Friction loss due to exit pipe to reservoir

AI Thread Summary
The discussion centers on the head loss formula k(V^2)/2g, specifically when water is discharged from a pipe into a reservoir versus the atmosphere. Participants agree that the coefficient k should remain equal to 1 in both scenarios, as there are no significant flow disruptions or eddy currents at the exit point. The question of whether head loss differs between discharging into a reservoir and the atmosphere is addressed, with consensus that k remains constant at 1. Clarity is provided for non-native speakers regarding the application of k in these contexts. Overall, the conclusion is that k equals 1 for both discharge scenarios.
foo9008
Messages
676
Reaction score
4

Homework Statement


we know that the formula of head loss is k(V^2) /2g , the k = 1 when the water is discharged from pipe to reservoir ...However , how if the water from pipe is released into atmosphere ? is it necessary to include the head loss due to exit ? if so , is the k still =1 ? in the example , the author take k=1 when water is released to atmosphere from pipe , is he correct ?

Homework Equations

The Attempt at a Solution

 

Attachments

  • 170.PNG
    170.PNG
    21.2 KB · Views: 650
  • 171.PNG
    171.PNG
    20.1 KB · Views: 524
  • 172.PNG
    172.PNG
    19.7 KB · Views: 522
  • 173.PNG
    173.PNG
    36.4 KB · Views: 645
foo9008 said:
No, can you try to answer and reply?
Greg's post is automated. It just bumps your thread back up to somewhere it is more likely to get noticed. He will not have seen your reply.
I see no reason why k should be anything but 1 here. k not equal to 1 represents losses caused by messy flows at the junction. On egress to a wide reservoir or to atmosphere, there is no reason to suppose any eddy currents.
 
haruspex said:
Greg's post is automated. It just bumps your thread back up to somewhere it is more likely to get noticed. He will not have seen your reply.
I see no reason why k should be anything but 1 here. k not equal to 1 represents losses caused by messy flows at the junction. On egress to a wide reservoir or to atmosphere, there is no reason to suppose any eddy currents.
Is the head loss due to water discahrged from pipe to reservoir same as head loss due to water discharged from pipe to atmosphere? The k is 1?
 
foo9008 said:
Is the head loss due to water discahrged from pipe to reservoir same as head loss due to water discharged from pipe to atmosphere? The k is 1?
I thought I just answered that. What in my answer did you not understand?
 
haruspex said:
I thought I just answered that. What in my answer did you not understand?
I am not native emglish speaker... Just to be sure, the k is. 1 when water discharged from pipe to atmosphere??
 
foo9008 said:
I am not native emglish speaker... Just to be sure, the k is. 1 when water discharged from pipe to atmosphere??
Yes.
 
  • Like
Likes foo9008
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top