# head coefficient and volume coefficient for centrifugal pump

If I need to calculate the head coefficient using $$c_h=\dfrac{gH}{N^2D^2}$$ and the volume coefficient using $$c_q= \dfrac{v}{ND^3}$$ while keeping $$N$$ (impeller rotational speed in rpm), how can I make the units cancel out and have a dimensionless answer?

I am measuring $$D$$ (impeller diameter in meters), $$g$$ (gravity m/s^2), $$H$$ (total pressure head in meters), $$V$$ (volumetric flow rate in m/s).

I made it dimensionless by changing RPM to 1/s, however I have been asked to keep RPM and change the other units, can someone plz help.

RPM is just another way to say $${min^{-1}}$$ so to make your equations dimensionless while still using RPM you need to convert your other measurements using seconds to minutes.

$$c_h=\dfrac{gH}{N^2D^2}$$
this means you need to convert $$g$$ from $$m.s^{-2}$$ to $$m.min^{-2}$$ (in other words multiply by 3600)

For the volume coefficient
$$c_q= \dfrac{v}{ND^3}$$
you need to convert volumetric flow into $$m^3.min^{-1}$$ from $$m^3.s^{-1}$$ (in other words multiply by 60)

For a general method, you would take your equation, substitute all the terms with their units and simplify by cancelling out, then see what you are left with and find a conversion factor that cancels out the remaining units. You then stick the conversion factor into the original equation and use your original units of measurement. For example in the head coefficient equation:

Units: $$\dfrac{(m.s^{-2})(m)}{(min^{-1})^2(m)^2}$$

This simplifies to $$\dfrac{min^2}{s^2}$$ so the conversion factor we need is $$\dfrac{(60s)^2}{min^2}$$ which you can then replace into the original equation and use the original units when measuring, i.e.

$$c_h=\dfrac{3600gH}{N^2D^2}$$

(remembering that the units of your conversion factor 3600 is $$s^2min^{-2}$$)

Volumetric flow rate is not m/s, it is m^3/s...

But if you want rpm in minutes then have the volumtric flow rate in m^3/minute.

This is just balancing units.

• Yes sorry my mistake, volumetric flow rate was measured in Liters/second. So if I convert l/s to m^3/min, do the units cancel out? – Shahad Feb 24 '19 at 9:28
• What is the difference between minutes and seconds? You should be able to cancel units... This is dimensional analysis 101... – Solar Mike Feb 24 '19 at 9:33
• I mean will 1/rpm cancel with minutes? Since I'm confused about the concept of rpm . I know its #of revolutions per minute. r/min. but do revolutions have units? – Shahad Feb 24 '19 at 10:17
• Think of rpm is (1/min) – Solar Mike Feb 24 '19 at 10:18
• Why not put all the terms into the equation, replace each with their units (ie Mass, Length or Time) and show they cancel out... – Solar Mike Feb 24 '19 at 10:22