Flow Rate of a Liquid: Calculating Mass & Pressure

[topcat123]

(a) FIGURE 2 shows two cylindrical tanks interconnected with a pipe
which has a valve that creates a constant resistance to flow of Rf
when fully open. The height of liquid (of density ρ) in the first tank
is hin and the second tank hout. The cross-sectional area of the first
tank is Ain m^2 and the second tank Aout m^2.

The flow rate of liquid through the valve is given by

Q=1/Rf (Pin - Pout)

where

Q = flow rate in m^3 s^-1
Pin = pressure due to hight of liquid in first tank (Pa)
Pout = pressure due to hight of liquid in second tank (Pa)

so my solution
Mass flow in = Mass flow out + accumulation (the change in mass within the tank)
Mass = density (ρ) x volume (V)
V = cross-sectional area (A) of the tank × the height of liquid (h) in the tank

ρiqvi = ρoqvo+accumulation

There is no gain in overall volume within the system.The Accumulation first tank = 0 - ρoqvo = - Q
The Accumulation second tank = ρiqvi - 0 = QSo the mass flow Q out of the first tank though the valve, mass flow into the second tank, is also equal to Q.
ρoqvo = ρiqvi = Aout dρhout/dt = Ain dρhin/dt

As the density and the change in time are the same, we can cancel them out.
Aout hout = Ain hin

so
hout = Ain hin / Aout

I am not sure about the time constant probable 63.3%

Any help will be apreciated
Thanks

 

[BvU]

1. What is he question ?
2. Use the template, don't erase it. You would have noticed the problem statement misses a problem.
2. You have a differential equation one moment, the next it is gone... ?

 

[topcat123]

Thank you for your reply.

2. Use the template, don't erase it. You would have noticed the problem statement misses a problem.

Sorry for not using the Template I have been informed by the moderator.

1. What is he question ?

Missed that bit off.
Produce a mathematical model of the process to determine the
change in height of fluid in the second tank when the valve is open.
(b) Determine the time constant for the system.

. You have a differential equation one moment, the next it is gone... ?

A_outδρh_out/δt = A_inδρh_in/δt
I am not sure after this point. I thought δt and δρ would cancel out, but looking at part (b) I will need δt

 

[topcat123]

I've had a bit of a rethink on how to solve this question!

pressure = force/area
pressure (p) = mg/A

mass = density (ρ) × volume (V)
so
p = ρVg/A

The volume
volume = cross-sectional area (A) of the tank × the hight of the liquid
meaning
p = ρAhg/A

Then we can cancel out (A) giving
p = ρhg

using the equation given Q = 1/R_f (p_in - p_out)

Q = 1/R_f (ρh_ing - ρh_outg)

R_fQ = ρh_ing - ρh_outg

h_out = (ρh_ing - R_fQ) / ρg

I think this is know correct.
I am still struggling with the time constant. As it is a rise in the second tank we can use

1 - e ^{- (t/T)}
Giving use a time constant of 63.2%

Any help is appreciated.

 

[Anthony McKenzie]

Hi Topcat

I am stuck in the very same question. Please could I ask how you got on?

Thanks

Jordan

 

[PaxFinnica96]

Hi Topcat

I am stuck in the very same question. Please could I ask how you got on?

Thanks

Jordan

Hi Anthony,

Did you manage to submit this answer for part a and was it correct?

Cheers,
PF

 

[jim mcnamara]

The original person asking/discussing the question has not been online at Pf for a long time. This thread has become meaningless. Thread closed.