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A geometrically similar model is made to the scale ratio of 1:14 to study the characterstics of spillway. If the energy dissipated per second in prototype is 15000 N-m, what will be the Energy dissipated per sec in model ________ N – m.
Correct answer is '20-21'. Can you explain this answer?
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A geometrically similar model is made to the scale ratio of 1:14 to st...
Concept: As it is the case of spillway, So Fraude model law is applicable
Pm = Power in model = ρQmgHm
Pp = Power in Prototype = ρQpgHp
Calculation:
Pm = 20.45 N – m per second
Hence power in model will be 20.45 N – m per second
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A geometrically similar model is made to the scale ratio of 1:14 to st...
The energy dissipated per second in the model can be found using the following formula:
Energy dissipated in model = (Scale factor)^3 x Energy dissipated in prototype
Here, the scale factor is 1:14, which means that every dimension in the model is 1/14th of the corresponding dimension in the prototype. Therefore, the scale factor is 1/14.
Using this information, we can calculate the energy dissipated in the model as follows:
Energy dissipated in model = (1/14)^3 x 15000 N-m
Energy dissipated in model = 0.0018 x 15000 N-m
Energy dissipated in model = 27 N-m
Therefore, the energy dissipated per second in the model is 27 N-m.
Energy dissipated in model = (Scale factor)^3 x Energy dissipated in prototype
Here, the scale factor is 1:14, which means that every dimension in the model is 1/14th of the corresponding dimension in the prototype. Therefore, the scale factor is 1/14.
Using this information, we can calculate the energy dissipated in the model as follows:
Energy dissipated in model = (1/14)^3 x 15000 N-m
Energy dissipated in model = 0.0018 x 15000 N-m
Energy dissipated in model = 27 N-m
Therefore, the energy dissipated per second in the model is 27 N-m.
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A geometrically similar model is made to the scale ratio of 1:14 to st...
Power ratio = density ratio * (length ratio)^(7/2)
Assuming density to be constant,
Power for model = power for prototype * (1/14)^3.5
= 150000* (1/14)^3.5
= 1.46 N-m
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A geometrically similar model is made to the scale ratio of 1:14 to study the characterstics of spillway. If the energy dissipated per second in prototype is 15000 N-m, what will be the Energy dissipated per sec in model ________ N – m.Correct answer is '20-21'. Can you explain this answer?
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