Fluid Mechanics Test 5

Concept:

If gravitational and inertial force are me only important forces, then the Frouds number must be the same in the model and prototype. Thus

\(\frac{{{V_m}}}{{\sqrt {g{L_m}} }} = \frac{{{V_p}}}{{\sqrt {g{L_p}} }}\)

Scale Ratio for length: \(L_r=\frac{L_p}{L_m}\)

From Froude Model Law:

\(\frac{{{V_m}}}{{\sqrt {g{L_m}} }} = \frac{{{V_p}}}{{\sqrt {g{L_p}} }} \Rightarrow \frac{{{V_m}}}{{\sqrt {{L_m}} }} = \frac{{{V_p}}}{{\sqrt {{L_p}} }}=\frac{V_p}{V_m}=\sqrt {\frac{L_p}{L_m}}=\sqrt {L_r}\)

Scale Ratio for Time: Time = Length/Velocity

\({T_r} = \frac{{{T_p}}}{{{T_m}}} = \frac{{{{\left( {\frac{L}{V}} \right)}_P}}}{{{{\left( {\frac{L}{V}} \right)}_m}}} = \frac{{{L_p}}}{{{L_m}}}.\frac{{{V_m}}}{{{V_p}}} = {L_r}.\frac{1}{{\sqrt {{L_r}} }} = \sqrt {{L_r}} \)

Scale ratio for Discharge:

Q = A × V = L² × L/T = L³/T

\({Q_r} = \frac{{{Q_p}}}{{{Q_m}}} = \frac{{{{\left( {\frac{{{L^3}}}{T}} \right)}_P}}}{{{{\left( {\frac{{{L^3}}}{T}} \right)}_m}}} = {\left( {\frac{{{L_p}}}{{{L_m}}}} \right)^3}.\frac{{{T_m}}}{{{T_p}}} = {L_r}^3.\frac{1}{{\sqrt {{L_r}} }} = {L_r}^{2.5}\)

Or

\({Q_r} = \frac{{{Q_p}}}{{{Q_m}}} = \frac{{{{\left( {A \times V} \right)}_P}}}{{{{\left( {A \times V} \right)}_m}}} = \frac{{{A_P}}}{{{A_m}}}.\frac{{{V_P}}}{{{V_m}}} = {\left( {\frac{{{L_p}}}{{{L_m}}}} \right)^2}.\sqrt {\frac{{{L_P}}}{{{L_m}}}} = {\left( {\frac{{{L_p}}}{{{L_m}}}} \right)^{2.5}} = {L_r}^{2.5}\)

Calculation:

As here only Length scale ratio and discharge is give so we will use Froude Model Law.

Q_r = L_r^2.5 = 202.5 = 1788.85

\({Q_r} = \frac{{{Q_p}}}{{{Q_m}}} \Rightarrow {Q_m} = \frac{{{Q_p}}}{{{Q_r}}} = \frac{4}{{1788.85}} = 0.0022\,{{\rm{m}}^{\rm{3}}}{\rm{/s}}\)

Concept:

For Laminar Boundary layer:

\(\frac{{\rm{\delta }}}{{\rm{X}}} = \frac{{\rm{K}}}{{\sqrt {{\rm{R}}{{\rm{e}}_{\rm{X}}}} }}\)

Where δ is the boundary layer thickness.

Re_x is Reynolds Number at a distance X from the leading edge.

K is any constant; generally it is equal to 5.

Now for Given X;

\(\delta \propto \frac{1}{{\sqrt {R{e_X}} }}\)

Calculation:

Given

R_{e x1} = 200 and R_{e x2} = 400

\(\frac{{{\delta _1}}}{{{\delta _2}}} = \sqrt {\frac{{R{e_{X2}}}}{{R{e_{X1}}}}} \)

⇒ \(\frac{{{\delta _1}}}{{{\delta _2}}} = \sqrt {\frac{{400}}{{100}}} = 2.0\)

Dimensionless profile shape factor (H) = Displacement thickness/Momentum thickness = δ*/θ

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

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

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

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

\(\theta =\mathop{\int }_{0}^{\delta }\left( \frac{y}{\delta }-\frac{{{y}^{2}}}{{{\delta }^{2}}} \right)dy\)

\(=\left\{ \frac{{{y}^{2}}}{2\delta }-\frac{{{y}^{3}}}{3{{\delta }^{2}}} \right\}_{0}^{\delta }\)

\(\Rightarrow \frac{\delta }{2}-\frac{\delta }{3}=\frac{\delta }{6}\)

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

Concept:

Specific Gravity or Relative Density: The ratio of the density of the fluid to the density of water—usually 1000 kg/m³ at a standard condition—is defined as Specific Gravity or Relative Density of fluids

∴ The density of fluid = Relative Density × Density of water

Poise = 0.1 Ns/m² = 0.1 kg/ms

Calculation:

Given data,

RD = 0.8 ⇒ ρ = 0.8 × 1000 = 800 kg/m3

(Re)_P = (Re)_m

\({{\left( \frac{ρ VD}{\mu } \right)}_{P}}={{\left( \frac{ρ VD}{\mu } \right)}_{m}}\)     ...(1)

\(\frac{{{\rho _m}}}{{{\rho _P}}}.\frac{{{V_m}}}{{{V_P}}}.\frac{{{D_m}}}{{{D_P}}}.\frac{{{\mu _P}}}{{{\mu _m}}} = 1\)

\(\frac{{1000}}{{800}} × \frac{{{V_m}}}{{{V_P}}} × \frac{{0.2}}{{1.8}} × \frac{{0.004}}{{0.001}} = 1\)

\(\frac{{{V_m}}}{{{V_P}}} = 1.8\)

\(\frac{{{Q_m}}}{{{Q_P}}} = \frac{{{A_m}{V_m}}}{{{A_P}{V_P}}} = {\left( {\frac{{{D_m}}}{{{D_P}}}} \right)^2} × \frac{{{V_m}}}{{{V_P}}}\)

\(\frac{{{Q_m}}}{{{Q_P}}} = {\left( {\frac{{0.2}}{{1.8}}} \right)^2} × 1.8 = 0.0222\)

Q_m = 0.0222 × Q_P = 0.0222 × 4 = 0.08888 m3/s

1 litre = 1000 cm³ = 10^-3 m³

∴ 1 m3/s = 1000 litre/s

Q_m = 0.08889 m³/s

∴ Qm = 88.89 litres/s

Concept:

For pipe flow, the dynamic similarity will be obtained if the Reynolds number in the model and prototype are equal.

(Re)_Model = (Re)_Prototype

\(\frac{{{\rho _m}{V_m}{D_m}}}{{{\mu _m}}} = \frac{{{\rho _p}{V_p}{D_p}}}{{{\mu _p}}}\)

\(\frac{{{V_m}{D_m}}}{{{\nu _m}}} = \frac{{{V_p}{D_p}}}{{{\nu _p}}}\)

Calculation:

Given: ν_oil = 1.0 × 10^-5 m²/s, v_w = 0.89 × 10^-6 m²/s, D_oil = 0.5 m, D_w = 20 mm = 0.02 m, V_oil = 10 m/s, V_w = ?

\({\left( {\frac{{VD}}{v}} \right)_P} = {\left( {\frac{{VD}}{v}} \right)_m}\)

\(\frac{{10 \times 0.5}}{{1.0 \times {{10}^{ - 5}}}} = \frac{{{V_w} \times 0.02}}{{0.89 \times {{10}^{ - 6}}}}\)

Concept:

The Reynolds number for flow over plate can be given as:

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

Where

X is the distance from leading edge.

U is the free mean velocity.

μ is the dynamic viscosity of fluid flowing.

ρ is the density of fluid flowing.

If Reynolds number is less than or equal to critical Reynold’s number than boundary layer formed over flat plate is laminar i.e.R_e ≤ R_eCr­

Calculation:

Given:

U = 1 m/s ; ρ = 1000 kg/m³ ; μ = 1 centipoise = 10^-3 N s/m²

R_{e cr} = 5 × 10⁵

Let ‘x’ is the distance from leading edge up to which Boundary layer formed is laminar,

Now,

R_ex ≤ R_{e cr}

\(\frac{{\rho Ux}}{\mu } \le 5 \times {10^5}\)

\(\frac{{{{10}^3} \times 1 \times x}}{{{{10}^{ - 3}}}} \le 5 \times {10^5}\)

x ≤ 0.5 m

∴ The maximum distance from leading edge of the plate up to which boundary layer is laminar is 0.50 m.