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Chapter 7: Problem 89

An axial-flow pump is required to deliver \(0.75 \mathrm{m}^{3} / \mathrm{s}\) of water at a head of \(15 \mathrm{J} / \mathrm{kg}\). The diameter of the rotor is \(0.25 \mathrm{m}\) and it is to be driven at 500 rpm. The prototype is to be modeled on a small test apparatus having a \(2.25 \mathrm{kW}, 1000\) rpm power supply. For similar performance between the prototype and the model, calculate the head, volume flow rate, and diameter of the model.

Short Answer

Expert verified
The values calculated are as follows: Head in the model (H')= \(30 \, \mathrm{J/kg}\), Volume flow rate in the model (Q') = \(1.5 \, \mathrm{m}^3/\mathrm{s}\), Diameter of the model (D') = \(0.5 \, \mathrm{m}\).

Step by step solution

01

Calculation for the head in the model

The head of the model can be calculated using the similitude between the model and the prototype. Due to dynamic similarity, the ratio of the head of the model to the head of the prototype is equal to the square of the ratio of their speeds (N/N'). Here, the speeds are given in revolutions per minute (rpm). As such, we need to convert the speed from rpm to rad/s by using the fact that 1 rpm = \(\frac{2 \pi}{60}\) rad/s before the calculation. Hence, \(H' = H \cdot (\frac{N}{N'})^2\).
02

Calculation for the volume flow rate in the model

The volume flow rate in the model can be calculated by using the relationship of kinematic similitude between the flow rates in the model and the prototype. The ratio of the flow rates (Q/Q') is equal to the cube of the ratio of their speeds (N/N'). Hence, \(Q' = Q \cdot (\frac{N}{N'})^3\).
03

Calculation for the diameter of the model

The diameter of the model pump can be determined using the principle of geometric similitude. The ratio of the diameters (D/D') is equal to the ratio of their speeds (N/N'). Hence, \(D' = D \cdot (\frac{N}{N'})\).

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