Calculating Mean Velocity: Is Re Above 4000 Really Turbulent?

[sci0x]

Homework Statement

Pipework between a buffer tank and filter is 100 mm diameter. Flow rate of beer is 400 hl h-1. Density of beer is 1010 kg m-3 and viscosity is 0.0025 Pa s. Calculare Reynolds no. For the flow through the pipework

Relevant Equations

Re = (density x mean velocity x diameter)/ viscosity

First calculate mean velocity:
400 hl h-1 = 40,000 L / h-1 = 666.67 L/min
1L/min = 10^ -3 m3/min
666.67 L/min = 0.666 m3/min
= 0.0111 m3/sec

Cross sec area = (3.14)(0.1/2)^2 = 7.853 x 10^-3 m2

0.0111/7.853 x 10^-3 = 1.413 m/s

Re = (1010)(1.413)(0.05)/0.0025
= 28,542.6

Am i right here?
Re above 4000 is turbulent which i thought was more for CIPs rather than beer transfers

 

[nrqed]

Homework Statement:: Pipework between a buffer tank and filter is 100 mm diameter. Flow rate of beer is 400 hl h-1. Density of beer is 1010 kg m-3 and viscosity is 0.0025 Pa s. Calculare Reynolds no. For the flow through the pipework
Relevant Equations:: Re = (density x mean velocity x diameter)/ viscosity

First calculate mean velocity:
400 hl h-1 = 40,000 L / h-1 = 666.67 L/min
1L/min = 10^ -3 m3/min
666.67 L/min = 0.666 m3/min
= 0.0111 m3/sec

Cross sec area = (3.14)(0.1/2)^2 = 7.853 x 10^-3 m2

0.0111/7.853 x 10^-3 = 1.413 m/s

Re = (1010)(1.413)(0.05)/0.0025
= 28,542.6

Am i right here?
Re above 4000 is turbulent which i thought was more for CIPs rather than beer transfers

Watch out, in your calculation of Re you seem to have used the radius instead of the diameter.
The rest looks good. I cannot judge if the result makes sense, maybe someone else with more knowledge in that area can chime in.

 

[sci0x]

Thanks.

Re =
Re = (1010)(1.413)(0.1)/0.0025
= 57,085.2

 

[sci0x]

The next Q is using a friction factor chart determine the value of the friction factor thru the pipe assuming the pipe has smooth internal surface.

Ive used this chart: Link
57,000 is approx midpoint between 10^4 and 10^5 on the bottom of the chart.
Drawing a line up to the smooth pipe, first curve and drawing a line left.
Im getting a friction factor of 0.014.
Am i correct here?

 

[gmax137]

I realize this is homework, so maybe my comment does not apply; but in the real world, when the pipe is specified as "100 mm" that does not mean that the inside diameter is 100 mm. You need to know the pipe schedule in addition to the nominal size, and then look up the ID. In this case, it doesn't matter much as I would assume sch40 pipe and I believe 100 mm sch 40 will have an ID of 102.3 mm.

So even though it makes little difference in this case, it is worth noting for future work.

 

[gmax137]

57,000 is approx midpoint between 10^4 and 10^5 on the bottom of the chart.
Drawing a line up to the smooth pipe, first curve and drawing a line left.
Im getting a friction factor of 0.014.
Am i correct here?

My moody chart shows friction factor of ~0.02 at Re=60,000 for 4-inch pipe

 

[sci0x]

Sorry yes, i read the chart wrong, approx 0.024

 

[sci0x]

Its a past exam paper Q.

The next part of the Q is: The beer level in the buffer tank is controlled to a depth of 3 metres. The maximum back pressure from the filter is 6 bar gauge. The inlet to the filter is at the same height as the outlet from the base of the buffer tank. Transfer of beer to filter is by centrifugal pump.

The pipe contains three 90 degree bends of standard radius. The 90 degree bends each have an equivilant length of 35 diameters.

Calculate the total pressure loss through the pipework.

Pressure drop: (4)(friction factor)(pipe length + fittings)(density)(velocity)^2 / diameter

Fittings: 35 x 0.1m = 3.5m
Pipe Length + fittings = 3m + (3x3.5) = 13.5 m

Pressure drop = (4)(0.024)(13.5)(1010(1.413)^2 / 0.1
= 26,134.28 Pa
= 0.26 bar

Is this correct?

 

[gmax137]

I'm not sure which Moody chart you are using (the link you gave doesn't lead to a single chart). But you need to keep track of the friction factor since there is the Fanning factor and the Darcy factor; the Darcy factor is four times the Fanning factor. So, using the Darcy factor you don't need the leading (4) in your calculation.

Pressure drop: (4)(friction factor)(pipe length + fittings)(density)(velocity)^2 / diameter

Normally this is written $f \frac{L} {D} \frac{\rho v^2} {2}$ so you can see how the different terms go together. I think you are missing the "divided by 2."

When you write out the calculation, include the units (meters, Pa, kg, etc) so that you can check that they all work out as well. That's usually the easiest way to find mistakes.

Finally I'm not sure where you get the pipe length of 3 meters (I thought you said that's how deep the tank is). Check the problem statement.

 

[sci0x]

Thanks gmax. They gave that pressure drop formula in the Q so i went with that.

Your right, the Q says the beer level in the buffer tank is controlled to a depth of 3.
They don't give a pipe lenth.
They do say the pipe contains three 90 degree bends of standard radius and the 90 degree bends each have an equivilant length of 35 diameters.

So maybe the pipe length is: 35 x 0.1m = 3.5m
3 x 3.5 = 10.5m

Pressure drop = (4)(0.024)(10.5m)(1010kgm-3)(1.413ms-1)^2 / 0.1m
= 20,326.67 Pa

 

[sci0x]

The next Q is to set out the Bernouilli equation that describes the whole system:

Info we have:
Applied pressure of CO2 is af 1 bar gauge
Beer level in buffer tank is 3m
Back pressure from filter is 6 bar gauge
Inlet of filter is same height as outlet from base of buffer tank
Darcys friction = 20,326.66 Pa
Beer density = 1010 kg m-3
Gravity acceleration = 9.81 m s-1

Bernouilli Eq: Applied CO2 pressure + Beer height in tank + Pump power = back pressure + Darcys friction

= 1 bar gauge + 3m + pump power = 6 bar gauge + 20,325.66 Pa

Convert 1 bar gauge to m = 100000 Pa/(1010 kgm-3)(9.81ms-1) = 10.092 m
Convert 6 bar gauge to m = 600000 Pa/(1010 kgm-3)(9.81ms-1) = 60.556 m
Convert 20,325.66 Pa to m = 20,325.66Pa/(1010 kgm-3)(9.81ms-1) = 2.03 m

= 10.092 m + 3m + pump power = 60.556m + 2.03 m

Have i included all the elements correctly for Bernouilli Eq?

 

[sci0x]

Or should the Bernouilli Eq for the system be:

1 bar gauge + Pump power = 6 bar gauge + Darcys friction

...
Bernouilli Eq long:
P1 + 1/2pv^2 + pgh +Pp = P2 + 1/2 pv2 + pgh2 + Frpg

 

[gmax137]

A little sketch would help. The elevation and velocity heads must be included at each end; if they are zero then they will drop out.

 

[sci0x]

Okay the velocity must be the same on both sides of the equation.
We do know the buffer tank height is 3m.

The final Q is to calculate the pump power:

So if I am correct here:
Bernouilli Eq: Applied CO2 pressure + Beer height in tank + Pump power = back pressure + Darcys friction

= 1 bar gauge + 3m + pump power = 6 bar gauge + 20,325.66 Pa

Convert 1 bar gauge to m = 100000 Pa/(1010 kgm-3)(9.81ms-1) = 10.092 m
Convert 6 bar gauge to m = 600000 Pa/(1010 kgm-3)(9.81ms-1) = 60.556 m
Convert 20,325.66 Pa to m = 20,325.66Pa/(1010 kgm-3)(9.81ms-1) = 2.03 m

= 10.092 m + 3m + pump power = 60.556m + 2.03m
Pump power head = 49.494 m

I need to convert to kW:
Ph(kW) = q ρ g h / (3.6x10^6)
Where q flow is in m3/hr
My flow is 0.666m3/min = 39.96 m3/hr

Ph(kw) = (39.96 m3/hr)(1010 kgm-3)(9.81ms-1)(49.494 m)/(3.6x10^6)

=19596044.4 / 3.6x10^6

=5.44kW

The pump works at 60% effeciency
So needs to run at 8.704 kW to work.

Answer should be 8.9kW
Am I correct in my method?

 

[Chestermiller]

The next Q is using a friction factor chart determine the value of the friction factor thru the pipe assuming the pipe has smooth internal surface.

Ive used this chart: Link
57,000 is approx midpoint between 10^4 and 10^5 on the bottom of the chart.
Drawing a line up to the smooth pipe, first curve and drawing a line left.
Im getting a friction factor of 0.014.
Am i correct here?

Like gmax137, I also get 0.02 for the friction factor from the Moody plot.

 

[gmax137]

I'm sorry, why do you think the velocity at the filter is zero?

 

[sci0x]

Not that its 0, but its the same on both sides of the equation?
So i'd have 1/2pv^2 on the left side of the equation and 1/2pv^2 on the right side of the equation whch would cancel each other out as velocity is the same on both sides.

Bernouilli Eq: Applied CO2 pressure + Beer height in tank + Pump power = back pressure + Darcys friction

= 1 bar gauge + 3m + pump power = 6 bar gauge + 20,325.66 Pa

 

[gmax137]

Normally when the system includes a "tank" the approach is to pick the liquid surface in the tank as one point (ie, one "side" of the equation), and then say the flow velocity at that point is essentially zero (since the tank cross-section area is normally much larger than the flow area in the connected pipe). Again, a sketch here would be helpful.