Froude Model Law MCQ Quiz in বাংলা - Objective Question with Answer for Froude Model Law - বিনামূল্যে ডাউনলোড করুন [PDF]

Given that Froude and Reynolds’s number is same for both model and prototype

\(\therefore {\left( {\frac{V}{{\sqrt {gL} }}} \right)_m} = {\left( {\frac{V}{{\sqrt {gL} }}} \right)_P}\)       … (i)

\( \Rightarrow \frac{{{V_m}}}{{{V_P}}} = \sqrt {\frac{{{L_m}}}{{{L_P}}}} = {\left( {\frac{{{L_m}}}{{{L_P}}}} \right)^{\frac{1}{2}}}\)

\({\left( {\frac{{\rho VL}}{\mu }} \right)_m} = {\left( {\frac{{\rho VL}}{\mu }} \right)_p}\)           … (ii)

\( \Rightarrow \frac{{{V_m}}}{{{V_P}}}.\frac{{{L_m}}}{{{L_P}}} = \frac{{{\mu _m}}}{{{\mu _P}}} \Rightarrow {\left( {\frac{{{L_m}}}{{{L_P}}}} \right)^{\frac{1}{2}}}.\frac{{{L_m}}}{{{L_P}}} = \frac{{{\mu _m}}}{{{\mu _P}}} \Rightarrow \frac{{{\mu _m}}}{{{\mu _P}}} = {\left( {\frac{{{L_m}}}{{{L_P}}}} \right)^{\frac{3}{2}}}\)

\(\frac{{{{\rm{\mu }}_{\rm{m}}}}}{{{{\rm{\mu }}_{\rm{p}}}}} = 0.0894\)

From (i) and (ii)

\(\frac{{{L_m}}}{{{L_p}}} = {\left( {\frac{{{\mu _m}}}{{{\mu _p}}}} \right)^{2/3\;}}\)

\( \Rightarrow \frac{{{{\rm{L}}_{\rm{m}}}}}{{{{\rm{L}}_{\rm{p}}}}} = {\left( 0.0894 \right)^{2/3}}=0.2=1:5\)

Model Scale:

L_m : L_p = 1 : 5

Note:

Froude number is defined as the square root of the ratio of inertia force of a flowing fluid to the gravity force.

\({F_e} = \sqrt {\frac{{{F_i}}}{{{F_g}}}} = \sqrt {\frac{{\rho A{V^2}}}{{\rho ALg}}} = \sqrt {\frac{{{V^2}}}{{Lg}}} = \frac{V}{{\sqrt {Lg} }}\)

Reynold Number is defined as the ratio of inertia force of a flowing fluid to the viscous force of the fluid.

\({{\mathop{\rm R}\nolimits} _e} = \frac{{{F_i}}}{{{F_v}}} = \frac{{\rho VL}}{\mu }\)

Concept:

Froude’s Model Law:

\(\frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt {{{\rm{L}}_{\rm{m}}}} }} = \frac{{{{\rm{V}}_{\rm{p}}}}}{{\sqrt {{{\rm{L}}_{\rm{p}}}} }}\)

Calculation:

Given:

V_p = 1 m/s, L_m: L_p = 1:25

Now, we know that

\(\frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt {{{\rm{L}}_{\rm{m}}}} }} = \frac{{{{\rm{V}}_{\rm{p}}}}}{{\sqrt {{{\rm{L}}_{\rm{p}}}} }}\)

\( \Rightarrow {{\rm{V}}_{\rm{m}}} = {{\rm{V}}_{\rm{p}}}\sqrt {\frac{{{{\rm{L}}_{\rm{m}}}}}{{{{\rm{L}}_{\rm{p}}}}}} = 1 \times \sqrt {\frac{1}{{25}}} = 0.2{\rm{\;m}}/{\rm{s}}\)

Important Points

Froude number:

Froude number is defined as the square root of the ratio of the inertia force of a flowing fluid to the gravity force.

\({F_e} = \sqrt {\frac{{{F_i}}}{{{F_g}}}} = \sqrt {\frac{{\rho A{V^2}}}{{\rho ALg}}} = \sqrt {\frac{{{V^2}}}{{Lg}}} = \frac{V}{{\sqrt {Lg} }}\)

\({l_r} = \frac{{{l_m}}}{{{l_p}}} = \frac{1}{{100}}\)

Froude model is applicable here because of influence of gravity force using Froude model law

\(\begin{array}{l} {V_r} = \sqrt {{l_r}} \\ \frac{{{F_m}}}{{{F_p}}} = \frac{{1kg}}{{{F_p}}} = \frac{{\rho l_m^2V_m^2}}{{\rho l_p^2V_p^2}} = l_r^3 = {\left( {\frac{1}{{100}}} \right)^3} \end{array}\)

or, F_P = 1000000 kg

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

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

Froude law is valid

\(\begin{array}{l} {f_r} = \frac{{{V_r}}}{{\sqrt {g{L_r}} }}\\ {V_r} = \sqrt {{L_r}} \\ {a_r} = \frac{{{V_r}}}{{{t_r}}} = \frac{{{V_r}}}{{\frac{{{L_r}}}{{{V_r}}}}}\\ = \frac{{V_r^2}}{{{L_r}}} = \frac{{{L_r}}}{{{L_r}}} = 1 \end{array}\)

For dynamic singularity, as per Froude law,

\(\begin{array}{l} {\left( {\frac{V}{{\sqrt {gL} }}} \right)_m} = {\left[ {\frac{V}{{\sqrt {gL} }}} \right]_p}\\ \therefore \frac{{{V_p}}}{{{V_m}}} = \sqrt {\frac{{{L_p}}}{{{L_m}}}} = \sqrt {60} \end{array}\)

Propulsive force of prototype

\(\begin{array}{l} {F_p} = {F_m}{\left( {\frac{{{V_p}}}{{{V_m}}}} \right)^2} \times {\left( {\frac{{{L_p}}}{{{L_m}}}} \right)^2}\\ = {F_m}{\left( {\sqrt {\frac{{{L_p}}}{{{L_m}}}} } \right)^2} \times {\left( {\frac{{{L_p}}}{{{L_m}}}} \right)^2} \end{array}\)

= 0.5 × 10 × (60) × 60²

As Froude No is equal So (Fe)_m = (Fe)_p

\( \Rightarrow \frac{{{V_m}}}{{\sqrt {{g_m}{L_m}} }} = \frac{{{V_P}}}{{\sqrt {{g_p}{L_p}} }}\)

Since g_m = g_p

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

\( \Rightarrow \frac{{{{\rm{V}}_{\rm{p}}}}}{{{{\rm{V}}_{\rm{m}}}}} = \sqrt {\frac{{{{\rm{L}}_{\rm{p}}}}}{{{{\rm{L}}_{\rm{m}}}}}} = \sqrt {{{\rm{L}}_{\rm{r}}}} \) 

Where L_r = Scale ratio for length

For open channel

\({\left. {{F_r}} \right)_p} = {\left. {{F_r}} \right)_m} \Rightarrow {\left. {\frac{{\frac{Q}{A}}}{{\sqrt L }}} \right)_p} = \left. {{{\left. {\frac{{\frac{Q}{A}}}{{\sqrt L }}} \right)}_m}} \right|\frac{{{L_p}}}{{{L_m}}} = 16\)

⇒Q_m = 4m³/s  c)