Page: Model Testing of centri fugal pump.
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Before manufacturing any large sized pumps, their model which are completely similar with actual pumps called as prototypes are made.

Then tests are conducted on the models and performance of the prototypes are predicted.

There would be a complete similarity between a model and actual pump only if the following conditions exist.

1] Specific speed of model = specific speed of prototype.

$(Ns)_m = (Ns)_p$

$(\frac{N \sqrt Q}{H_m^{3/4}})_m = \frac{N\sqrt Q}{H_m^{3/4}})_p$------(1)

2] Tangential velocity (u)

$u = \frac{\pi DN}{60} or u \alpha \sqrt H$

$\therefore$ $\sqrt{H_n} \alpha DN$

$\therefore$ $\frac{\sqrt{H_n}}{DN} = constant$

or $(\frac{\sqrt {H_m}}{DN})_m = (\frac{\sqrt {H_m}}{H_m}_p$

3] From equation $(Q \alpha D^2 \times vf)$

$\therefore$ $Q \alpha D^2 \times vf$

$\alpha D^2 \times D \times N$

$\alpha D^3 \times N$

$\therefore$ $\frac{Q}{D^3N} = constant$

or $(\sqrt {Q}{D^3N})_m = (\frac{Q}{D^3N})_p$

4] Power of the pump.

$p = \frac{p \times 9 \times Q \times H_m}{75}$

$p \alpha Q \alpha H_m$

$\alpha D^3 \times N \times H_m$

$\alpha D^3 N \times D^2 N^2$

$\alpha D^5 N^3$

$\therefore$ $\frac{P}{D^5 N^3} = constant$

or $(\frac{P}{D^5 N^3})_m = (\frac{P}{D^5 N^3})_p$

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 modified 2 days ago by Abhishek Tiwari ★ 50 written 9 weeks ago by RB ♦♦ 100