Single Pipe/Parallel Pipes

breadmonkey

I was doing a Hydraulics question which asks you to determine whether a single pipe of a given diameter, or twin pipes of a different diameter is more efficient for carrying a given discharge, Q.

I did this using the Darcy-Weisbach equation for frictional head loss and comparing the two (hf = λLV^2/2gD). (This is what I'm supposed to do).

My question is this: When considering twin pipes, I split the discharge so that Q/2 is flowing through each pipe but for some reason I'm not supposed to multiply this answer by 2 before comparing with hf for the single pipe. Why is this?

godtabh

breadmonkey wrote:

I was doing a Hydraulics question which asks you to determine whether a single pipe of a given diameter, or twin pipes of a different diameter is more efficient for carrying a given discharge, Q.

I did this using the Darcy-Weisbach equation for frictional head loss and comparing the two (hf = λLV^2/2gD). (This is what I'm supposed to do).

My question is this: When considering twin pipes, I split the discharge so that Q/2 is flowing through each pipe but for some reason I'm not supposed to multiply this answer by 2 before comparing with hf for the single pipe. Why is this?

Is this a pumped network?

If it is could you look at the power required to pump the system and compare it that way?

breadmonkey

No, there are no pumps, it's just two reservoirs with a height difference between their top water levels. The way I'm doing it is correct (as far as my course is concerned, it may be too simple for real world application).

godtabh

Just noticed (and I think this is the answer) that the V (i assume its velocity its been a while since i actually used this) is squared so its not a one is to one relationship i.e you cant just multiply by 2.

Reading over this I'm not sure if this is correct but I think it is.

Ask some one in the maths forum as I think it comes down to the formula why you dont multiply by 2

breadmonkey

Ok thanks. Can a mod please move this to the maths forum?

Dundhoone

breadmonkey wrote:

My question is this: When considering twin pipes, I split the discharge so that Q/2 is flowing through each pipe but for some reason I'm not supposed to multiply this answer by 2 before comparing with hf for the single pipe. Why is this?

Im not sure I understand your question. Are you asking why isnt double the headloss in one pipe in the two pipe scenario the same as with flow through one larger pipe? If so then compare the cross sectional area of pipes involved in both scenarios and you will see why!

breadmonkey

I can't really follow that!! I'm asking why isn't hf for the single pipe half that of the hf in one of the parallel pipes?

godtabh

breadmonkey wrote:

I can't really follow that!! I'm asking why isn't hf for the single pipe half that of the hf in one of the parallel pipes?

Again I think its down to the formula. Its down to the slope i.e 1:x^2 as opposed to 1:x

Ghost Train

I think it's because the two things arn't directly related, even though thinking about it you'd think they are

eg
head loss due to friction for the 1" pipe will be x

head loss due to friction of 1/2" diameter pipe carrying half amount fluid will be y

adding a second 1/2" pipe beside the first 1/2" pipe is not going to increase the head loss due to friction, you've worked out the head loss due to friction (per unit volume of fluid leaving the tanks) for the two situations, so no need to multiply by 2

you divide Q by two because there are two pipes compared to one in the first case, but you don't multipy the head loss by two because you have two pipes

I'm more an elec engineer, but I think thats the reason, hope it helps

godtabh

eolhc wrote:

I think it's because the two things arn't directly related, even though thinking about it you'd think they are

eg
head loss due to friction for the 1" pipe will be x

head loss due to friction of 1/2" diameter pipe carrying half amount fluid will be y

adding a second 1/2" pipe beside the first 1/2" pipe is not going to increase the head loss due to friction, you've worked out the head loss due to friction (per unit volume of fluid leaving the tanks) for the two situations, so no need to multiply by 2

you divide Q by two because there are two pipes compared to one in the first case, but you don't multipy the head loss by two because you have two pipes

I'm more an elec engineer, but I think thats the reason, hope it helps

The forumlas are the same so the two things are directly related.

The only thing that changes are the inputs Q.

The velocity of the system is dependednt on Q and the diameter of the pipe .

Cant remember what L is but everything else is constant so then it comes down to the fact that hf is proportional to V^2.

Ghost Train

kearnsr wrote:

The forumlas are the same so the two things are directly related.

The only thing that changes are the inputs Q.

The velocity of the system is dependednt on Q and the diameter of the pipe .

Cant remember what L is but everything else is constant so then it comes down to the fact that hf is proportional to V^2.

yep your right they're are related i worded it wrong
the two situations being comparing are

frictional head loss one pipe with discharge Q.... a full system

frictional head loss one pipe with discharge Q/2... 1/2 flow rate 1/2 system capacity (1 of 2 pipes)

you have to divide the system in two because the formula is for one pipe length of certain diameter

if you add another pipe and increase the flow to Q frictional head loss stays the same
frictional head loss for one pipe and Q/2=frictional head loss for two pipes and Q

Add a second pipe does not increase the frictional head loss, it will reduce it, pipe velocity will be slower for same total flow rate

godtabh

eolhc wrote:

Add a second pipe does not increase the frictional head loss, it will reduce it, pipe velocity will be slower for same total flow rate

The velocity is dependent on the diameter.

If q is constant then the flow will increase with a decrease in diameter and vice versa.

Assuming L & D is constant then the only variable is V

Ghost Train

kearnsr wrote:

The velocity is dependent on the diameter.

If q is constant then the flow will increase with a decrease in diameter and vice versa.

Assuming L & D is constant then the only variable is V

Yep thats right
I was saying something like
1 pipe situation q=5l/s
add second pipe in parallel q=2.5l/s in each pipe but total flow of system is still 5l/s

The situation mentioned in the first post states that the twin pipe situation is with pipes of different diameter to the one pipe situation. So you can't relate the calculation for the two seperate hf calculations

Back to the original post if you multipy hf by two for the twin pipe situation it would be like doubling the length of the pipe, not adding a second pipe in parallel

breadmonkey

I just can't get my head around this.

There is a certain volume of water in the 1st parallel pipe. The interaction between the water and the internal surface of the pipe caues a frictional head loss. The same thing is happening in the 2nd parallel pipe so why.......ahhhhhhhhhhhh

So I can have n parallel pipes?

Ghost Train

breadmonkey wrote:

I just can't get my head around this.

There is a certain volume of water in the 1st parallel pipe. The interaction between the water and the internal surface of the pipe caues a frictional head loss. The same thing is happening in the 2nd parallel pipe so why.......ahhhhhhhhhhhh

So I can have n parallel pipes?

you have a small pipe and have a system flow rate of 10l/s the frictional headloss is high cause the fluid is fast

you add a second pipe and the flow rate gets divided between the 2 pipes, 5l/s in each pipe the frictional headloss for the system gets lower because the fluid only goes at half the speed.

If the fluid goes down pipe 1 the frictional head loss is hf, if the fluid goes down pipe two the head loss is hf. Fluid only has to go through one of the pipes not both of the pipes to get where it's going

by dividing the system flow rate q by the number of pipes to get hf of the system

you'd have to divide the flow rate by n then for n pipes

godtabh

eolhc wrote:

you have a small pipe and have a system flow rate of 10l/s the frictional headloss is high cause the fluid is fast

you add a second pipe and the flow rate gets divided between the 2 pipes, 5l/s in each pipe the frictional headloss for the system gets lower because the fluid only goes at half the speed.

If the fluid goes down pipe 1 the frictional head loss is hf, if the fluid goes down pipe two the head loss is hf. Fluid only has to go through one of the pipes not both of the pipes to get where it's going

by dividing the system flow rate q by the number of pipes to get hf of the system

you'd have to divide the flow rate by n then for n pipes

Thats about sums it up.

Your effieiency will go down the faster the water flows in the pipe I think as HF will be higher