Minor loss due to expansion

• foo9008
In summary, the formula for minor loss due to contraction can be expressed as k[(mean velocity)^2]/2g, while the formula for minor loss due to expansion is also k[(mean velocity)^2]/2g, with the mean velocity being the velocity averaged over the cross section of the pipe. The old thread (post #1) considers a special case where V = 0. In the current thread, there is no contraction at CD and V1 plays the role of Vc, while V1 and V2 are linked through continuity. Therefore, you can get a kL for V2, but also a different kL for V1. The mean velocity is not suitable and should be replaced by V1 (mean velocity

Homework Statement

, 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 ??

The Attempt at a Solution

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is it feasible ?

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

foo9008
BvU said:
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

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in the old thread , how can one of the v = 0 ? it is [(Vc -V2)^2]/ 2g

foo9008 said:
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 ?

foo9008 said:
At which region?
in the reservoir
foo9008 said:
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...

foo9008
foo9008 said:
in the old thread , how can one of the v = 0 ? it is [(Vc -V2)^2]/ 2g
I meant the one in post #1

foo9008
BvU said:
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??

BvU said:
it doesn't state v = 0 , it just stated mean velocity in 165

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 ) .

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

foo9008
BvU said:
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 ) .

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 ?

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BvU said:
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 ) .

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 ?

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