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Head loss in pipe

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  • #1
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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

  • #2
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The head loss is ##\frac{(P_1-P_2)}{\rho g}##
 
  • #3
678
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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
19,919
4,095
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
678
4
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 ??
 
  • #6
19,919
4,095
yes , [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss is it correct ??
No. There should be a minus sign in front of head loss.
 
  • #7
678
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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
19,919
4,095
The head loss is defined as the decrease in head from the upstream point to the downstream point.
 
  • #9
678
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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
678
4
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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4,095
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
678
4
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
19,919
4,095
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
678
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.
can you help how to solve this question ?
 
  • #15
678
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
678
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.
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
19,919
4,095
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
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,
$$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
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
19,919
4,095
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
19,919
4,095
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?
 

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