wongkf
 
文章: 16
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 于 2004-03-11 14:54    
It's true that your theory can be used to estimate the maximum flow rate of 去水 pipe, but substitue "D" = H is wrong.
You should substitue the height of "level A" in the formula, that is, the water level of the falling chamber of main tank.
The maximum water level allowed is equal to the height of the falling chamber of main tank. So assume your last chamber height is 1m and your pipe size is 0.025m. Then the maximum circulation flow rate allowed would be 7200 L/H. (Assume no friction loss, the falling pipe connect to sump tank without elbow)
So,
if circulation rate > >7200 L/H, the water level of the whole tank rise
If circulation rate = 7200 L/H, the falling chamber would full of water
if circulation rate < 7200, the falling chamber would not full.
Assume "level A" of falling chamber is 0.2m , then, the actual circulation rate is about 3500 L/H.
So, by measuring the height of "level A" and your pipe size, you can estimate your existing circulation flow rate.
But, for the above example, is it mean that we cannot buy a pump of larger than 7200L/H? The answer is no, we still can have pump larger than 7200L/H.
Why? It is because 7200L/H is only the specified flow rate of the pump. It mean that under no resistance, it can output 7200L/H.
But it's true that the total flow rate of the two outlets cannot be larger than 7200L/H.
Example xx)
To size the pump , you need full hydraulic calculation.
Q in L/min
P in bar (assume 1 bar = 10m)
Refer to the drawings
Assume you want to have about 2000 L/H, that is, about 1000L/H from nozzle 1, Q1 = 17 L/min
assume k factor of nozzle 1 = 100, (just assumption, usually check from table)
then, pressure at 1 (P1):
Q1 = 100 x sqrt (P1)
17 = 100 x sqrt(P1)
thus, P1 = 0.0289 bar
neglect length of 1-2, assume supply pipe = 20mm
equivalent lenght (EL) of 20 mm elbow = 0.3m (usually from table),
assume length of 1-3 = 1m
total EL from 1 to 3 = 1 + 0.3 = 1.3
k factor for 20mm pipe, k20 = 3.8 x 10^-5
then pressure loss across 1-4, P14 = k20 x EL x Q13 ^2,
(actual P = k x EL x Q^1.85, but I assume Q^2 for simplification)
P14 = 3.8 x 10 ^ -5 x 1.3 x 17^2
P14 = 0.014 bar
thus, P at 4, P4 = P1 + P14 = 0.0289 + 0.014 = 0.043 bar,
neglect the T joint at 4 for simplification (actually, you cannot neglect that)
Assume k factor for nozzle 1 = nozzle 3 = 100
Then Q3 = 100 x sqrt (0.043)
Q3 = 20.7 L/min
So, total flow rate from 5 to 4, Q5 = 17 + 20.7 = 37.7 L/min,
Assume length of 5 to 4 = 0.5,
then P loss from 5 to 4, P45 = k20 x Q5 ^ 2 x EL, (EL = 0.5 + 0.3 = 0.8)
thus, P45 = 3.8 x 10^ -5 x 37.7^2 x 0.8 = 0.043 bar
thus P5 = P4 + P45 = 0.043 + 0.043 = 0.086 bar
Assume length of 5-6 = 1.5 m
P6 = P5 + static pressure + loss
P6 = 0.086 + 0.15 (static) + 3.8 x 10^-5 x 1.5 x 37.7^2 ( system characteristic)
P6 = 0.317 bar
P6 = 3.17 m
So, at 6 , P = 3.17m, 2262 L/H
When you plot this point on the pump curve, if the point is within the pump curve, then your pump can provide more than the flow you need, if it is outside the pump curve, the pump cannot provide sufficient flow.
Actually, if you want to check the exact flow rate, you can substitue, different flow at the above system characteristic, e.g. when Q = 0, P6 = 0.236 bar,
when Q = 10, P6 = 0.241, then the system characteristic curve can be plot on the pump curve. The intersection point is the operating point.
Please note all the above k factors are not true but just to illustrate the calculation. You have to refer to some book for the true value.
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