Mike11s
Civil/Environmental
- Oct 29, 2010
- 3
Just one more thought on the water hammer thread - it's true that the correct and standard method is a highly professional task, separating harmonic DE's as per the method of characteristics, but there is an easy approximation that does just as well:
For a slow control operation such as closing a valve, the time of closure tc has to be greater than the return time tr = 2L/c where L is the length of pipe and c is the celerity. This ensures that the positive pressure wave is reduced by the returning negative wave.
You can use this to design a closure time tc > tr that reduces the positive pressure surge to whatever your design safety pressure is, say Pdesign = 100 kPa.
The idea is to assume a constant flow deceleration from your start flow velocity V1 down to zero, in the interval of your closure time tc - this is not too far from reality and you can experiment with closure rates to get it close to a constant rate of pressure increase, which is the effect you need.
The deceleration is then approximately dV/dt = V1/tc (tc still unknown) and constant. The rate of pressure increase is then constant as per compression-continuity, ie
dP/dt = (density x c) x dV/dt = (density x c) x V1/tc
(This is from DP = (d x c) x V1 - just differentiate wrt time)
OK, so you have a pressure increase rate. The pressure will increase up to the moment when the return wave arrives, after which it is effectively reduced by the negative pressure going in the other direction. The maximum pressure occurs at the return time, tr = 2L/c, and has a value of
Pmax = dP/dt x tr
= (d x c x V1 / tc) x (2L/c)
= 2d x V1 x L/tc
Which is nicely free of the large celerity c. Assuming a design pressure Pdesign that your system can't exceed, and a safety factor (say 2 to be really safe), you get
Pmax = 2 x Pdesign = 2d x V1 x L/tc
Which allows you to solve for the closure time tc:
tc = d x V1 x L / Pdesign
eg for Pdesign = 100 kPa, V1 = 2 m/s, L = 1000 m -> tc = 20 seconds
- I think this what Jeffvalve had in mind when he was trying to recall the method - I think I'm just reminding him.
Regards, Mike Eleven
thread408-81043
For a slow control operation such as closing a valve, the time of closure tc has to be greater than the return time tr = 2L/c where L is the length of pipe and c is the celerity. This ensures that the positive pressure wave is reduced by the returning negative wave.
You can use this to design a closure time tc > tr that reduces the positive pressure surge to whatever your design safety pressure is, say Pdesign = 100 kPa.
The idea is to assume a constant flow deceleration from your start flow velocity V1 down to zero, in the interval of your closure time tc - this is not too far from reality and you can experiment with closure rates to get it close to a constant rate of pressure increase, which is the effect you need.
The deceleration is then approximately dV/dt = V1/tc (tc still unknown) and constant. The rate of pressure increase is then constant as per compression-continuity, ie
dP/dt = (density x c) x dV/dt = (density x c) x V1/tc
(This is from DP = (d x c) x V1 - just differentiate wrt time)
OK, so you have a pressure increase rate. The pressure will increase up to the moment when the return wave arrives, after which it is effectively reduced by the negative pressure going in the other direction. The maximum pressure occurs at the return time, tr = 2L/c, and has a value of
Pmax = dP/dt x tr
= (d x c x V1 / tc) x (2L/c)
= 2d x V1 x L/tc
Which is nicely free of the large celerity c. Assuming a design pressure Pdesign that your system can't exceed, and a safety factor (say 2 to be really safe), you get
Pmax = 2 x Pdesign = 2d x V1 x L/tc
Which allows you to solve for the closure time tc:
tc = d x V1 x L / Pdesign
eg for Pdesign = 100 kPa, V1 = 2 m/s, L = 1000 m -> tc = 20 seconds
- I think this what Jeffvalve had in mind when he was trying to recall the method - I think I'm just reminding him.
Regards, Mike Eleven
thread408-81043
