# Pumps' geometric similarity

Given two pumps that are similar :

$$H_1=70-1.41x10^{-3}Q_1^2$$

$$H_2=100-1.1x10^{-3}Q_2^2$$

Find the geometric scale :

$$\frac{D1}{D2}= ?$$

I am also given maximum efficiency = 1.

I tried solving but I got stuck I don’t know which pi terms should I use for instance:

I tried $$\pi_{gh},\pi_Q,\pi_{\nu}$$ and got stuck with hard complicated algebra.

Here is the result in the denominator it should be $$Q_2^{2}$$. Left side of the lower (2nd) equation. • I am willing to offer bounty because I have been sitting on this question for 2 Hours Apr 21 at 13:12
• Check out the dimensionless groups - a good textbook will have them. Apr 21 at 13:29
• I did also checked from lecture all good but I get weird result Apr 21 at 13:31
• @Mather: In that case, please edit your question to show us precisely what you tried and what results you got, not just that they were weird.
– Wasabi
Apr 21 at 15:56
• I have updated the result , from the 3 pi groups I got the equation in the picture , I don’t know how to solve it Apr 23 at 12:50

The key principle is that, when expressed in terms of the non-dimensional head coefficient $$K_H := gH/\left(D\omega\right)^2$$ and the non-dimensional flow coefficient $$K_Q := Q/\left(D^3\omega\right)$$, there is a single head-flow characteristic for the whole family of geometrically similar pumps. That is to say, when expressed in terms of those quantities, the given dimensional head-flow characteristics of the two pumps will both collapse onto the same equation.

So the first given head-flow characteristic $$H_1 = A_1-B_1Q_1^2$$ $$\Rightarrow \frac{g_1H_1}{\left(D_1\omega_1\right)^2} = \frac{g_1A_1}{\left(D_1\omega_1\right)^2}-\frac{g_1B_1Q_1^2}{\left(D_1\omega_1\right)^2} = \left(\frac{g_1A_1}{\left(D_1\omega_1\right)^2}\right)-\left(g_1D_1^4B_1\right)\left(\frac{Q_1}{D_1^3\omega_1}\right)^2\textrm{,}$$ i.e. $$K_H = \left(\frac{g_1A_1}{\left(D_1\omega_1\right)^2}\right)-\left(g_1D_1^4B_1\right)K_Q^2\textrm{.}$$ Similarly, the second given head-flow characteristic $$H_2 = A_2-B_2Q_2^2$$ leads to $$K_H = \left(\frac{g_2A_2}{\left(D_2\omega_2\right)^2}\right)-\left(g_2D_2^4B_2\right)K_Q^2\textrm{.}$$

In order for both head-flow characteristics to have collapsed to the same equation as required, we must have $$\frac{g_1A_1}{\left(D_1\omega_1\right)^2} = \frac{g_2A_2}{\left(D_2\omega_2\right)^2}$$ and $$g_1D_1^4B_1 = g_2D_2^4B_2\textrm{.}$$

The second of those two equations rearranges to $$\frac{D_1}{D_2} = \left(\frac{g_2}{g_1}\frac{B_2}{B_1}\right)^{1/4}\frac{\omega_2}{\omega_1}\textrm{;}$$ If we assume both pumps are on the same planet $$g_1 = g_2$$, then $$\frac{D_1}{D_2} = \left(\frac{B_2}{B_1}\right)^{1/4} = \left(\frac{1.1}{1.41}\right)^{1/4} = 0.94\textrm{.}$$

I observe in passing that, in the dimensional head-flow characteristic equations as they appear at the start of the question, someone was very naughty in forgetting to provide units on the numerical $$A_i$$ and $$B_i$$ values, thus making the equations dimensionally inconsistent; this probably makes it harder than it needs to be to see the way forward with the problem.

• No-one noticed that I messed up the definition of $K_Q$; all corrected now. Apr 26 at 9:24

We know $$Q = AV$$, and $$A = \piD^2/4$$, let's rearrange the terms and get to the root "D":

$$Q_1^2 = (A_1V_1)^2 = (\piD_1^2V_1/4)^2$$ = $$(70-H_1)/1.41 * 10^{-3}$$

$$D_1^2 = (4/V_1\pi)\sqrt{(70 - H_1)/1.41 * 10^{-3})}$$

$$D_1 = \sqrt{(4/V_1\pi)\sqrt{(70 - H_1)/1.41 * 10^{-3})}}$$

You can do the same for $$D_2$$ and get $$D_1/D_2$$. (First to check my operation for mistakes)

• I solved the question it needed 2 pi terms not 3 pi groups as I thought , I was close but the algebra was hidden thanks anyway Apr 23 at 17:54
• I've added a "close parenthesis" to the term 70-H1. Also, you can simplify the process by setting the entire term (70-H1)/1.41x10-3 as a constant, then plug it back to the final expression for D. It is more clear, thus causes less chance of making mistakes.
– r13
Apr 23 at 18:21
• Two problems with this. Firstly, there's no particular reason to suppose that the cross-sectional area $Q/V$ through which the fluid flows is the same as the area $\mathrm{\pi}D^2/4$ swept out by the impeller blades. Secondly, you're still left with too many unknowns (heads, velocities, diameter ratio) for the number of equations you've got. Apr 25 at 16:16
• @Daniel Hatton My effort was to tip the OP over the hump of algebra operations, not to do his homework. Cong that he has selected your answer,
– r13
Apr 25 at 20:45
• @r13 Yeah, the risk occurred to me, but I figured I was late enough to the party that it probably wasn't a big problem any more. Apr 25 at 21:16