Example 5:
Calculate water flow for pipe draining under gravity with valves, fittings and elevation change in fully developed Turbulent Flow


Background

The examples provide a comparison of AioFlo results with published data from well known and respected references that are generally accessible to engineers. This will allow prospective AioFlo users to validate its accuracy against a range of typical calculations. The worked examples can also be run by new users as part of their learning process. To learn more about AioFlo click on "Home" in the menu above.

Description

This is a more complex problem than the simple calculation for water flow illustrated in Example 4. This example calculates the mass flow rate of water draining under gravity as it falls from a given elevation to a lower one. The source and destination tanks are both open to atmosphere, so there is no pressure driving force to overcome the friction pressure drop of the flowing water other than the static head due to the elevation change. The pipe is of known diameter and length, and includes a globe valve and 90 degree bends. The exit (acceleration) loss is taken into account, but the entrance loss is ignored.


Problem Reference

Coulson and Richardson's Chemical Engineering, Vol 1, 6th Ed, (1999), Page 92, Example 3.8

Fluid Details

Fluid : Water
Phase : Liquid (incompressible)
Density : 1000 kg/m≥
Viscosity : 0.001 Pa.s (1.0 cP)

Pipe Details

Inside diameter : 150.0 mm (5.906")
Roughness : 1.5 mm
Length : 105 m
Elevation change : -10 meter
Fittings : 1 x globe valve
2 x weld elbow (r/d=1.5)
1 x exit (acceleration) loss)

To be Calculated

The mass flow rate of water that would give a pressure drop equal to the static head caused by the elevation difference.

Download Link

You can run this example in AioFlo by downloading and then opening the data file in AioFlo.

Comparison of Results

Calculated Item Reference AioFlo
Reynolds Number 367000 (*) 363573
Flow Regime Complete turbulence Complete turbulence
Friction factor (Stanton) 0.0045 0.00476
Flow rate (kg/s) 43.3 (*) 42.83
Velocity (m/s) 2.45 (*) 2.40

(*) The 1999 printing of this reference contains a mistake. If the velocity head term in the Darcy-Weisbach equation is expressed as (velocity≤)/(2g) and the Moody friction factor is used then the integer constant in the D-W equation should be 8. This is correctly stated in Equation 3.20 on pg 68 of this edition. However, when re-written in this example the 2 in the denominator of the velocity head term is cancelled into the integer constant to make it 4. There is no problem with this, but in the 9th line of the solution on page 93 the 2 reappears in the velocity head term. The numbers shown in the table above have been corrected to compensate for this mistake.

Discussion

As in Example 4, this problem cannot be solved directly because when the Darcy-Weisbach equation is recast to calculate the flow rate it is found that the result depends on the friction factor, which in turn depends on the flow rate (which is not yet known). The C&R solution uses the iterative procedure that is also used in AioFlo. The example solution stops after the first iteration because the new friction factor calculated from the results of the first iteration is "sufficiently close" to the initial guess for the friction factor. The initial guess was obviously selected to keep the example solution simple and to avoid too much calculation.

If a second iteration were to be done on the reference example numbers the flow rate would become 42.5 kg/s, which is indeed a small change. A third iteration would not change the flow rate within the first 3 significant digits. The reason for this very rapid convergence is that the flow is fully turbulent, and in this regime the friction factor is a function only of the relative roughness of the pipe and is independent of changes in the Reynolds Number. The example has been forced into the fully turbulent regime by the high roughness of the pipe (1.5 mm) which lifts the operating point on the Moody chart well away from the Smooth Flow line.

The results above show very close agreement between the corrected solution of C&R and the solution from AioFlo, and the agreement would be even closer if the example had continued with a second iteration.

If you are using this example for learning AioFlo then you should take note that in the screen shot below the Total Pressure Loss is given as zero because both tanks are under atmospheric pressure, but the Elevation Difference is given as -10 m. In the Calculation results it can be seen that the Static Head and Friction Loss are exactly equal in magnitude, but opposite in sign. The static head supplies the driving force and the friction loss consumes it.

AioFlo showing Input Data and Results for Example 5

AioFlo running Example 5