Head loss in pipe

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Homework Statement


estimate the energy head lost along a short length of pipe suddenly enlarging a diameter of 350mm to 700mm which discharges 0.7(m^-3) of water per second .
the solution given is Q1= Q2 = 0.7(m^3)/s
0.7 = pi ((350x10^-3)^2 ) V1 / 4 , V1= 7.28m/s
0.7 = pi ((700x10^-3)^2 ) V2 / 4 , v2 = 1.82m/s
[(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss
head loss = 2.53m

Homework Equations




The Attempt at a Solution


the bernoulli's equation is P1/ ρg + (V1^2) /2g + z1 = P2/ ρg + (V2^2) /2g + z2, by writing [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss , the author assume P1 = P2 , how could it be possible ? since the inlet is smaller than outlet , the pressure at inlet should be lower , right ? the velocity at inlet is smaller
 

Answers and Replies

  • #3
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The head loss is ##\frac{(P_1-P_2)}{\rho g}##
but i dont have P1 and P2.... how to solve this ?
 
  • #4
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but i dont have P1 and P2.... how to solve this ?
You're solving algebraically for the entire quantity ##\frac{(P_1-P_2)}{\rho g}## (which is the head loss), without having to know either P1 or P2 individually. You are aware that the pipe is assumed to be horizontal so that z1 = z2, correct? And you are aware of the definition of head, correct?
 
  • #5
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You're solving algebraically for the entire quantity ##\frac{(P_1-P_2)}{\rho g}## (which is the head loss), without having to know either P1 or P2 individually. You are aware that the pipe is assumed to be horizontal so that z1 = z2, correct? And you are aware of the definition of head, correct?
yes , [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss is it correct ??
 
  • #7
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No. There should be a minus sign in front of head loss.
the energy flow from point 1 to 2 , so we should add the head loss at point 2 , right ?
 
  • #8
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The head loss is defined as the decrease in head from the upstream point to the downstream point.
 
  • #9
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The head loss is defined as the decrease in head from the upstream point to the downstream point.
ya , point 1 is upstream , point 2 is downstream , right ?
 
  • #10
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The head loss is ##\frac{(P_1-P_2)}{\rho g}##
why the head loss is ##\frac{(P_1-P_2)}{\rho g}##?
shouldn't it be P1/ ρg + (V1^2) /2g + z1 = P2/ ρg + (V2^2) /2g + z2 + head loss ???? we cancel our z1 and z2 , since they are the same
 
  • #11
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why the head loss is ##\frac{(P_1-P_2)}{\rho g}##?
shouldn't it be P1/ ρg + (V1^2) /2g + z1 = P2/ ρg + (V2^2) /2g + z2 + head loss ???? we cancel our z1 and z2 , since they are the same
What is your understanding of the term head, and what is your understanding of the term head loss?
 
  • #12
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What is your understanding of the term head, and what is your understanding of the term head loss?
A portion of energy is lost to the resistance to flow....
 
  • #13
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A portion of energy is lost to the resistance to flow....
That is not the general definition of head loss and, in this problem, the resistance to flow is implicitly assumed to be zero. More generally, head loss is the decrease in ##P/(\rho g)+z## between inlet and outlet.
 
  • #14
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That is not the general definition of head loss and, in this problem, the resistance to flow is implicitly assumed to be zero. More generally, head loss is the decrease in ##P/(\rho g)+z## between inlet and outlet.
can you help how to solve this question ?
 
  • #15
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4
That is not the general definition of head loss and, in this problem, the resistance to flow is implicitly assumed to be zero. More generally, head loss is the decrease in ##P/(\rho g)+z## between inlet and outlet.
sorry , i missed out part of thge question . the full question given is
estimate the energy head lost along a short length of pipe suddenly enlarging a diameter of 350mm to 700mm which discharges 0.7(m^-3) of water per second .. If the pressure at entrance is 105N/ (m^2) . Find the pressure at exit
 
  • #16
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That is not the general definition of head loss and, in this problem, the resistance to flow is implicitly assumed to be zero. More generally, head loss is the decrease in ##P/(\rho g)+z## between inlet and outlet.
so the equation should be (105x10^3)/ (1000x9.81) + (7.28^2) / (2x9.81) +z1 = P2 / (1000x9.81) + (1.82^2) / (2x9.81) + head loss , am i right ?
cancel off z1 and z2 , we still have unknown P2 , how to find the head loss?
 
  • #17
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so the equation should be (105x10^3)/ (1000x9.81) + (7.28^2) / (2x9.81) +z1 = P2 / (1000x9.81) + (1.82^2) / (2x9.81) + head loss , am i right ?
cancel off z1 and z2 , we still have unknown P2 , how to find the head loss?
No. It's not correct.

$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
 
  • #18
678
4
No. It's not correct.

$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
pls refer to the link diagram 6.5 , the formula fo hf is :
CjUGeji.png

http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Examples/eg6_ans.htm
 
  • #19
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4
No. It's not correct.

$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
upload_2016-4-15_12-31-7.png
 
  • #20
678
4
No. It's not correct.

$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
View attachment 99132
 
  • #21
678
4
No. It's not correct.

$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
is it because when we find the head loss , we pick 2 points which have the same velocity and same pressure.... so we can have only P1 and P2 (2 unknowns) ?
in my calculation , $$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$ , i have chose the wrong points , which have different pressure and different velocity , so i am having problems of finding the head loss?
 
  • #23
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is it because when we find the head loss , we pick 2 points which have the same velocity and same pressure.... so we can have only P1 and P2 (2 unknowns) ?
No. We pick points at the inlet and outlet.
in my calculation , $$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$ , i have chose the wrong points , which have different pressure and different velocity , so i am having problems of finding the head loss?
No. This equation is correct. To get the head loss, you don't have to know P1 and P2 individually. You just have to determine their difference. Do you know how to do the algebra to get their difference?
 
  • #24
678
4
No. We pick points at the inlet and outlet.

No. This equation is correct. To get the head loss, you don't have to know P1 and P2 individually. You just have to determine their difference. Do you know how to do the algebra to get their difference?
as you said in the previous post , The head loss is ##\frac{(P_1-P_2)}{\rho g}## ? so ,
$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$ ??
 
  • #25
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5,008
as you said in the previous post , The head loss is ##\frac{(P_1-P_2)}{\rho g}## ? so ,
$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$ ??
Is there a question here?
 
  • #26
678
4
No. We pick points at the inlet and outlet.

No. This equation is correct. To get the head loss, you don't have to know P1 and P2 individually. You just have to determine their difference. Do you know how to do the algebra to get their difference?
as you said in the previous post , The head loss is ##\frac{(P_1-P_2)}{\rho g}## ? so ,
No. It's not correct.

$$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
$$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
Is there a question here?
that's the correct working for this question , right ? then , i think i understand already , thanks!
 
  • #27
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as you said in the previous post , The head loss is ##\frac{(P_1-P_2)}{\rho g}## ? so ,


that's the correct working for this question , right ? then , i think i understand already , thanks!
Yes. I said that this is the correct working for the question previously. I think your issue is more with terminology than with the actual application of the methodology.
 
  • #29
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https://en.wikipedia.org/wiki/Bernoulli's_principle#Incompressible_flow_equation
head loss is also known as piezometric loss , which has formula of h = P / ρg + z ?????
wHO4fWj.png


but , in my question , it said energy loss , shouldn't it look like this( involve (V^2) /2g ) ?
Xaz1qVO.png
Since you never gave the full exact statement of your problem, I would have to be a clairvoyant to know what it was. Maybe you want to do that now? It might have helped quite a bit if you have given the exact statement from the outset.
 
  • #30
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SUf8eu4.png

here it is , it should be 105N/(m^2)
 
  • #31
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5,008
SUf8eu4.png

here it is , it should be 105N/(m^2)
Well, it's hard to know what they are referring to by "energy head lost." It isn't clear whether that is the same as "head lost." It also isn't clear whether the "given solution" is correct, because there is actually a head gain calculated , not a head loss.

Also, have you been learning about discharge coefficients and frictional head loss? If so, then the given solution is incorrect. There should be a frictional head loss included in the equation that is calculated from a discharge coefficient, based on the velocity at the 350 mm location (and and the diameter ratio at the sudden enlargement). So, have you been learning about discharge coefficients and, if so, what does your book give for the discharge coefficient for a sudden expansion to twice the diameter?
 
  • #32
678
4
Well, it's hard to know what they are referring to by "energy head lost." It isn't clear whether that is the same as "head lost." It also isn't clear whether the "given solution" is correct, because there is actually a head gain calculated , not a head loss.

Also, have you been learning about discharge coefficients and frictional head loss? If so, then the given solution is incorrect. There should be a frictional head loss included in the equation that is calculated from a discharge coefficient, based on the velocity at the 350 mm location (and and the diameter ratio at the sudden enlargement). So, have you been learning about discharge coefficients and, if so, what does your book give for the discharge coefficient for a sudden expansion to twice the diameter?
no , i havent learn about discharge coefficient , but , according to the book , the loss due to expansion is given by [(V1 -V2)^2 /] 2g
 
  • #33
678
4
Well, it's hard to know what they are referring to by "energy head lost." It isn't clear whether that is the same as "head lost." It also isn't clear whether the "given solution" is correct, because there is actually a head gain calculated , not a head loss.

Also, have you been learning about discharge coefficients and frictional head loss? If so, then the given solution is incorrect. There should be a frictional head loss included in the equation that is calculated from a discharge coefficient, based on the velocity at the 350 mm location (and and the diameter ratio at the sudden enlargement). So, have you been learning about discharge coefficients and, if so, what does your book give for the discharge coefficient for a sudden expansion to twice the diameter?
but , if the solution look like the previous post ( your working ) , teh question should be rephrased as estimate the head loss , but not estimate the total head loss/ total energy loss ?
 
  • #34
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but , if the solution look like the previous post ( your working ) , teh question should be rephrased as estimate the head loss , but not estimate the total head loss/ total energy loss ?
For calculating the frictional head loss at a sudden expansion, see Table 7.5-1 of Transport Phenomena by Bird, Stewart, and Lightfoot.
 

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