The flow in river is 1500 cumecs. A distorted model is build with

SOLUTION

Concept:

Distorted model:

A model is said to be distorted if it is not geometrically similar to its prototype. For a distorted model different scale ratios for the linear dimensions are adopted.

Advantages of distorted model:

1. The vertical dimension of the model can be measured accurately.

2. The cost of the model can be reduced.

3. Turbulent flow in the model can be maintained.

Scale ratio for horizontal dimension, (L_r)_H = \(\frac{{{L_P}}}{{{L_m}}} = \frac{{{B_P}}}{{{B_m}}} = \frac{{Linear\;horizontal\;dimension\;of\;prototype}}{{Linear\;horizontal\;dimension\;of\;model}}\)

Scale ratio for vertical dimension, (L_r)_v = \(\frac{{{h_p}}}{{{h_m}}} = \frac{{Linear\;vertical\;dimension\;of\;prototype}}{{Linear\;vertical\;dimension\;of\;model}}\)

Then the scale ratio of velocity, area of flow, discharge etc., in terms of (L­_r)_H and (L_r)_V can be obtained for distorted models as given below:

1. Scale ratio for velocity

Let  V_P = Velocity in prototype

V_m = Velocity in model

Then \(\frac{{{V_P}}}{{{V_m}}} = \frac{{\sqrt {2g{h_p}} }}{{\sqrt {2g{h_m}} }} = \sqrt {\frac{{{h_p}}}{{{h_m}}}} = \sqrt {{{\left( {{L_r}} \right)}_V}} \)

2. Scale ratio for area of flow

Let A_P = Area of flow in prototype = B_P × h_p

A_m = Area of flow in model = B_m × h_m

\(\therefore \;\;\frac{{{A_P}}}{{{A_m}}} = \frac{{{B_P} \times {h_P}}}{{{B_m} \times {h_m}}} = \frac{{{B_P}}}{{{B_m}}} \times \frac{{{h_P}}}{{{h_m}}} = {\left( {{L_r}} \right)_H} \times {\left( {{L_r}} \right)_V}\)

3. Scale ratio for discharge

Let Q_P = Discharge through prototype = A_P × V_P

Q_m = Discharge through model = A_m × V_m

\(\therefore \;\frac{{{Q_P}}}{{{Q_m}}} = \frac{{{A_P} \times {V_P}}}{{{A_m} \times {V_m}}} = {\left( {{L_r}} \right)_H} \times {\left( {{L_r}} \right)_V} \times \sqrt {{{\left( {{L_r}} \right)}_V}} = {\left( {{L_r}} \right)_H} \times {\left[ {{{\left( {{L_r}} \right)}_V}} \right]^{3/2}}\)

Calculation:

\({\left( {{L_r}} \right)_H} = \frac{1}{{150}},\;\;{\left( {{L_r}} \right)_V} = \frac{1}{{25}},\;{Q_m} = 1500\;{m^3}/s\)

\(\left( {{Q_r}} \right) = \frac{1}{{150}} \times {\left( {\frac{1}{{25}}} \right)^{3/2}} = \frac{1}{{18750}}\)

\(\frac{{{Q_M}}}{{{Q_P}}} = \frac{1}{{18750}} \Rightarrow \;{Q_m} = \frac{{1500}}{{18750}} = 0.08\;{m^3}/{s^{ - 1}}\)