Fluid Mechanics Test 1

Concept:

\({\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} \Rightarrow + {\rm{ve\;}} \Rightarrow {\rm{\;flow\;is\;separated}}\)

\({\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} \Rightarrow 0{\rm{\;}} \Rightarrow {\rm{\;on\;verge\;of\;separation}}\)

\({\left( {\frac{{\partial u}}{{\partial u}}} \right)_{y = 0}} \Rightarrow - {\rm{ve\;}} \Rightarrow {\rm{\;stick\;to\;surface}}\)

Calculation:

\(\frac{u}{U} = 2{\left( {\frac{y}{\delta }} \right)^2} - {\left( {\frac{y}{\delta }} \right)^3}\)

\(\frac{{\partial u}}{{\partial y}} = U\left[ {4\left( {\frac{4}{\delta }} \right) - 3{{\left( {\frac{y}{\delta }} \right)}^2}} \right]\)

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

∴ Flow is on the verge of separation.

Concept:

Newton’s law of viscosity

\(\tau = \mu \frac{{du}}{{dy}}\)

Calculation:

τ = 1.2 × 10^-4 × 1000

∴ τ = 0.12 N/m²

Concept:

The Laminar boundary layer for flat plate given by Blausius equation is :

\(\delta =\frac{5x}{\sqrt{Re}}\)  

\(Re = \frac{\rho Ux}{\mu} \Rightarrow Re\propto x\)

\(\delta =\frac{5x}{\sqrt{Re}} \Rightarrow \delta \propto \sqrt x\)

\(\Rightarrow \frac{{{\delta }_{1}}}{{{\delta }_{2}}}= \sqrt {\frac{x_1}{x_2}} = \sqrt{\frac{R{{e}_{1}}}{R{{e}_{2}}}}\)

Calculation:

Re₂ = 225 and Re₁ = 121

\(\frac{{{\delta }_{1}}}{{{\delta }_{2}}}=\sqrt{\frac{R{{e}_{1}}}{R{{e}_{2}}}}=\sqrt{\frac{121}{225}}=0.733\)

Concept:

\({R_e} = \frac{{\rho Vd}}{\mu }\)

∵ (V = V_average)

\({V_{average}} = \frac{{{V_{max}}}}{2}\)

Calculation:

\({V_{avg}} = \frac{4}{2} = 2\;m/s\)

\({R_e} = \frac{{\rho {V_{avg}}d}}{\mu } = \frac{{800 \times 2 \times 0.05}}{{0.2}}\)

R_e = 400 < 2000

∴ Flow is Laminar

\(\therefore f = \frac{{64}}{{{R_e}}}\;\;\;\because\left( {flow\;is\;laminar} \right)\)

\(\therefore f = \frac{{64}}{{400}}\)

∴ f = 0.16

Concept:

Specific weight or weight density (ω) of a fluid is the ratio between the weight (W) of a fluid to its volume (V).

ω = W/V

\(\Rightarrow \omega = \frac{{mg}}{V} = \left( {\frac{m}{V}} \right)g = \rho g\)

Specific gravity is the ratio of the specific weight of the liquid to the specific weight of a standard fluid. In the case of liquids standard fluid is taken as water with specific gravity 1.

\(Specific\;gravity = \frac{{Density\;of\;liquid}}{{Density\;of\;water}}=\frac{\rho}{1000}\)

Calculation:

Given: SG = 0.7

\(SG= \frac{\rho }{{1000}} = 0.7\)

\(\Rightarrow {\rho _{petrol}} = 700\;kg/{m^3}\)

ω = ρg = (9.81)(700) = 6867 N/m³ = 6.867 kN/m³

Concept:

Discharge through orifice-meter is given by;

\(Q=~\frac{{{C}_{d}}{{A}_{1}}{{A}_{0}}}{\sqrt{A_{1}^{2}-A_{0}^{2}}}.\sqrt{2gh}\)

A₁ = area of pipe, A₀ = area of the orifice

Calculation:

\(h=\frac{\text{ }\!\!\Delta\!\!\text{ }p}{\rho g}=\frac{\left( 14.715-9.81 \right)\left( {{10}^{4}} \right)}{\left( 9.81 \right)\left( {{10}^{3}} \right)}=5~m~of~water\)

\({{A}_{1}}=\frac{\pi }{4}.{{\left( 0.30 \right)}^{2}}=0.07068~{{m}^{2}}\)

\({{A}_{0}}=\frac{\pi }{4}.{{\left( 0.15 \right)}^{2}}=0.01767{{m}^{2}}\)

\(Q=\frac{\left( 0.6 \right)\left( 0.07068 \right)\left( 0.01767 \right)}{\sqrt{{{\left( 0.07068 \right)}^{2}}-{{\left( 0.01767 \right)}^{2}}}}~\sqrt{\left( 2 \right)\left( 9.81 \right)\left( 5 \right)}\)

Q = 0.108452 m³/s = 108.452 litres/s

Concept:

Expression for capillary rise or fall

\(h = \frac{{4\sigma \cos \theta }}{{\rho gd}}\)

Calculation:

Given: θ = 130°, σ = 0.52 N/m, d = 2.5 × 10^-3m

\(h = \frac{{4\; \times \;0.52\; \times\; \cos 130^\circ }}{{13600\; \times \;9.81\; \times \;2.5\; \times \;{{10}^{ - 3}}}}\)

∴ h = - 0.4 cm

Note: - ve sign indicates capillary depression (or full)

Concept:

The shape factor for the boundary layer is defined as \(H=\frac{{{\delta }^{*}}}{\theta }\), where δ*= displacement thickness and θ = momentum thickness.

\({{\delta }^{*}}=\mathop{\int }_{0}^{\delta }\left( 1-\frac{u}{U} \right)dy\)

\(\theta =\mathop{\int }_{0}^{\delta }\frac{u}{U}\left( 1-\frac{u}{U} \right)~dy\)

Calculation:

\(\frac{u}{U}=\frac{y}{\delta }\)

\({{\delta }^{*}}=\mathop{\int }_{0}^{\delta }\left( 1-\frac{u}{U} \right)dy=\mathop{\int }_{0}^{\delta }\left( 1-\frac{y}{\delta } \right)~dy~=~\frac{\delta }{2}\)

\(\theta =\mathop{\int }_{0}^{\delta }\frac{u}{U}\left( 1-\frac{u}{U} \right)~dy=\mathop{\int }_{0}^{\delta }\frac{y}{\delta }\left( 1-\frac{y}{\delta } \right)~dy~=\frac{\delta }{6}\)

So, Shape factor:

\(H=\frac{{{\delta }^{*}}}{\theta }=\frac{\frac{\delta }{2}}{\frac{\delta }{6}}=3\)

Concept:

Bernoulli’s equation is given as,

\(\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + Z = constant\)

\(\frac{P}{{\rho g}} = Pressure\;head,\;\;\frac{{{V^2}}}{{2g}} = velocity\;head,\;\;z = datum\;head\)

Now,

Piezometric head = Pressure head + datum head

\({\rm{Piezometric\;head\;}} = \frac{P}{{\rho g}} + z\)

\({\rm{Piezometric\;head\;}} = \frac{{19.62\; \times \;{{10}^3}}}{{800\; \times \;9.81}} + 3.5\)

∴ Piezometric head = 6 meters of oil

Concept:

For any type of flow, the shear stress at wall,

\(\tau = - \left( {\frac{{dp}}{{dx}}} \right)\frac{R}{2}\;\)

\(\therefore \tau = \frac{{\rho g{h_L}}}{L} \times \frac{R}{2}\)

\(\tau = \frac{{1000 \times 9.81 \times 10}}{{100}} \times \frac{{0.1}}{4}\)

∴ τ = 24.525 Pa

Explanation:

Reynold’s model law is applicable for:

(i) Pipe flow

(ii) Resistance experienced by sub-marines, airplanes, etc.

Froude model law is applicable for:

(i) free surface flows such as spillways, channels, etc.

(ii) Sea waves

Euler model law is applicable for:

Where pressure forces are dominant like flow taking place in closed pipe, cavitation phenomenon etc.

Weber model is applicable for:

(i) Capillary rise in narrow passage

(ii) Capillary movement of water in soil

Concept:

\(\frac{P}{{{D^2}{H^{3/2}}}} = {\rm{\;constant}}\)

Calculation:

\(\frac{{{P_1}}}{{D_1^2H_1^{3/2}}} = \frac{{{P_2}}}{{D_2^2H_2^{3/2}}}\)

\(\therefore \frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{{H_2}}}{{{H_1}}}} \right)^{\frac{3}{2}}}\)

\(\frac{{{P_2}}}{{200}} = {\left( {\frac{{20}}{{40}}} \right)^{3/2}}\)

P₂ = 70.71 kW

\({\rm{\% \;decrease\;in\;Power\;}} = \frac{{{P_1} - {P_2}}}{{{P_1}}} \times 100\)

∴ % decrease in Power = 64.64 %

Concept:

The wall shear stress is defined as:

\({\tau _w} = {C_f} \times \frac{1}{2}\rho {V^2}\)

Here, C_f is the Fanning’s friction factor and is related to Darcy friction factor f as:

f = 4C_f

\({\tau _w} = \frac{f}{8}\rho {V^2}\;or\;V = \sqrt {\frac{{8{\tau _w}}}{{\rho f}}} \)

And discharge Q = A.V

Calculation:

τ_w = 50 Pa, ρ = 700 kg/m³, f = 0.05, D = 0.6 m

\(V = \sqrt {\frac{{8{\tau _w}}}{{\rho f}}} \)

\(V = \sqrt {\frac{{8 \times 50}}{{700 \times 0.05}}} = 3.38\;m/s\)

Concept:

To check possibility of fluid flow, continuity equilibrium should be verified, i.e.

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

Angular velocity about z-axis,

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

Calculation:

\(u = 2x + \frac{{{y^3}}}{3} - {x^2}y\)

\(v = x{y^2} - 2y - \frac{{{x^3}}}{3}\)

\(\frac{{\partial u}}{{\partial x}} = 2 - 2xy\;\;\;;\;\frac{{\partial u}}{{\partial y}} = {y^2} - {x^2}\)

\(\frac{{\partial v}}{{\partial x}} = {y^2} - \frac{{3{x^2}}}{3} = {y^2} - {x^2}\)

\(\frac{{\partial v}}{{\partial y}} = 2xy - 2\)

For 2-D flow, continuity equation,

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

(2 – 2xy) + (2xy - 2) = 0

Hence, it represents possible case of fluid flow.

Now,

\({\rm{Angular\;velocity\;}}\left( {{\omega _z}} \right) = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right)\)

\({\omega _z} = \frac{1}{2}\left[ {\left( {{y^2} - {x^2}} \right) - \left( {{y^2} - {x^2}} \right)} \right] = 0\)

Concept:

Both velocity potential function (ϕ) and stream function (ψ) must satisfy Cauchy-Reimann equations:

\(\frac{{\partial \phi }}{{\partial x}} = \frac{{\partial \psi }}{{\partial y}}\)

\(\frac{{\partial \phi }}{{\partial y}} = - \frac{{\partial \psi }}{{\partial x}}\)

Calculation:

ϕ = 4x (3y - 4)

ϕ = 12xy – 16x

\(\Rightarrow \frac{{\partial \phi }}{{\partial x}} = 12y - 16\)

\(\frac{{\partial \psi }}{{\partial y}} =\frac{{\partial \phi }}{{\partial x}} = 12y - 16\)

Integrating w.r.t y keeping x constant,

ψ = 6y² – 16y + c(x)

ϕ = 12xy – 16x

\(\frac{{\partial \phi }}{{\partial y}} = 12x = - \left[ {c'\left( x \right)} \right]\)

\( \frac{{\partial \psi }}{{\partial x}}=-\frac{{\partial \phi }}{{\partial y}} \Rightarrow c'(x)=-12x\)

c(x) = -6x²

ψ = 6y² – 16y – 6x² = 6(y² – x²) – 16y

ψ = 6(y² – x²) – 16y

At x = 2, y = 3

ψ = 6{9 - 4} – (16)(3)

ψ = 30 – 48

ψ = -18

Explanation:

First we have to check whether the flow is laminar are turbulent.

Given: ρ = 1150 kg/m³, μ = 1.2 Pa-s, v = 8 m/s,, D = 15 cm = 0.15 m, L = 25 m

Now,

\({\rm{Reynolds\;number\;}}\left( {Re} \right) = \frac{{\rho VD}}{\mu }\)

\(Re = \frac{{\left( {1150} \right)\left( 8 \right)\left( {0.15} \right)}}{{1.2}}\)

∴ Re = 1150

It is less than 2000, hence flow is laminar.

For laminar flow, head loss

\({h_f} = \frac{{32\;\mu vL}}{{\rho g{D^2}}}\)

\({h_f} = \frac{{32\left( {1.5} \right)\left( 8 \right)\left( {25} \right)}}{{{{10}^3}\; \times \;9.81\; \times \; {{0.15}^2}}}\)

∴ h_f = 34.794 m

\(Q = A.V = \frac{\pi }{4}\left( {{{0.15}^2}} \right)\left( 8 \right)\)

∴ Q = 0.1413 m³/s

Now,

Power loss P = ρQ_g h_f

P = 1000 × 0.1413 × 9.81 × 34.794

P = 48229.8 W

∴ P = 48.2 kW