Question

The pipe in question is part of a straight pipeline water flow system. The flow within this system is due to ΔP (Pressure drop) across the length of a pipe (L=300m), carrying water, with diameter (D=0.2m), flowing at a rate of 1.7 m/s. The pipe is made out of cast iron (e=0.25mm)
The following data needs to be applied:
• The losses due to entrance of the pipe (k=0.5)
• The losses due to exit of the pipe (k=0.3)
• NRE = ρVD/μ
Determine the head losses in pipeline flow;
1. due to friction (both minor and major)
2. due to entrance and exit
3. pressure head losses within the system

Answer 1

I can help you with this...what have you tried so far?

Answer 2

I have tried calculating the friction factor by dividing 0.25mm by 200mm as the diameter is 0.2m, I am unsure where to use this friction factor figure?

Answer 3

The friction factor is obtained from the Moody chart. Are you familiar with that?

Answer 4

I have to run but will be back....

Answer 5

First step is to calculate the Reynolds number for the flow...will be back in a couple of hours.

Answer 6

\[RE=((0.2*1.7*1000)\div0.001)=340,000 \]

Answer 7

I have tried this so far but my lecturer only ever gave us one example of this equation so I cannot be sure if 340,000 is a reasonable answer?

Answer 8

That look ok...now go to the Moody chart and look up the friction factor using:
your Reynolds number and e/d...gotta run now...will be back

Answer 9

I am unsure how to use the reynolds number at this point but I think e/d = 0.00125 so it seems the figure would be 2.3 on the moody chart, does this sound correct?

Answer 10

So Re = 3.4 E5
e/d = .25mm/200 mm = .00125
That is correct....now the friction factor where these intersect is about .02

Answer 11

Let me now if you see where I got .02 for the friction factor off the Moody chart

Answer 12

Yes I can see, it is on the left on my graph

Answer 13

Great...now you need to apply your equations for head loss for:
1) straight pipe length
2) entrance loss
3) exit loss
Check your notes to see if you have those equations

Answer 14

\[-f (L \div D) * ( v^2 \div 2 ) * \rho\]

Can I use this equation to solve the straight pipe length?

Answer 15

h = f*L/D*V^2/2....where di d you get the rho from?

Answer 16

Its just because I have a variant of the equation I think, it ends in (v^2/2g)*pg but the g's cancel out. Must be for something else. Anyway....

\[0.02\times(300\div0.2)\times(0.7^{2}\div2) = 7.35\]

Does this seem like a reasonable figure?

Answer 17

V = 1.7 not 0.7 ...be careful

Answer 18

sorry....my bad v = 1/7 m/s = .143 m/s

Answer 19

Better check the problem ...is V = 1.7 m/s or 1/7 m/s??

Answer 20

Sorry about that it was 1.7^2

0.02*(300/0.2)*(1.7^2/2) = 43.35

Answer 21

ok...just make sure the problem said 1.7 m/s

Answer 22

Now you have to calculate the losses at entrance and exit...were you provided with formula for that?

Answer 23

Okay I just checked and it is definitely 1.7 m/s. No I wasn't given a formula for this but I have been looking at using...

\[Kentrance(V ^{2}\div2g)\]

And the other K factor the exit loss,

Does this seem acceptable?

Answer 24

h loss = K*V^2/2

Answer 25

That definitely looks familiar...

\[k \times (v ^{2}\div(2\times9.81)) = \]

Entrance
0.5 * (1.7^2/(2*9.81)) = 28.35

Exit
0.3 * (1.7^2/(2*9.81)) = 17.01

Am I doing this correctly?

Answer 26

I don't have g in my equation....

Answer 27

Check your text to see what equations you should be using...some have g

Answer 28

I would rather trust your equation, all of mine include g but my lecturer has messed up on many equations

Answer 29

Ok my answers are

Entrance
\[k=0.5\times(0.7^{2}\div2)=0.1225\]

Exit
\[k=0.3\times(0.7^{2}\div2)=0.0735\]

Is that a better result?

Answer 30

No.... I mean 1.7^2

k0.5 = 0.7225

k0.3 = 0.4335

Answer 31

I guess these are the pressure head losses in the entrance and exit? Am I to designate Pa as their unit or KPa if I am to do this in S.I units?

Answer 32

Just got back ...give me a few minutes and we can finish this up

Answer 33

if your instructor is writing the Bernoulli equation in the following manner, then the head loss equations he/she provided to you are correct:
P + 1/2*rho*v^2 + rho*g*h

Answer 34

I was wondering whether I need to use the 'h' because it is just a straight pipe, is that correct?

\[(\frac{ 1 }{ 2 }\times \rho \times) v ^{2} + (\rho \times g)\]

Can I use the Bernoulli equation as shown above?

Answer 35

what you have above is part of the Bernoulli equation...compare it to mine to see what you are missing

Answer 36

Use the equations provided by your instructor and put in all the units to ensure your calculations for the head loss is in the correct units.

Answer 37

The Bernoulli equation that I was provided with is shown below...

\[\Delta ( \frac{ \rho }{ \rho g } + \frac{ v ^{2} }{ 2g } + z)\]

But I have been told that if the pipe is straight horizontal then we do not need to use the 'z' as it is for gradient.

Answer 38

I think maybe I need to use...

\[( \frac{ p _{1} }{ \rho } ) + (\frac{ v ^{2} }{ 2 } )\]

For entrance and...

\[( \frac{ p _{2} }{ \rho } ) + (\frac{ v ^{2} }{ 2 } )\]

For exit? If I use the figures generated from the K factor equations as P1 and P2 would this give me a feasible answer?