Pipe Flow Problems

There are three typical problems encountered in pipe flows, depending upon what is known and what is to be found. These are the ''unknown driving force,'' ''unknown flow rate,'' and ''unknown diameter'' problems, and we will outline here the procedure for the solution of each of these for both Newtonian and non-Newtonian (power law and Bingham plastic) fluids. A fourth problem, perhaps of even more practical interest for piping system design, is the ''most economical diameter'' problem, which will be considered in Chapter 7.

We note first that the Bernoulli equation can be written


DF represents the net energy input into the fluid per unit mass (or the net ''driving force'') and is the combination of static head, pressure difference, and pump work. When any of the terms in Eq. (6-69) are negative, they represent a positive ''driving force'' for moving the fluid through the pipe (positive terms represent forces resisting the flow, e.g., an increase in elevation, pressure, etc. and correspond to a negative driving force). In many applications the kinetic energy terms are negligible or cancel out, although this should be verified for each situation.

We will use the Bernoulli equation in the form of Eq. (6-67) for analyzing pipe flows, and we will use the total volumetric flow rate (Q) as the flow variable instead of the velocity, because this is the usual measure of capacity in a pipeline. For Newtonian fluids, the problem thus reduces to a relation between the three dimensionless variables:

A. Unknown Driving Force

For this problem, we want to know the net driving force (DF) that is required to move a given fluid (p . p) at a specified rate (Q) through a specified pipe (D, L, e). The Bernoulli equation in the form DF = ef applies.

1. Newtonian Fluid

The "knowns" and "unknowns" in this case are

All the relevant variables and parameters are uniquely related through the three dimensionless variables f, NRe, and e/D by the Moody diagram or the Churchill equation. Furthermore, the unknown (DF = ef) appears in only one of these groups (f). The procedure is thus straightforward:

1. Calculate the Reynolds number from Eq. (6-70).

2. Calculate e/D.

3. Determine f from the Moody diagram or Churchill equation [Eq. (6-41)]; (if NRe < 2000, use f = 16/NRe).

4. Calculate ef (hence DF) from the Bernoulli Equation, Eq. (6-68).

From the resulting value of DF, the required pump head (—w/g) can be determined, for example, from a knowledge of the upstream and downstream pressures and elevations using Eq. (6-69).

2. Power Law Fluid

The equivalent problem statement is

Note that we have an additional fluid property (m and n instead of p), but we also assume that pipe roughness has a negligible effect, so the total number of variables is the same. The corresponding dimensionless variables are f, NRe pl, and n [which are related by Eq. (6-47)], and the unknown (DF = ef) appears in only one group (f). The procedure just followed for a Newtonian fluid can thus also be applied to a power law fluid if the appropriate equations are used, as follows.

1. Calculate the Reynolds number (NRe pl), using Eq. (6-45) and the volumetric flow rate instead of the velocity, i.e.,

3. Bingham Plastic

The problem statement is

The number of variables is the same as in the foregoing problems; hence the number of groups relating these variables is the same. For the Bingham plastic, these are f, NRe, and NHe, which are related by Eq. (6-65) [along with Eqs. (6-60) and (6-63)]. The unknown (DF = ef) appears only in f, as before. The solution procedure is similar to that followed for Newtonian and power law fluids.

1. Calculate the Reynolds number.

2. Calculate the Hedstrom number:

D pto

3. Determine f from Eqs. (6-65), (6-63), and (6-60). [Note that an iteration is required to determine f from Eq. (6-60).]

B. Unknown Flow Rate

In this case, the flow rate is to be determined when a given fluid is transported in a given pipe with a known net driving force (e.g., pump head, pressure head, and/or hydrostatic head). The same total variables are involved, and hence the dimensionless variables are the same and are related in the same way as for the unknown driving force problems. The main difference is that now the unknown (Q) appears in two of the dimensionless variables ( f and NRe), which requires a different solution strategy.

1. Newtonian Fluid

The problem statement is

The strategy is to redefine the relevant dimensionless variables by combining the original groups in such a way that the unknown variable appears in one group. For example, f and NRe can be combined to cancel the unknown (Q) as follows:

Thus, if we work with the three dimensionless variables fNRe, NRe, and e/D, the unknown (Q) appears in only NRe, which then becomes the unknown (dimensionless) variable.

There are various approaches that we can take to solve this problem. Since the Reynolds number is unknown, an explicit solution is not possible using the established relations between the friction factor and Reynolds number (e.g., the Moody diagram or Churchill equation). We can, however proceed by a trial-and-error method that requires an initial guess for an unknown variable, use the basic relations to solve for this variable, revise the guess accordingly, and repeat the process (iterating) until agreement between calculated and guessed values is achieved.

Note that in this context, either f or NRe can be considered the unknown dimensionless variable, because they both involve the unknown Q. As an aid in making the choice between these, a glance at the Moody diagram shows that the practical range of possible values of f is approximately one order of magnitude, whereas the corresponding possible range of NRe is over five orders of magnitude! Thus, the chances of our initial guess being close to the final answer are greatly enhanced if we choose to iterate on f instead of NRe. Using this approach, the procedure is as follows.

1. A reasonable guess might be based on the assumption that the flow conditions are turbulent, for which the Colebrook equation, Eq. (6-38), applies.

2. Calculate the value offNRe from given values.

3. Calculate f using the Colebrook equation, Eq. (6-38).

4. Calculate NRe = (fNRe/f )1/2, using f from step 3.

5. Using the NRe value from step 4 and the known value of e/D, determine f from the Moody diagram or Churchill equation (if NRe < 2000, use f = 16/NRe).

6. If this value of f does not agree with that from step 3, insert the value of f from step 5 into step 4 to get a revised value of NRe.

7. Repeat steps 5 and 6 until f no longer changes.

8. Calculate Q = nDpNRe/4p.

2. Power Law Fluid

The problem statement is

The simplest approach for this problem is also an iteration procedure based on an assumed value of f:

1. A reasonable starting value for f is 0.005, based on a ''dart throw'' at the (equivalent) Moody diagram.

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