# Hazen Williams Eq'n

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### #1 tymr

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Posted 27 January 2009 - 08:01 PM

The Hazen-Williams Eq'n to calculate head loss as presented in CERM 10 is stated:
hf = (10.44*L*Q^1.85)/(C^1.85*d^4.87), for US units.

However, in the Standard Handbook for Civil Engineers the Hazen-Williams Eq'n to calculate head loss is stated:
hf = (4.727/d^4.87)L(Q/C)^1.85, for US units. NCEES seems to support this eq'n instead of the Lindberg eq'n but I cannot figure out why there is such a significant difference between the value of the constant between the eq'ns. What am I missing?

### #2 Chucktown PE

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Posted 27 January 2009 - 08:56 PM

QUOTE (tymr @ Jan 27 2009, 03:01 PM) <{POST_SNAPBACK}>
The Hazen-Williams Eq'n to calculate head loss as presented in CERM 10 is stated:
hf = (10.44*L*Q^1.85)/(C^1.85*d^4.87), for US units.

However, in the Standard Handbook for Civil Engineers the Hazen-Williams Eq'n to calculate head loss is stated:
hf = (4.727/d^4.87)L(Q/C)^1.85, for US units. NCEES seems to support this eq'n instead of the Lindberg eq'n but I cannot figure out why there is such a significant difference between the value of the constant between the eq'ns. What am I missing?

I have always used 10.44 but as always it depends on how you derive the equation and what units you are using. If Q is in gpm, L is in feet, and d is in inches then use 10.44. My guess is the 4.727 comes from using a different unit for one of those variables.

### #3 tymr

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Posted 28 January 2009 - 04:09 PM

Thanks. I figured it was a conversion. Units is what took care of it. The 4.727 is when Q is ft^3/sec and d is in ft. Life, as I know it, can go on.

### #4 Chucktown PE

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Posted 28 January 2009 - 04:38 PM

QUOTE (tymr @ Jan 28 2009, 11:09 AM) <{POST_SNAPBACK}>
Thanks. I figured it was a conversion. Units is what took care of it. The 4.727 is when Q is ft^3/sec and d is in ft. Life, as I know it, can go on.

I had a fluid mechanics professor in grad school that really hated Hazen Williams and Mannings equations because they weren't dimensionally homogenious. However, he knew there was nothing better. He preferred for us to use Reynold's numbers to solve for headloss.

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