1. Apr 14, 2016

foo9008

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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

2. Apr 14, 2016

Staff: Mentor

The head loss is $\frac{(P_1-P_2)}{\rho g}$

3. Apr 14, 2016

foo9008

but i dont have P1 and P2.... how to solve this ?

4. Apr 14, 2016

Staff: Mentor

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. Apr 14, 2016

foo9008

yes , [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss is it correct ??

6. Apr 14, 2016

Staff: Mentor

7. Apr 14, 2016

foo9008

the energy flow from point 1 to 2 , so we should add the head loss at point 2 , right ?

8. Apr 14, 2016

Staff: Mentor

The head loss is defined as the decrease in head from the upstream point to the downstream point.

9. Apr 14, 2016

foo9008

ya , point 1 is upstream , point 2 is downstream , right ?

10. Apr 14, 2016

foo9008

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. Apr 14, 2016

Staff: Mentor

12. Apr 14, 2016

foo9008

A portion of energy is lost to the resistance to flow....

13. Apr 14, 2016

Staff: Mentor

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. Apr 14, 2016

foo9008

can you help how to solve this question ?

15. Apr 14, 2016

foo9008

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. Apr 14, 2016

foo9008

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. Apr 14, 2016

Staff: Mentor

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. Apr 14, 2016

foo9008

pls refer to the link diagram 6.5 , the formula fo hf is :
http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Examples/eg6_ans.htm

19. Apr 14, 2016

foo9008

20. Apr 14, 2016

foo9008

View attachment 99132