe 1 and 3 are in parallel ⇒ h_f1 = h_f3

Concept:

To solve this, apply the Bernoulli Equation between points A and Summit.

For Point A

The pressure at A is atmospheric i.e. Gauge pressure = 0

Velocity head at A = 0

Elevation head at A = 0 (assumed datum at the water level of A) 

For point S

The pressure at S = P (Let)

Velocity head at S = 1 m of water (given)

Elevation head at S = 10 m (see figure)

Calculation:

Apply Bernoulli equation b/w (A) and (S)

$\frac{P_A}{\rho g}+\frac{V_A^2}{2g}+z_A=\frac{P_S}{\rho g}+\frac{V_S^2}{2g}+z_S+h_{AS}$

$0+0+0=\frac{P}{\rho g}+1+10+1.5$

$\Rightarrow \frac{P}{\rho g}atS=-12.5mof{H}_{2}O$

The separation point S is determined from the condition

${\left(\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\right)}_{\mathrm{y}=\mathrm{o}}=0$

So, If

${\left(\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\right)}_{\mathrm{y}=0}<0\Rightarrow \mathrm{f}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{e}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}$ 

$\begin{array}{l}{\left(\frac{\partial u}{\partial y}\right)}_{y=0}\Rightarrow \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{f}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{o}\mathrm{n}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{e}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\\ {\left(\frac{\partial u}{\partial y}\right)}_{y=0}>0\Rightarrow \mathrm{F}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{s}\mathrm{e}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\end{array}$

So, $\frac{u}{v}=\frac{5}{2}\left(\frac{y}{\delta }\right)-\frac{1}{2}{\left(\frac{y}{\delta }\right)}^2+\frac{2}{3}{\left(\frac{y}{\delta }\right)}^3$

$\begin{array}{l}\frac{\partial u}{\partial y}=\frac{5v}{2\delta }-\frac{yv}{{\delta }^2}+\frac{2y^2}{{\delta }^3}\\ {\left(\frac{{\partial }_u}{{\partial }_y}\right)}_{y=0}=\frac{5}{2\delta }v>0\end{array}$

⇒ The flow will not separate.

Due to surface tension, the pressure difference will occur at a curved interface i.e. air-water interface and this pressure difference is given by:

$P_i-P_o=\frac{2\sigma }{R}$

Where,

R is the radius of air bubble

σ is the surface tension of the air bubble

P_i is pressure inside the air bubble.

P_o is pressure outside air bubble

Since the air bubble is surrounded by water, therefore, the outside pressure will be equal to the hydrostatic pressure at level of air bubble due to water. It is given by

P₀ = ρgh

Where

h is the height of air bubble from top of container

ρ is the density of water.

Calculations:

Refer the following figure:

Air bubble is formed at a height of 5 cm from the bottom of the container, hence the height of the air bubble from the top of the container, h = 15 - 5 = 10 cm.

Outside pressure, P₀ = ρgh  = 1000 × 9.81 × 0.1 = 981 N/m².

Now,

$P_i-P_o=\frac{2\sigma }{R}$

$\Rightarrow P_i-981=\frac{2\times 0.072}{0.5\times {10}^{-3}}$

$P_i-981=288$

∴P_i = 1269 N/m2

 The pressure inside air Bubble, P_i = 1269 N/m².

Explanation:

Turbulent flow – Turbulent flow occurs at relatively larger velocities and is characterized by chaotic behaviour, irregular motion, large mixing and eddies. For such flows, inertial effects are more pronounced than the viscous effects.

• Mathematically the velocity field of turbulent flow is represented as$V=\overline{V}+V^{\prime }$ or the velocity fluctuates at small time scales around a large time-averaged velocity.
• Similarly,  $P=\overline{P}+P^{\prime },T=\overline{T}+T^{\prime }$etc.
• Parameter Reynolds number $\left(\frac{\rho vd}{\mu }\right)$ is used to characterize Laminar and Turbulent flow.
• If R_e < 2100 for pipe flow the flow is laminar and if R_e > 10⁴ the flow is turbulent.

For Turbulent flow, the velocity or pressure field may not be exactly (Analytically) represented. So we will solve it using dimensional analysis.

   $\mathrm{\Delta }P=\mathrm{\Delta }P\left(D,L,\overline{V,}\rho ,\mu ,\epsilon \right)$      where ε = Surface roughness

$\frac{\mathbf{\Delta }P}{\rho \overline{V^2}}=f\left(\frac{\rho vd}{\mu },\frac{L}{D},\frac{\epsilon }{D}\right)=f\left(R_e,\frac{L}{D},\frac{\epsilon }{D}\right)$,  the turbulent flow through the pipe turbulent pressure drop is a function of the square of velocity.

• The experimental observations suggest that pressure drop in the pipe flow under turbulent conditions depends upon Reynolds number and surface roughness.

Concept

Let H be the height at which water level in the tank becomes steady.

At steady state:

Rate at which water enters to tank = Rate at which water flow from orifice = 10 l/sec (given)

Now,

Flow through orifice, $Q=a_o{C}_d\sqrt{2gH}$

Where ; a_o = area of orifice

H is steady state head

Calculation

Given: C_d = 0.6, d_o = 10 cm, Q = 10 l/sec

$10\times {10}^{-3}=\frac{\pi }{4}\times {\left(0.1\right)}^2\times 0.6\times \sqrt{2\times 9.81\times H}$

Concept:

Hydraulic efficiency of the Pelton wheel is given by:

${\eta }_h=\frac{Runnerpower}{Kineticenergypersecond}=\frac{2\times \left(V_1-u\right)\times (1+k\cos \varphi )u}{V_1^2}$

For maximum efficiency,

When u = V₁/2

$({\eta }_h{)}_{max}=\left(\frac{1+k\cos \varphi }{2}\right)$

k = friction factor for blades

Calculation:

Given:

ϕ = 60°

As friction is neglected.

Maximum efficiency is:

$({\eta }_h{)}_{max}=\left(\frac{1+k\cos \varphi }{2}\right)$

$({\eta }_h{)}_{max}=\left(\frac{1+\cos {60}^{\circ }}{2}\right)=\frac{1+\frac{1}{2}}{2}=\frac{3}{4}\times 100=75\mathrm{\%}$

Concept:

For a Newtonian fluid,

$Shearstress\left(\tau \right)=dynamicvisocity\left(\mu \right)\times rateofshearstrain\left(\frac{du}{dy}\right)$

For the above situation, 

$\tau =\mu \times \frac{V}{y}$

$\frac{mg\sin \theta }{A}=\mu \times \frac{V}{y}$

where, A = Area in contact of a Newtonian fluid, V = Terminal velocity of the block

Calculation:

Given:

Edge of the block, a = 100 mm = 0.1 m , weight, W = 1 kN, θ = 30°, y = 0.02 mm, $\mu =0.2\frac{Ns}{m^2}$

Shear stress acting at the face of the cube in contact with fluid,

$\tau =\mu \times \frac{V}{y}$

$\frac{mg\sin \theta }{A}=\mu \times \frac{V}{y}$

$\frac{1000\times 0.5}{0.1\times 0.1}=0.2\times \frac{V}{0.02\times {10}^{-3}}$

V = 5 m/s

Concept:

Hydraulic press works on the principle of Pascal’s law,

The pressure intensity on plunger will be transmitted equally as pressure intensity at ram.

Since no frictional losses, W_in = W_output ⇒ F_p × h_P = F_R × h_R

Where h is the movement of plunger/ram.

Mechanical advantage = F_output/F_input = F_R/F_P

Calculation:

Given:

D_P = 5 cm = 0.05 m; D_R = 40 cm = 0.4 m; P_base ≤ 260 kPa; h = 1 m (from base);

A_P = π × 0.05²/4 = 1.96 × 10^-3 m²; A_R = π × 0.4²/4 = 0.1256 m²

The maximum weight to be lifted is limited by the base of press as the pressure intensity applied at plunger will equally transmit to the base also.

⇒ P_plunger + ρgh = P_base ⇒ P_plunger + ρgh ≤ 260 kPa

⇒ P_plunger + 10³ × 10 × 1 ≤ 260 × 10³ ⇒ P_plunger ≤ 250 kPa

∴ Maximum pressure that can be applied at plunger = 250 kPa

⇒ P_ram (max) = 250 kPa

⇒ F_ram (max) = W_max = P_ram (max) × A_R = 250 × 0.1256 = 31.4 kN (Option 1)

Now,

Always P_plunger = P_ram

⇒ $\frac{F_P}{A_P}=\frac{F_R}{A_R}\Rightarrow \frac{F_R}{F_P}=\frac{A_R}{A_P}$

⇒ Mechanical advantage = 0.1256/00196 = 64 (Option 2 is wrong)

Now,

If F_P = 400 N, F_R = M.A × F_P = 400 × 64 = 25.6 kN (Option 3)

Since the fluid is enclosed, the volume displaced in plunger side will be equal to volume displaced in ram side.

⇒ A_P h_P = A_R h_R ⇒ h_R = h_P / M.A = 20/64 = 3.12 mm (Option 4)

Concept:

A Stream line is an imaginary curve drawn through a flowing fluid in such a way that tangent to it any point gives the direction of instantaneous velocity at that point.  The equation of stream line for 2 D flow is given as:

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{u}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{v}}$ 

Where, u and v are the x and y components of the velocity.

Calculation:

Given:

V = (5x) î - (5y) ĵ

u = 5x and v = -5y

The equation of streamline is given as

$\frac{dx}{5x}=\frac{dy}{-5y}\Rightarrow \int \frac{dy}{y}=-\int \frac{dx}{x}$

⇒ In y = -In x + C

⇒ In (xy) = C’ ⇒ xy = λ                                   [C, C’, λ are constants]

If the stream line pases through point (1,1)  i.e. x = 1 and y = 1 ⇒ λ = 1

∴ The equation of sream line is xy = 1

Concept:

When a body is submerged fully or partially in any fluid it will experience an upward force, called buoyant force, which is equal to weight of fluid displaced by that body.

The cube will experience two buoyant forces.

One is due to liquid having specific gravity 8 and second one is due to liquid having SG of 4. These two buoyant forces will act upward.

Cube is floating in water, this means that total upward buoyant force on portion of cube submerged in fluids is balanced by downward weight of cube. i.e.

F₁ + F₂ = Mg

Where M is the mass of cube

F₁ = ρ₁g V₁ ,F₂ = ρ₂ g V₂

V₁ = Displaced Volume due to liquid of SG = 4

V₂ = Displaced Volume due to liquid of SG = 8

Calculation:

F₁ + F₂ = mg

4 × ρ_w × a × (1)² × g + 8 × ρ_w × b × (1)² g = (6 × ρ_w) × (a + b) (1)² × g

⇒ 4a + 8b = 6a + 6b ⇒ 2b = 2a

$\Rightarrow a=bor\frac{a}{b}=1$

Concept

Applying the principal of manometry

$P_A+{\rho }_{oil}\times g\times h_{oil}=P_B+{\rho }_{water}\times g\times h_{water}+{\rho }_{mercury}\times g\times h_{mercury}$

Calculations

Given,

ρ_w = 1000 kg/m³, ρ_mercury = 13600 kg/m³

ρ_oil = 900 kg/m³

i) Initial condition

$P_A+{\rho }_{oil}\times g\times h_{oil}=P_B+{\rho }_{water}\times g\times h_{water}+{\rho }_{mercury}\times g\times h_{mercury}$

$P_A+900\times 9.81\times 4=P_B+1000\times 9.81\times 2+13600\times 9.81\times 0.4$

$P_A-P_B=37,670.4Pa$

= 37.67 kPa

ii) When mercury levels are made equal on both limbs

For levels to be equal, Mercury level in right limb has to move down by 0.2 m down so in the left limb mercury level rises by 0.2 m.

$P_A+{\rho }_{oil}\times g\times h_{oil}=P_B+{\rho }_{water}\times g\times h_{water}$

$P_A+900\times 9.81\times \left(3.8\right)=P_B+1000\times 9.81\times \left(2.2\right)$

$P_A-P_B=-11,968.2Pa$

= - 11.968 kPa

∴ Additional pressure to make equal level is

= (37.67 – (- 11.968))

= 49.638 kPa

Concept :-

In case of rotating cylindrical vessels or U-Tube vessels there is parabolic water surface level developed after rotation of the vessels. Water level decreases at the center of the vessels and increases at the vessels wall in case of cylinder. U-tube vessels has also the same physics behind the rotation but at the center we can only assume decrease in water level hypothetically. 

Water surface elevation above the vertex at a radial distance ‘r’ is given as:

$y=\frac{r^2{w}^2}{2g}$

Where, ω = radial speed/rotational speed.

1)

As M, N and S are mid points of vertical limes and horizontal limbs respectively. After rotation

h_M = h_N = H

h_s = H – y

2)

For clear understanding of the above case of U-Tube rotation, we can assume a rotational speed such that, vertex of parabola touches the point S in the horizontal Limb.

∴ y = H

⇒ h_s = H – y = H – H = 0

h_M = h_N = H

$\begin{array}{l}y=H=\frac{r^2{w}_{ar}^2}{2g}\\ w_{ar}^2=\frac{2gH}{r^2}\\ W_{cr}=\frac{\sqrt{2gH}}{r}\end{array}$

3)

If we further increase the value of rotational speed such that ω > ω_cr, then water will split form point S and a negative pressure is develop at point S.

h_s = H – y

h_M = h_N = H  (till there is water remains in the limb of U-Tube.)

Calculation :-

$\omega =\frac{2\pi N}{60}\Rightarrow \frac{2\pi \times 120}{60}=4\pi rad/s$

During rotation, there will be apparent depression in water column at centre by ‘y’.

$y=\frac{r^2{w}^2}{2g}=\frac{{0.15}^2\times {\left(4\pi \right)}^2}{2\times 9.81}\Rightarrow 0.181m=18.1cm$

∴ h_s = 25 – 18.1 = 6.9 cm

So, pressure at M, N & S are

P_M = P_N = 9.81 × 1.25 × 0.25 = 3.06 kPa

P_S = 9.81 × 1.25 × 0.069 = 0.84 kPa

Concept:

According to Newton’s law of viscosity, shear stress is given by:

$\tau =\mu \times \frac{du}{dy}=\mu \times \frac{d\theta }{dy}$

τ = shear stress, μ = coefficient of viscosity or absolute viscosity (or dynamic viscosity)

$\frac{du}{dy}=Velocitygradient$

$\frac{d\theta }{dt}=RateofangulardeformationorRateofshearstrain$

Calculation:

Given:

Velocity profile 

$u\left(y\right)=U_{\mathrm{\infty }}\times \sin \left(\frac{\pi y}{2\delta }\right),0\le y\le \delta$

$\tau =\mu \times \frac{du}{dy}$

$\tau =\mu \times \frac{du}{dy}=\mu \times U_{\mathrm{\infty }}\times \frac{\partial }{\partial y}\left(\sin \left(\frac{\pi y}{2\delta }\right)\right)$

$=\mu \times U_{\mathrm{\infty }}\times \cos \left(\frac{\pi y}{2\delta }\right)\times \frac{\pi }{2\delta }$

As we have to find the shear stress at the wall,

y = 0

${\tau }_{wall}=\mu \times U_{\mathrm{\infty }}\times \frac{\pi }{2\delta }=\frac{\pi \mu U_{\mathrm{\infty }}}{2\delta }$

Given data; L = 5 m, B = 2 m, V = 4 m/s, ρ = 1.25 kg/m³

ν = 1.5 × 10^-5 m²/s

$R{e}_L=\frac{\rho VL}{\mu }=\frac{VL}{\nu }=\frac{\left(4\right)\left(5\right)}{1.5\times {10}^{-5}}=13.33\times {10}^5$

Re_L = 13.33 × 10⁵ > 5 × 10⁵

⇒ Flow is turbulent.

Average drag coefficient (C_D) $=\frac{1}{L}{\int }_0^L{C}_{D,x}dx$

$\Rightarrow C_D=\frac{1}{L}{\int }_0^L\frac{0.059}{R{e}_x^{1/5}}dx=\frac{0.07375}{R{e}_L^{1/5}}$

$C_D=\frac{0.07375}{{(13.33\times {10}^5)}^{1/5}}=0.004393$

$C_D=\frac{F_D}{\frac{1}{2}\rho A{V}^2}\Rightarrow F_D=\left(0.004393\right).\frac{1}{2}.\left(1.25\right)\left(10\right)\left(16\right)$

Concept:

Velocity triangle of the centrifugal pump is:

The tangential velocity of the impeller at the outlet can be calculated by:

$u_2=\frac{\pi \times D_2\times N}{60}$

The vane angle (ϕ) at the outlet can be calculated by:

$\tan \varphi =\frac{V_{f2}}{\left(u_2-V_{w2}\right)}$

Manometric efficiency is given by:

${\eta }_{mano}=\frac{g\times H_m}{V_{w2}\times u_2}$

Discharge at the outlet is given by:

Q = π × D₂ × B₂ × V_f2

where D­₂ = diameter of the impeller at the outlet, B₂ = outlet width, N = speed,

H_m = net head,vV_w2 = whirl velocity at outlet, V_f2 = flow velocity at outlet

Calculation:

Given:

H_m = 16 m, N = 900 rpm, ϕ = 30°, D₂ = 360 mm = 0.36 m, B­₂ = 40 mm = 0.04 m, η_mano = 80% = 0.8

The tangential velocity of the impeller at the outlet is:

$u_2=\frac{\pi \times D_2\times N}{60}=\frac{\pi \times 0.36\times 900}{60}=16.96m/s$

Manometric efficiency is:

${\eta }_{mano}=\frac{g\times H_m}{V_{w2}\times u_2}$

$0.8=\frac{9.81\times 16}{V_{w2}\times 16.96}\Rightarrow V_{w2}=11.56m/s$

From the outlet velocity triangle, we have

$\tan \varphi =\frac{V_{f2}}{\left(u_2-V_{w2}\right)}$
$\tan {30}^{\circ }=\frac{V_{f2}}{\left(16.96-11.56\right)}\Rightarrow V_{f2}=5.4\times \tan {30}^{\circ }=3.117m/s$

Discharge is:

Q = π × D₂ × B₂ × V_f2

Q = π × 0.36 × 0.04 × 3.117 = 0.14 m³/s

Concept

Rotation components in a flow are given by

${\omega }_x=\frac{1}{2}\left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z}\right)$ 

${\omega }_y=\frac{1}{2}\left(\frac{\partial u}{\partial z}-\frac{\partial w}{\partial x}\right)$ 

${\omega }_z=\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)$ 

Where u is velocity component in x direction

v is the velocity component in y direction

w is the velocity component in z direction

If ω_z = 0. Flow is irrotational.

ω_z ≠ 0 . Flow is rotational  

Shear force developed on plate is given by

Shear force = shear stress × Area of plate

From newton’s law of viscosity,

$shearstress\left(\tau \right)=\mu \times \frac{du}{dy}$

Calculations

Given,

h = 6 mm, μ = 2 × 10^-4 pa-sec

Area of plate = 0.33 m²

Velocity profile $u=V\times \frac{y}{h}$

$u=V\times \frac{y}{h}$ , v = 0, w = 0

∴ Flow is 1-D

V = 3 cm/sec

i) Rotation components in a flow are given by

${\omega }_x=\frac{1}{2}\left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z}\right)$ 

$\frac{1}{2}\times \left(0-0\right)$

ω_x = 0.

${\omega }_y=\frac{1}{2}\left(\frac{\partial u}{\partial z}-\frac{\partial w}{\partial x}\right)$ 

$=\frac{1}{2}\left(\frac{\partial }{\partial z}\left(V\frac{y}{h}\right)-0\right)$

= 0

${\omega }_z=\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)$ 

$=\frac{1}{2}\times \left(0-\frac{\partial }{\partial y}\left(V\frac{y}{h}\right)\right)$

$=-\frac{V}{2h}$

∴ ${\omega }_z=-\frac{V}{2h}$

∴ Flow is rotational as ω_{z ≠} 0

ii) Shear force

Shear force = shear stress × Area of plate

V = 3 cm/sec

h = 6 mm, μ = 2 × 10^-4 pa-sec

$shearstress\left(\tau \right)=\mu \times \frac{du}{dy}$ 

$\frac{du}{dy}=\frac{d}{dy}\left(V\frac{y}{h}\right)=\frac{V}{h}$

$\tau =\mu \times \frac{V}{h}$

$=2\times {10}^{-4}\times \frac{3\times {10}^{-2}\frac{m}{s}}{6\times {10}^{-3}m}$

$={10}^{-3}\frac{N}{m^2}$

$shearforce=\tau \times Area$

$={10}^{-3}\times 0.33$

= 33 × 10^-5 N

Concept:

(a) Series systems – the flow rate through the entire system remains constant, the total head loss in this case is equal to the sum of the head losses in individual pipes

(b) Parallel system – head loss is the same in each pipe, and the total flow rate is the sum of the flow rates in individual pipes

Calculation:

$h_f=\frac{fL{V}^2}{2gd}=\frac{fL}{2gd}{V}^2$

$\frac{fL}{2gd}{\left(\frac{4Q}{\pi d^2}\right)}^2=\frac{8f{Q}^2}{{\pi }^{2}g}\left(\frac{L}{d^5}\right)\Rightarrow h_f\propto \frac{Q^{2}L}{d^5}$

$\frac{Q^2{L}_{eq}}{d_{eq}^5}=\frac{{\left(\frac{\mathrm{Q}}{3}\right)}^{2}L}{{\mathrm{d}}^5}$

$\Rightarrow \frac{1}{{\mathrm{d}}_{\mathrm{e}\mathrm{q}}^5}=\frac{1}{9{\mathrm{d}}^5}$

$\Rightarrow {\mathrm{d}}_{\mathrm{e}\mathrm{g}}={\left(9\right)}^{\frac{1}{5}}\mathrm{d}=1.55\mathrm{d}$

Concept:

The total pressure force acting on the curved surface has horizontal and vertical components.

The horizontal component (F_h) is equal to the projection of the curved surface on the vertical plane and acts at the center of the pressure of the projected area.

The vertical component of the pressure force (F_v­) is equal to the weight of liquid supported by the curved surface up to the free surface of the liquid and it acts at the centroid of liquid

F_h = γ A h̅

A is the projected area.

h̅ is the Centroid of the projected area.

F_v = γ × Volume of liquid held by curved surface up to the free surface.

Calculation:

The horizontal component of Force, F_h :-

F_h = γ A h̅

A = 2 × 3 = 6m²

$\overline{h}=\frac{2}{2}=1m$

F_h = 9.81 × 6 × 1 = 58.86 kN

Location of F_H: At the center of pressure of the projected area calculated as

$h_{cp}=\overline{h}+\frac{I}{A\overline{h}}\left({\sin }^2\theta \right)$                       (I = MOI about C.G of the projected area)

$h_{cp}=\frac{2}{2}+\frac{\frac{3\times 2^3}{12}}{\left(2\times 3\right)\times \left(1\right)}\times \left({\sin }^2{90}^{\circ }\right)where,\left[\begin{array}{c}\theta ={90}^{\circ }\\ \overline{h}=1\\ A=2\times 3\end{array}\right]$

$h_{cp}=1+\frac{1}{3}=\frac{4}{3}m$

∴ F_h = 58.86 kN and acts at 4/3 from the top surface.

Vertical components of force : (F_V) ( Acting downwards )

F_V = γ_w × Volume of water in portion (O A B)

${\mathrm{F}}_{\mathrm{V}}=\left(9.81\right)\times \frac{1}{4}\pi {\left(2\right)}^2\times 3=92.41kN$

Location of F_V : At centroid of OAB. It is located at distance  $\frac{4r}{3\pi }$ from OB.

$\overline{y}=\frac{4r}{3\pi }=\frac{4\times 2}{3\pi }=0.849m$

∴ F_V = 92.41 kN and acts at 0.849 m from OB.

Now considering moment equilibrium of the gate about hinge ‘O’,

F_H × 2/3 + F_V × 0.849 = F × 2

⇒ $58.86\times \frac{2}{3}+92.41\times 0.849=F\times 2$

⇒ F = 58.84 kN