• Support PF! Buy your school textbooks, materials and every day products Here!

Minor loss due to expansion

  • Thread starter foo9008
  • Start date
  • #1
678
4

Homework Statement


from the previous thread , https://www.physicsforums.com/threads/minor-loss-in-pipe.869148/
, i know that formula of loss due to contraction can also be expressed as k[( mean velocity )^ 2 ] / 2g
how about the formula of minor loss due to expansion , can we expressed as k[( mean velocity )^ 2 ] / 2g ??

Homework Equations




The Attempt at a Solution

 

Attachments

Last edited:

Answers and Replies

  • #2
678
4
is it feasible ?
 
  • #3
BvU
Science Advisor
Homework Helper
2019 Award
13,033
3,012
  • #4
678
4
What mean velocity did you have in mind ? ##V_1+ V_2\over 2\ \ ## ?

Or ##k_1 ## in combination with ##V_1## ?

The old thread is a 'special case' with one of the V = 0
Really, the old thread has one of the v =0 ? At which region?
For the loss due to expansion, I mean (V1 +V2 ) /2
 
Last edited by a moderator:
  • #5
678
4
in the old thread , how can one of the v = 0 ? it is [(Vc -V2)^2]/ 2g
 
  • #6
678
4
Really, the old thread has one of the v =0 ? At which region?
For the loss due to expansion, I mean (V1 +V2 ) /2
What do you mean by combination of k and v1 ???
 
  • #7
BvU
Science Advisor
Homework Helper
2019 Award
13,033
3,012
At which region?
in the reservoir
For the loss due to expansion, I mean (V1 +V2 ) /2
well, the link gives the expressions. You can try to wiggle ##\ \
\left (V_1+ V_2\over 2\right )^2 \ \ ## into this but i doubt that succeeds...
 
  • #8
BvU
Science Advisor
Homework Helper
2019 Award
13,033
3,012
  • #9
678
4
in the reservoir
well, the link gives the expressions. You can try to wiggle ##\ \
\left (V_1+ V_2\over 2\right )^2 \ \ ## into this but i doubt that succeeds...
So , its not possible to change the to change the original formula of head loss due to expansion into
k[( mean velocity )^ 2 ] / 2g??
 
  • #10
678
4
  • #11
BvU
Science Advisor
Homework Helper
2019 Award
13,033
3,012
Ah ! I was mistaken in post #3. In 165 there is no counterpart of the ##V_1## in 166 -- so I wanted to consider ##V_1## as zero in picture 165. But it plays no role in either case: the ##v## involved are ## V_c## and ##V_2## in both cases, so 165 and 166 say the same (as haru indicated and you Ok'd)

(we are now talking about pictures in another thread - never mind :smile:) .

In 166 you see a ## k_L = \left ( {\displaystyle A_2\over \displaystyle A_c} - 1\right )^2##.

In the current thread the idea is that there is no contraction at CD and that means ##V_1## plays the role of ##V_c## (and ##A_1## is ##A_c##).

##v_1## and ##v_2## are linked through continuity (##V_1 A_1 = V_2 A_2##) so -- analogous to 166 -- you get a ##k_L## for ##V_2## but you can also get a (different) ##k_L## for ##V_1##.

'Mean velocity in the pipe' may have caused confusion: what is meant is volume flow divided by area (i.e. the velocity averaged over the cross section -- so not lengthwise). In other words: there is no place for (V1 +V2 ) /2
 
  • #12
678
4
Ah ! I was mistaken in post #3. In 165 there is no counterpart of the ##V_1## in 166 -- so I wanted to consider ##V_1## as zero in picture 165. But it plays no role in either case: the ##v## involved are ## V_c## and ##V_2## in both cases, so 165 and 166 say the same (as haru indicated and you Ok'd)

(we are now talking about pictures in another thread - never mind :smile:) .

In 166 you see a ## k_L = \left ( {\displaystyle A_2\over \displaystyle A_c} - 1\right )^2##.

In the current thread the idea is that there is no contraction at CD and that means ##V_1## plays the role of ##V_c## (and ##A_1## is ##A_c##).

##v_1## and ##v_2## are linked through continuity (##V_1 A_1 = V_2 A_2##) so -- analogous to 166 -- you get a ##k_L## for ##V_2## but you can also get a (different) ##k_L## for ##V_1##.

'Mean velocity in the pipe' may have caused confusion: what is meant is volume flow divided by area (i.e. the velocity averaged over the cross section -- so not lengthwise). In other words: there is no place for (V1 +V2 ) /2
ok , i understand that the velocity is average velocity across the area , but can still i say that the head loss due to contraction expansion is k[( mean velocity^2] / 2g ???
 
Last edited:
  • #13
678
4
Ah ! I was mistaken in post #3. In 165 there is no counterpart of the ##V_1## in 166 -- so I wanted to consider ##V_1## as zero in picture 165. But it plays no role in either case: the ##v## involved are ## V_c## and ##V_2## in both cases, so 165 and 166 say the same (as haru indicated and you Ok'd)

(we are now talking about pictures in another thread - never mind :smile:) .

In 166 you see a ## k_L = \left ( {\displaystyle A_2\over \displaystyle A_c} - 1\right )^2##.

In the current thread the idea is that there is no contraction at CD and that means ##V_1## plays the role of ##V_c## (and ##A_1## is ##A_c##).

##v_1## and ##v_2## are linked through continuity (##V_1 A_1 = V_2 A_2##) so -- analogous to 166 -- you get a ##k_L## for ##V_2## but you can also get a (different) ##k_L## for ##V_1##.

'Mean velocity in the pipe' may have caused confusion: what is meant is volume flow divided by area (i.e. the velocity averaged over the cross section -- so not lengthwise). In other words: there is no place for (V1 +V2 ) /2
sorry , pls ignore my several posts before this . do you mean the mean velocity is not suitable and should be replaced by v1(mean velocity at region 1 ) or v2( mean velocity at region 2 ) to avoid confusion ?
 
Last edited:

Related Threads on Minor loss due to expansion

  • Last Post
Replies
8
Views
3K
Replies
3
Views
740
Replies
2
Views
548
Replies
2
Views
587
  • Last Post
Replies
17
Views
662
  • Last Post
Replies
3
Views
462
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
10
Views
90K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
7
Views
665
Top