# Flow rate of a liquid

1. Mar 24, 2016

### topcat123

• members are reminded that the 3 header template is required in the homework forums
(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 = Q

So the mass flow Q out of the first tank though the valve, mass flow in to 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

#### Attached Files:

• ###### fig2.png
File size:
1.9 KB
Views:
93
2. Mar 24, 2016

### 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... ?

3. Mar 25, 2016

### topcat123

Thank you for your reply.

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

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.

Aoutδρhout/δt = Ainδρhin/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

4. Mar 25, 2016

### 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/Rf (pin - pout)

Q = 1/Rf (ρhing - ρhoutg)

RfQ = ρhing - ρhoutg

hout = (ρhing - RfQ) / 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.

5. Aug 15, 2017 at 1:40 PM

### Anthony McKenzie

Hi Topcat

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

Thanks

Jordan

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted