Minor loss due to expansion

[foo9008]

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

 

[foo9008]

is it feasible ?

 

[BvU]

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]

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

 

[foo9008]

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

 

[foo9008]

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 ?

 

[BvU]

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...

 

[BvU]

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]

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??

 

[foo9008]

I meant the one in post #1

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

 

[BvU]

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]

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 ?

 

[foo9008]

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 ?