1.16. Problem 5: Corresponding Model Quantities

A proposed model of a river stretch of 15 kilometres is to have horizontal scale of 1 / 200 and a vertical scale of 1 / 400. What does this question say? I mean, what does this indicate? This indicates that we are dealing with distorted scales because we have a different vertical scale, sorry, a different, we have a different horizontal scale and a different vertical scale.

In this modeling scenario, we are working with a river stretch of 15 kilometers, and the scales for the model are different in horizontal and vertical dimensions. The horizontal scale of 1/200 means that for every 200 units in the model, it corresponds to 1 unit in the real (prototype) river. Similarly, the vertical scale of 1/400 indicates that for every 400 units in the model's vertical representation, 1 unit corresponds to the actual river's vertical dimension. This discrepancy indicates that we are creating a distorted model.

Think of a cake model that you want to make look like a real bakery cake. If you use a large plate as your base (the horizontal scale), you might decide that the height (the vertical scale) should be much smaller to fit the display. The height of your doll cake model might be different from the actual height of the cake it represents, just like how our river model has different scales for width and depth.

So, Q_m / Q_p we have already found out, it was L_r into h_r to the power 3 / 2. So, implies Q_m is going to be Q_p into L_r into h_r to the power 3 / 2 and Q_p is 152 meter cube per second multiplied by 1 / 200 and h_r is 1 / 40 to the power 1.5 and this will give 0.03 meter cube per second in model.

To find the model discharge (Q_m), we use the relationship derived from the Froude model law, which states that the model discharge is proportional to the prototype discharge (Q_p) multiplied by the horizontal and vertical scales. In this case, the prototype discharge is given as 152 cubic meters per second. We combine the ratios for length (1/200) and the vertical scale (1/40 raised to the power of 1.5) to compute the model discharge. This calculation leads us to determine that the model discharge will be 0.03 cubic meters per second.

Imagine using a watering can (the model) to pour water that represents the flow of a river (the prototype). If you know how much water flows in the real river, you can calculate how much water needs to come out of the smaller can to mimic that flow, taking into account how much smaller the can is compared to the actual river.

Second thing is depth, y_m / y_p is h_r, simple. So, this implies y_m is equal to y_p into h_r or y_p here given is, so y_p the depth was actually, so I should have actually first. So, I will also write down the known quantities. Q prototype was given as, 152 meter cube per second, width was given as, 90 meter and y_p was given as, 2 meter. So, y_p was 2 meter into h_r was 1 / 40 and it comes out to be 0.05 meter. So, that means in model it is 0.05 meter, y_m and Q_m was 0.03 meter cube per second.

To find the model depth (y_m), we establish a direct relationship with the prototype depth (y_p) using the vertical scale (h_r). Given that the prototype depth y_p is 2 meters, we multiply this by the vertical scaling factor (1/40) to get the model depth. This calculation indicates that the model depth will be 0.05 meters, making it 40 times smaller than the actual depth of the river.

If you were transitioning a swimming pool into a model for a toy city, you would scale down the depth of the pool in the model accordingly. So if the real pool is 2 meters deep, your model swimming pool might only be 0.05 meters deep, just like our river model reflects a smaller version of the actual river's depth.

Now, the third thing is width, B_m we have to find. So, B_m / B_p is the length ratio. So, this implies, B_m is, so width is B_p basically. So, B_m is L_r into B_p. So, L_r 1 / 200 into B_p was 90 meters. So, B_m comes out to be 0.045 meters.

The width of the model (B_m) is calculated by multiplying the prototype width (B_p) by the horizontal scale (L_r). Given that the prototype width of the river is 90 meters and applying the scale factor of 1/200, we find that the model width will be 0.045 meters. This step helps us ensure that both the physical and proportional representations of the river are maintained in the model.

This is similar to designing a model airplane. If the real airplane is 90 meters wide, but you're building a model that's 1/200 the size, you'd need to adjust its wingspan to be much smaller. Thus, your model airplane might only be 0.045 meters wide, maintaining the same proportions as the real aircraft.

So, Manning's n, n_m / n_p, you have to find out the formula but it comes out to be, h_r to the power 2 / 3 divided by L_r to the power half. This implies, model n_m is Manning's number in prototype into h_r to the power 2 / 3 L_r to the power half. So, in the known quantities we were also told the prototype roughness was given, that was 0.025. So, this is 0.025 into h_r was 1 / 40 to the power 2 / 3 divided by 1 / 200 to the power half. So, eta_m comes out to be 0.03. So, prototype roughness was 0.025 and the model roughness is 0.03. So, model here, has to be rougher then the prototype.

We determine the Manning's roughness coefficient (n_m) for the model using a derived formula that accounts for both the horizontal and vertical scales of the model. Plugging in the known prototype roughness (0.025) and the scaling factors yields a model roughness of 0.03. This indicates that, while other dimensions decrease, the roughness must increase in the model to accurately simulate the flow characteristics of the river.

Consider a different scenario where different surfaces affect a toy car's movement over a tabletop. If the prototype has a smooth road surface that represents low roughness, the model may use a rough-textured surface to ensure the toy car reflects what would happen on an actual road. This ensures the model behaves like a real-life scenario, albeit in a smaller, controlled environment.

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