
#1
Aug2807, 09:16 AM

P: 4

Hi all,
I'm sure a lot of you know about the head loss due to sudden expansion: Hl = (1/2g)*(v1v2)^2 This equation can be derived from Bernoulli, continuity and momentum balans equations. The underlying assumption in deriving this equation is that the pressure just after the expansion (say, p0) is equal to the pressure before the expansion (p1). Now, I was wondering how accurate is this assumption? And what is the underlying physics behind this assumption? best regards, Arman 



#2
Aug2807, 09:30 AM

HW Helper
PF Gold
P: 1,198

[tex] \frac{p_a  p_b}{\rho} = \frac{a_b v_b^2  a_a v_a^2}{2} + h_e[/tex] If you disregard the correction factors and use the momentum and continuity equation, you'll end up with [tex] h_e = \frac{(v_av_b)^2}{2}[/tex] 



#3
Aug2807, 10:01 AM

P: 4

Well, the g should be there, otherwise the dimensions of your equation does not match:
m (n.e.) m^2/s^2 (how do you post an equation in a thread by the way?) About the pressure: yes the (static) pressure is not the same between a and b. What I mean is the (static) pressure at border between small area and big area (so at position before b but just after a) is assumed to be equal to the (static) pressure at a. This is the underlying assumption in the momentum balans equation to get to the head loss equation I post earlier (it is far easier to explain if I can post a picture, but I do not know how). best regards, Arman 



#4
Aug2807, 11:01 AM

HW Helper
PF Gold
P: 1,198

head loss due to sudden expansion 



#5
Aug2807, 11:24 AM

Sci Advisor
P: 5,095

The expression for a sudden expansion that I am used to seeing is
[tex]h_L=K_L \left(\frac{V_1^2}{2g}\right)[/tex] where [tex]K_L = \left(1\frac{A_1}{A_2}\right)^2[/tex] According to my references, the sudden expansion is the only case in which the simple analysis actually matches actual measurements. 



#6
Aug2807, 01:09 PM

HW Helper
PF Gold
P: 1,198

In fact, I did this very experiment in my momentum transfer lab this week, and the experimental value of k_l matched well with the analysis. 



#7
Aug2907, 07:42 AM

Sci Advisor
P: 5,095





#8
Aug2907, 10:17 AM

P: 4

Bernoulli equation with pressure loss:
[tex] p_{1}+0.5\rho v_{1}^{2}=p_{2}+0.5\rho v_{2}^{2}+p_{L}[/tex] With [tex]p_{L}[/tex] the pressure loss. The head loss, [tex] h_{L} [/tex] can be computed by dividing the equation with [tex] \rho g [/tex], and after rearranging: [tex] h_{L} =\frac{p_{1}p_{2}}{\rho g }+\frac{v_{1}^{2}v_{2}^{2}}{2g }[/tex] Applying the momentum balans on the Control Volume between position 1 and 2 (see attachment): [tex] p_{1}A_{1}+p_{0}\left(A_{2}A_{1}\right)p_{2}A_{2}=\rho \left(A_{2}v_{2}^{2}A_{1}v_{1}^{2}\right)[/tex] And here when you make an assumption that the pressure [tex]p_{0}[/tex] equals the pressure [tex]p_{1}[/tex]. And after using continuity equation the equation becomes: [tex] \frac{p_{2}p_{1}}{\rho g }=\frac{v_{2}\left(v_{1}v_{2}\right)}{g}[/tex] Fill this in to the expression of the head loss (second equation), voila, you get the head loss due to sudden expansion: [tex] h_{L} =\frac{\left(v_{1}v_{2}\right)^{2}}{2g }[/tex] Now, this expression of the head loss is derived after we make an above mentioned assumption regarding the pressures. How accurate is this assumption then? I mean if the two pressures are not the same, the equation will have different form. best regards, Arman 



#9
Jul1610, 11:26 PM

P: 3

actually, the "head" has dimensions of length ...




#10
Mar1511, 04:33 AM

P: 1

hi all..
i wanted to know the value of "k" for practical uses. i was calculating head loss in expansion after main inlet valve in hydro plant. Going by penstock manual from Central water commission, New Delhi, the value of k =(1a1/a2)sin(theta) where ; theta=one half of flare angle, a taper of 1:10 is recommended. Some colleague say the value of k is to be taken from practical values or practical charts. help 



#11
Mar1911, 08:21 AM

P: 1

can any one derive for me the darcy weisbach formular for head loss




#12
May2113, 05:54 AM

P: 1

Long time passed since your question, Dear previah
This is for the people who is concerned to head loss in sudden expansioning. I am sure you're misunderstanding the underlying pressure equality assumption. This assumption says "the pressure at the annular area, A2A1,is equal to the pressure in A1" for the calculation of pressure difference perpendicular to control volume. This annular area contacts turbulent eddies. So, as you say, " the pressure just after the expansion (say, p0) is equal to the pressure before the expansion (p1)" has completly different meaning. 


Register to reply 
Related Discussions  
Minor loss due to Sudden contraction (help needed)  Mechanical Engineering  13  
Fluids  Head Loss complicated  Engineering, Comp Sci, & Technology Homework  5  
Head Loss  Materials & Chemical Engineering  2  
Effect on flow after sudden expansion  General Engineering  3  
Head loss  General Engineering  6 