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.
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.
We know $Q = AV$, and $A = \pi$$D^2/4$, let's rearrange the terms and get to the root "D":
$Q_1^2 = (A_1V_1)^2 = (\pi$$D_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)
