how do you calculate the head loss for this pipe

johnnyc

I have the following question it deals with the sudden contraction of a pipe... I have the following answers.but I am looking how do you calculate the head loss
v1=1.783m/s
v2=3.164m/s
a1=0.0314m2 (pie*.1^2)
a2=.0177m2 (pie*.075^2)

Q=.056m3/s
a2/a1=.5636

johnnyc

Anybody have any idea please, i am stuck....

Philistine

Not sure if this is actually what you're looking for but its a good starting point.


http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm

johnnyc

Ok the first section of the pipe is 200mm in width and it suddenly goes to 150mm in width for the next segment.... I am trying to first calculate the coefficent... This is what i am getting i can be corrected if i am wrong
KL=(1-A2/A1)^2
(1-.5625)^2=0.191


(BUT I KNOW I AM WRONG HERE AS I HAVE THE LECTURER ANSWER WHICH IS 0.107)
HEAD LOSS=KL(V2^2/2G)
>.191(3.164/2*9.81)
>.191(10.011/19.62)
>.191(.161)
>0.031

i HAVE TRY TO FIGURE IT OUT BY MYSELF TO NO AVAIL

mawk

lmtd?
http://en.wikipedia.org/wiki/Log_mean_temperature_difference



try the last formula

Nukem

Don't you have to have pipe lengths and friction factors for this calculation? (please note been a while since i did fluids)

johnnyc

Thanks mawk and nukem for your responses....Mawk the temperature is the same between the two segments, the pipe width is 200mm and suddenly contracts to 150mm pipe width.... With the calculations above you should be able to get the coefficent?

Granny Lover

you could use bernoullis equation
(P1/ρg)+((v1^2)/2g)+z1+hp=(P2/ρg)+((v2^2)/2g)+z2+hf
Where:
P= pressure
ρ=density (water= 1000)
g=gravity (9.81)
v=velocity
Z=change in height
Hp=pump head
Hf= head loss due to friction

most cades something in the equation will cancel such as z if there is no change in height this formula only works for fluids for compressed air there is another formula
if you cant work it out from this look up darcys formula.
that is the way we calculate it out in college

sme

As this is pipe contraction A1 doesn't really matter, a vena contracta forms immediately after the contraction and the headloss is due to the expansions which follows. Empirically the area contracts by about 40% so you get Kl = 0.44. Although this still doesn't get the same answer as your lecturer.

hl = (0.44V2^2)/2g = (0.44*3.164^2)/2g = 0.225m

Given that you posted this 3 weeks ago, this probably won't help!

mawk

johnnyc wrote: »

Thanks mawk and nukem for your responses....Mawk the temperature is the same between the two segments, the pipe width is 200mm and suddenly contracts to 150mm pipe width.... With the calculations above you should be able to get the coefficent?

Weird.. I just saw own response in this thread and neither remembered writing it nor knew why i brought lmtd up..

I think i must have read head loss as heat loss early in the morning stupids sometimes..