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Consider I have a pump sized to deliver 200 gpm of a fluid with a specific gravity of 0.7 and a total differential head of 60 ft of water. What is the hydraulic horse power?

To calculate the hydraulic horse power from the following equation: $$ \text{Hyd HP} = Q \rho g h $$ Wouldn't I need to convert the "60 ft of water" to the equivalent head of my fluid (SG=0.7)? Something like $$ P=\frac{\rho g h_{water}}{g_c} = \frac{62.4\cdot 60\cdot 32}{32} = 3744 \, \text{psf}\\ h_{0.7}= \frac{P g_c}{\rho g} = \frac{3744\cdot 32}{0.7\cdot 62.4 \cdot 32} \approx 85 \, \text{ft} $$

Or am I overthinking this?

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  • $\begingroup$ The question title maybe should have been "hydraulic hp" instead of "pump power". I understand that the difference between the two is the pump efficiency. $\endgroup$ – Nukesub Oct 25 '20 at 21:58
  • $\begingroup$ you are overthinking it. h is there to account for the equivalent of potential energy, like lifting a rock. Density is already in the equation. $\endgroup$ – Tiger Guy Oct 26 '20 at 5:07
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    $\begingroup$ @TigerGuy But the density in the equation is for water, OP needs to use density of the fluid being pumped. $\endgroup$ – Algo Oct 26 '20 at 6:27
  • $\begingroup$ @Algo, I suppose of h is using units of water instead of units of the fluid, you are right. I wouldn't express head in inches of water if I weren't pumping water, but there's nothing saying you can't. $\endgroup$ – Tiger Guy Oct 26 '20 at 7:42
  • $\begingroup$ Right - the problem statement specifically says "total differential head of 60 ft of WATER" so as @Algo pointed out, I think that means i need to convert it to an equivalent column of my fluid (SG=0.7). Right? $\endgroup$ – Nukesub Oct 26 '20 at 17:26

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