parallel piping system

I'm not an electrical engineer but I think your equation is wrong. It should be

1/Rp=(1/R1+1/R2+......1/Rn)

To be honest I'm not sure if running parallel pipes follows this relationship. Thinking out loud, if I have a header tank with one pipe coming out the bottom and have X flow then I put in another identical pipe I'd get 2X flow. This would mean that the resistance would have halved if the relationship between flow and resistance is linear. Now I seem to remember that it is not so I'd suggest the relationship isn't as simple as the above equation suggests.

Thats my thinking of it. Can someone prove (or disprove) this with calcs?

thank u for your comment. yes i was wrong earlier. i maen to say 1/Rp= (1/R1+1/R2+...1/Rn). while designing a parallel piping system with different flow rate in different branch, how would i calculate the pressure differnce. should i use this formula to calculate equivalent pressure loss?

No, you can not do like that.

Every pipe branch point have the same pressure.pressure drop in a branch is same as much in another pipe.

if A is the branch point which divided into 2 branch, and B is the branch point that 2 branch mentioned can become unity again. The pressure drop is pd A minus pd B, this value will be equal with Pressure Drop on every branch.

Yes, basic principals apply to both systems.

The sum of the flow at a node is zero. Current is analogous to flow. KCL.

The voltage around a loop is zero. KVL.

Replace V=IR with Q=VA and you are good to go.

Use Bernoulli equation to balance potential and kinetic energy plus friction losses.

Did your circuit also contain reactive components (i.e. capacitance and inductance)? You did not say.

It's not that simple. In elec case, voltage difference is proportional to current, but for fluid flow pressure difference is proportional flow² in nearly all practical cases. Also V=IR is Ohm's law, but Q=VA is just velocity x area and doesn't say anything about pressure difference.

But there is a formula analogous to 1/Rp = 1/R1+1/R2+......1/Rn

Calling R1 = ΔP/q1², R2 = ΔP/q2², etc (analogous to R1 = V/I1, R2 = V/I2), q1 = √(ΔP/R1), q2 = √(ΔP/R2) etc. To calculate R1, R2 etc, just take an arbitrary (but reasonable) figure for q, find ΔP by usual hydraulic methods, and find ΔP/q². Doesn't have to be the common ΔP for the overall system.

q_tot = √(ΔP/Rp) = q1 + q2 + ...... + qn = √(ΔP/R1) + √(ΔP/R2) + ....... + √(ΔP/Rn)

So, dividing by √ΔP, 1/√Rp = 1/√R1 + 1/√R2 + ....... + 1/√Rn

Cheers......Codey

thanks codey 4 your answer. but i am little confused about the relation Q1=√(ΔP/R1) since the unit of ΔP & R1 are same and is expressed in ft of water in case of a hydronic system designing. please clarify it.

Hello khurshid

No, R has units ΔP/Q². So ΔP/R has units Q².

E.g. to take arbitrary figures, if ΔP = 10000 Pa at flow 100 m³/h, R = 1000/100² = 0.1 Pa/(m³/h)².

Come back if you need more.......Codey

I have many times tried to use electrical/hydraulic analogies looking at circuits of each. For the most part they are equivalent.

However, if you want to stretch you mind, examine the units of your electrical or mechanical equations. It is then you may decide on personal limits of inquisitiveness.