Question

An opening of area 0.252 cm2 in an otherwise closed beverage keg is 40.1 cm below the level of the liquid (of density 1.01 g/cm3) in the keg. What is the speed of the liquid flowing through the opening if the gauge pressure in the air space above the liquid is (a) zero and (b)0.532 atm?

Answer 1

@IrishBoy123 can you help me?

Answer 2

toricelli

Answer 3

start here
https://en.wikipedia.org/wiki/Torricelli%27s_law

Answer 4

i know how to do in letter a
however, in a letter b i can't seem to get the answer :(

Answer 5

ok

so we may have to go back to Bernoulli, again from that Wiki page:

\[gz+{p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\]

but here we seem to have different pressures so

\[gz+{p_{\color{red}{inside}}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\]

sove for v from there

with a) \(p_{inside} = 0\)

and b) \(p_{inside} = ????\) where ??? is (0.5* atmospheric pressure) in SI units

Answer 6

1 thing to be careful about

it's gauge pressure

so inside for a \(P = P_{atm}\)

for b), i think you add a further 0.5 atm

been a while

Answer 7

i think gauge pressure is the deviation from atmospheric pressure..

Answer 8

This question is kinda of confusing

Answer 9

I have the same problem also and seems can't get the answer

Answer 10

thank you guys for helping me :)

Answer 11

yes, ganesh, my recollection too, and i have had a check as well

so for internal gauge = 0 you get
\(gz+{p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\)
\(\implies gz={v^2 \over 2}\) and hence Toricelli

for the second where internal gauge = 0.532 you get
\(gz+{1.532p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\)
\(\implies gz+{0.532p_{atm}\over\rho}={v^2 \over 2}\)
so you plug the numbers [SI uits] into a fuller Bernoulli equation to solve

Answer 12

oh i't not equal to zero :)) but the answer is the same as my a

Answer 13

Hey, still here ?