Fluid Mechanics Test 5

Concept:

laminar flow Friction factor ⇒ \(f = \frac{{64}}{{Re}}\) 

Where Re is the Reynolds number

For laminar flow friction factor depends only upon the Reynolds number of the flow  

Calculation:

Given, f = 0.05

\(0.05 = \frac{{64}}{{Re}} \Rightarrow Re = \frac{{64}}{{0.05}} = 1280\)

Boundary Layer:  When a fluid of ambient velocity flows over a flat stationary plate, the bottom layer of fluid directly contacts with the solid surface and its velocity reaches to zero. Due to the cohesive forces between two layers, the bottom layer offers resistance to the adjacent layer and due to this reason, the velocity gradient develops in a fluid. A thin region over a surface velocity gradient is significant, known as the boundary layer.

The thickness of the boundary layer is given by

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

\(\delta =\frac{5x}{\sqrt{Re_x}}\)  

\( \Rightarrow \delta \propto (Re_x)^{-1/2}\)

Note:

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

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

Where, x = distance where the boundary layer is to be found, Re = Reynolds no, ρ = density of the fluid, V = velocity of the fluid

µ = dynamic viscosity fluid

Important Point:

The Turbulent boundary layer for flat plate given

For turbulent flow  Re < 107

\(\begin{array}{l} \delta = \frac{{0.379x}}{{{{\left( {Re_x} \right)}^{\frac{1}{5}}}}}\\ \delta \propto {x^{1 - \frac{1}{5}}}\\ \delta \propto {x^{\frac{4}{5}}} \end{array}\)

Explanation:

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

If \({{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}=0,\) flow is on the verge of separation

If \({{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}=-ve,\) flow has separated

If \({{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}=+ve,\) flow will not separate

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

Concept:

As the velocity of oil at any point in the fluid is constant, hence the flow is laminar. Shear stress at any distance ‘r’ from the centre of the pipe is given by.

\(\tau = \frac{{ - r}}{2}\left( {\frac{{\partial p}}{{\partial x}}} \right)\)

At r = R, i.e. at the, pipe wall, shear stress is maximum and is given by

\({\tau _{max}} = \frac{{ - R}}{2}\left( {\frac{{\partial p}}{{\partial x}}} \right)\)

Where, R = Radius of the pipe

\(\frac{{\partial p}}{{\partial x}}\) = Pressure gradient over the length of the pipe.

Equation of velocity profile:

\(u = \frac{{ - 1}}{{4u}}\left( {\frac{{\partial p}}{{\partial x}}} \right)\left( {{R^2} - {r^2}} \right)\)

At, r = 0, u = u_max

\({u_{max}} = \frac{{ - 1}}{{4u}}\left( {\frac{{\partial p}}{{\partial x}}} \right)\left( {{R^2}} \right)\)

Also,

\({u_{max}} = 2\bar u\)

Where, u̅ = average velocity

Calculation:

Diameter = 15 cm = 0.15 m

Radius of pipe = 0.075 m

μ (viscosity of oil) = 0.8 Pa-s

\({\tau _{max}} = 200\;N/{m^2}\)

\(200 = - 0.075\left( {\frac{{\partial p}}{{\partial x}}} \right)\)

\(\frac{{\partial p}}{{\partial x}} = \frac{{ - 200 \times 2}}{{0.075}}\)

\(\frac{{\partial p}}{{\partial x}} = - 5333.33\frac{N}{{{m^2}}}/m\)

\({u_{max}} = \frac{{ - 1}}{{4 \times 0.8}}( - 5333.33){\left( {0.075} \right)^2}\)

u_max = 9.37 m/s

Concept:

θ = momentum thickness

\({\rm{\theta \;}} = \mathop \smallint \limits_0^\delta \frac{u}{U}\left[ {1 - \frac{u}{U}} \right]dy\)

Calculation:

\(\theta = \mathop \smallint \limits_0^\delta \frac{u}{U}\left[ {1 - \frac{u}{U}} \right]dy\)

\(\theta = \;\mathop \smallint \limits_0^\delta \frac{y}{{2\delta }}\left[ {1 - \frac{y}{{2\delta }}} \right]dy\)

\(\theta = \;\mathop \smallint \limits_0^\delta \left[ {\frac{y}{{2\delta }} - \frac{{{y^2}}}{{4{\delta ^2}}}} \right]dy\)

\(\theta = \left[ {\frac{{{y^2}}}{{4\delta }} - \frac{{{y^3}}}{{12{\delta ^2}}}} \right]_0^\delta \)

\(\theta = \frac{\delta }{4} - \frac{\delta }{{12}}\)

\(\therefore \frac{\theta }{\delta } = \frac{1}{6}\)

Concept:

For turbulent flow,

\(\frac{{{v_{max}} - v}}{{{v_*}}} = 5.75\;{\log _{10}}\left( {\frac{R}{y}} \right)\)        ---(1)

\({{\rm{V}}_{\rm{*}}} = {\rm{\;shear\;velocity\;}} = \sqrt {\frac{{{\tau _w}}}{\rho }} \)

Calculation:

Given:

D = 130 mm = 0.13 m, v_max = 2.5 m/s, v = 1.5 m/s, r = 35 mm = 0.035 m

Now,

\(y = R - r = \left( {\frac{{0.13}}{2}} \right) - \left( {0.035} \right)\)

y = 0.03

Using the relation from equation (1)

\(\frac{{2.5 - 1.5}}{{{v_*}}} = \left( {5.75} \right){\log _{10}}\left( {\frac{{0.065}}{{0.03}}} \right)\)

∴ V_* = 0.5179 m/s

Now,

\({v_*} = \sqrt {\frac{{{\tau _w}}}{\rho }}\)

\({v_*} = 0.5179 = \sqrt {\frac{{{\tau _w}}}{{{{10}^3}}}} \)

Concept:

\(u = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2} - {r^2}}}{{4\mu }}} \right]\)

Calculation:

Given: D = 0.5 mm ⇒ R = 0.25 mm, L = 2m, Q = 48 cm³/min = 8 × 10^-7 m³/sec,

μ = 1.92 centipoise = 1.92 × 10^-3 Ns/m²

Now,

Q = A × V_avg

\(8 \times {10^{ - 7}} = \frac{\pi }{4} \times {\left( {0.5 \times {{10}^{ - 3}}} \right)^2} \times {V_{avg}}\)

∴ V_avg = 4.074 m/s

\(\therefore {u_{max}} = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2}}}{{4\mu }}} \right]\)

And u_mean = V_avg u_max

\(\therefore {V_{avg}} = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2}}}{{8\mu }}} \right]\)

\(\therefore {\rm{\Delta }}P = \frac{{{V_{avg}} \times L \times 8\mu }}{{{R^2}}}\)

∴ ΔP = 2 MPa

Points to Remember:

\({\rm{\Delta }}P = \frac{{32\;\mu \;{V_{avg}}\;L}}{{{D^2}}}\)

Concept:

Friction drag on plate,

\(F = \frac{1}{2}\rho A{V^2}\;{C_D}\)

For laminar boundary layer,

\({C_D} = \frac{{1.328}}{{\sqrt {Re} }}\)

Calculation:

Given:

ℓ = 0.8 m, b = 0.3 m, v = 3 m/s, μ = 9 × 10^-4 Pa-s

We have to find the distance from leading edge (= x) from where flow changes into turbulent flow.

\(R{e_c} = 5 \times {10^5} = \frac{{\rho vx}}{\mu }\)

\(5 \times {10^5} = \frac{{{{10}^3}\; \times \;3\; \times \;x}}{{9\; \times \;{{10}^{ - 4}}}}\)

∴ x = 0.15 m

\({C_D} = \frac{{1.328}}{{\sqrt {5\; \times \;{{10}^5}} }} = 0.001878\)

\(F = \frac{1}{2} \times {10^3} \times \left( {0.15 \times 0.3} \right)\left( {{3^2}} \right) \times 0.001878\)

∴ F = 0.38 N

Mistake Point:

This is drag force on one side of plate only. (option (a))

So, for both sides

F_total = 2F = 0.76 N