1.A turbine is rated at 650 hp when the flow of water through it is 0.85 m3/sec. Assuming an efficiency of 84%, what is the head acting on the turbine.

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• Welcome to Math.SE! We like to see questions that include context. However, I would like to point out that this is not a math problem. Yes, you use math to solve the problem, but it is not a problem about mathematics. Instead, it is a physics problem... and certainly not a group-theory problem. – apnorton Aug 12 '15 at 4:46
• @apnorton: This is a Physics question. You are correct that it will be closed in a heartbeat if migrated. It will only land back here if OP is intransigent. I participate on both sites and prefer our homework policy, but this question supports the motivation for theirs-they get many more plug and chug than we do. – Ross Millikan Aug 12 '15 at 4:51
• @RossMillikan Ohh.... I forgot that failed migrations result in closure on the original site (I was thinking they just "bounced" back here for some reason). Thanks for reminding me. – apnorton Aug 12 '15 at 4:53
• Hopefully the people that follow fluid dynamics won't murder me for sending this question there (I don't know anything about fluid dynamics...). Although hopefully this question wont be around long enough for them to find out, they didn't even remove the question numbering. (I got to this question thinking it would be some cool question connecting thermodynamics and group theory :(, I was mistaken) – Paul Plummer Aug 12 '15 at 4:55
• Thank you Harish Chandra Rajpoot it really helps me. And sorry about this question guys, Im having really a hard time here because of this question and no one to help. Im almost done, but Im having hard time because of this question. Sorry guys – Christopher Balmes Aug 13 '15 at 0:51

This is a nice question on $\color{blue}{\text{Hydraulic Turbine}}\ \color{red}{ \text{(FLUID MECHANICS)}}$
Given $$\text{rated power}, P_{\text{rated}}=650 \ H.P.=650\times 746=484900 \ W$$ $$\text{discharge}, Q=0.85\ m^3/sec$$ $$\text{efficiency }, \eta=84\ \text{%}$$
Let, $H$ be head under which the turbine is working
$$\eta=\frac{\text{Output power of turbine}}{\text{Rated power of turbine}}$$ $$=\frac{\rho g Q H}{P_{\text{rated}}}$$ Where, $\rho=\text{density of water}=1000\ kg/m^3$ & $g=9.81\ m/sec^2$ $$H=\frac{\eta P_{\text{rated}}}{\rho gQ}$$ Now, setting the corresponding values, we get $$H=\frac{0.84\times 484900}{1000\times 9.81\times 0.85}$$ $$H\approx 48.8476\ m$$