Ratio of Discharge in Prototype to Discharge in Model

To understand the given problem, we need to analyze the concept of geometric similarity between a prototype and a model. Geometric similarity states that if two structures are geometrically similar, all their corresponding dimensions are proportional to each other.

In this case, we have a prototype spillway and a geometrically similar model spillway. Let's assume that the scale factor between the prototype and the model is 'n'. This means that every dimension of the model is 'n' times smaller than the corresponding dimension of the prototype.

Ratio of Discharge in Prototype to Discharge in Model:

The discharge in a spillway is determined by various factors such as the cross-sectional area, the velocity of flow, and the coefficient of discharge. Since the prototype and the model are geometrically similar, the cross-sectional area and the velocity of flow should also be proportional to each other.

Let's assume the discharge in the prototype spillway is Qp and the discharge in the model spillway is Qm. According to the given problem, the ratio of Qp to Qm is 35000.

Qp/Qm = 35000

Cross-sectional Area and Velocity of Flow:

The cross-sectional area of flow is directly proportional to the square of the linear dimensions, while the velocity of flow is inversely proportional to the linear dimensions. Using these relationships, we can express the dimensions of the model in terms of the prototype dimensions.

Let's assume the linear dimensions of the prototype spillway are Lp, Bp, and Hp (length, width, and height respectively), and the linear dimensions of the model spillway are Lm, Bm, and Hm.

Lm = n * Lp
Bm = n * Bp
Hm = n * Hp

Since the cross-sectional area is proportional to the square of the linear dimensions, we have:

Am = (Lm * Bm) = (n * Lp) * (n * Bp) = n^2 * (Lp * Bp)

Since the velocity of flow is inversely proportional to the linear dimensions, we have:

Vm = (Qp/Am) = (1/n^2) * (Qp/ (Lp * Bp))

Ratio of Discharge:

Now, let's substitute the expressions for the cross-sectional area and velocity of flow into the ratio of discharge equation:

Qp/Qm = (Qp/Am) / (Qm/Am)
35000 = (1/n^2) * (Qp/ (Lp * Bp)) / (Qm/ (Lm * Bm))

Simplifying the equation:

35000 = (1/n^2) * (Qp/Qm) * ((Lm * Bm)/ (Lp * Bp))

Since Qp/Qm = 35000, we have:

35000 = (1/n^2) * 35000 * ((Lm * Bm)/ (Lp * Bp))

Cancelling out the common terms:

1 = (1/n^2) * ((Lm * Bm)/ (Lp * Bp))

Simplifying further:

n^2 = (Lm * Bm)/ (Lp * Bp)

Therefore, n = sqrt((Lm * Bm)/ (Lp * Bp))

As the scale of the model