Estimation of internal pipe diameter

 This problem has been adapted from the technical book Flow of fluids through valves, fittings and pipe. Crane Co.

This problem deserves attention since at first view it seem as having no solution. You may think that there is a missing data. The key is in the proper mathematical undrstnading of the equations to be solved.

The case of the unknown pipe diameter

Take the case of 138 L/min being transported through a Sch. 40 commercial steel pipe with $\Delta P=P_1-P_2=85$ psi (as shown in the sketch below). The transported fluid is water at 25 $^\circ$C with s.w. = 9.77 kN/m$^3$ and dynamic viscosity of 0.89 mPa s. The pipe has ND = 2 inch from point 1 to the size reducer and from this point to 2 the diameter is unknown.

For this case, use the Colebrook equation to find the friction factor.

Fig. 01 Sketch of a pipe split by a reducer

Solution

First, all available data must be written to give some sort of big picture (even if it is incomplete). In this way, we may proceed to know what we do not know.

Fig. 02 Fluid flowing through a pipe with a reducer

Available data

• Pressure gradient: $\Delta P=P_1-P_2=85$ psi = $5.86\times 10^{5}$ Pa
• Flow rate: $Q=138$ L/min $= 2.3 \times 10^{-3}$ m$^3$/s

For the pipe section: point  1 to pipe reducer:

• pipe length: $L_{1-R}=50$ m,
• pipe material: schedule 40 commercial steel,
• nominal diameter: $ND_{1-R}=2$ inch,
• internal diameter: $D_{1-R}=52.5$ mm = $5.25 \times 10^{-2}$ m (search in tables),
• pipe roughness: $\epsilon$ = 0.045 mm = $4.5\times 10^{-5}$ m (search in tables).

For the pipe section: pipe reducer to point 2

• pipe length: $L_{R-2}=70$ m
• pipe material: schedule 40 commercial steel,
• nominal diameter: $ND_{R-2}=??$ inch,
• internal diameter: $D_{R-2}=??$ m,
• pipe roughness: $\epsilon$ = 0.045 mm = $4.5\times 10^{-5}$ m (search in tables).

For the working fluid

• specific weight: $\gamma$ = 9.77 kN/m$^3$,
• dynamic viscosity: $\mu$ = 0.89 mPa s = $0.89 \times 10^{-3}$ Pa s,
• denisty: $\rho$ = 997 kg/m$^3$ (search in tables).
• kinematic viscosity: $\mu = 8.93\times 10^{-7}$ m$^2$/s.

Understanding the procedure to follow on the estimation of the unknown pipe diameter

There is plenty of data but, what can we do with it?

First. The only available set of equations are:

• the general energy balance,

$\dfrac{P_1}{\gamma} +  z_1 - h_L + \dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma} +  z_2 + \dfrac{v_2^2}{2g}$        Eq. (01)

where $P$, $z$ and $v$ indicate the pressure, pipe height and fluid velocity at points 1 and 2, respectively. $h_L$ is the head loss and $\gamma$ the specific weight.

Follow this link for a physical explanation of this equation.

• the Darcy equation for head loss $h_L$,

$h_L=f_F\dfrac{L}{D}\dfrac{v^2}{2g}$        Eq. (02)

where $f_F$ represents the friction factor causing a head loss $h_L$ in a fluid travelling at a speed $V$ through a pipe of length $L$ and internal diameter $D$.

• and the continuity equation,

$Q_1=Q_2$        Eq. (03)

where $Q$ represents the volumetric flow rate.

Be aware that the Colebrook equation:

$\dfrac{1}{\sqrt{f_F}}=-2\log_{10}\left( \dfrac{\epsilon}{3.7D} + \dfrac{2.51}{N_{Re}\sqrt{f_F}} \right)$        Eq. (04)

is also available as tool to solve find the unknown intermal diameter but this equation do not increase or decrease the number of unknowns!

Second. The process is quite straight. You may isolate the unknown internal diameter $D_{R-2}$ directly from the energy balance Eq. (02). 

However, one may say that there is an obstacle on the way: the friction factor $f_{F,R-2}$ for the pipe section $R-2$ depends on the internal diameter $D_{R-2}$ of this pipe. Well, this is false since the Colebrook Eq. (04) is a trascendental equation for $f_F$ to  be substituted into Eq. (01) so that the friction factor is in fact a spurious unknown.

The steps to follow

Step 1. Estimate, numerically, the Reynolds number $N_{Re,1-R}$ for the pipe section 1-R.

Step 2. Estimate, analitically, the Reynolds number $N_{Re,R-2}$ for the pipe section R-2 in terms of the internal pipe diameter $D_{R-2}$.

For this case, you have to be sure that the only unknown is $D_{R-2}$.

Step 3. Estimate, numerically, the head loss $h_{L,1-R}$ from the Darcy Eq. (02) for the pipe section 1-R.

Step 4. Estimate, analitically, the head loss $h_{L,R-2}$ from the Darcy Eq. (02) for the pipe section R-2 in terms of the internal pipe diameter $D_{R-2}$.

For this case, you have to be sure that the only unknown is $D_{R-2}$. However, since the Colebrook Eq. (04) is trascendental the friction factor can not be isolated but numerically determined. This means that the corresponding Colebrook equation must be solved, numerically, simultaneously with the general energy equation for the $R-2$ pipe section.

Do not forget to include minor losses.

Step 5. Make an overall mechanical balance for the whole pipe including $1-R$ and $R-2$ pipe sections. In other words, make use of the energy balance Eq. (01).

For short, you will arrive at a system of two equations to be solved for the internal pipe diameter  $D_{R-2}$ and the friction factor $f_{F,R-2}$.

Step 6. Solve numerically. You may use a trial-error approach by using an initial guess internal diameter similar to that for $1-R$ pipe section. Next, solve the corresponding Colebrook equation to determine $f_{F,R-2}$. Next, check on the corresponding energy equation by substitution of the guess $D_{R-2}$ and the found $f_{F,R-2}$.

If the energy equation zeroes or as close as you desire, then you have solved (almost) the problem. If not, you should increse or decrease your diameter guess.

Step 7. Use the found internal diameter $D_{R-2}$ to find the nominal diameter pipe with internal diameter as close as possible to it.

Step 8. The calculated diameter $D_{R-2}$ is a theoretical estimation and the nominal diameter $ND_{R-2}$ provides the real diameter.

Therefore, the internal diameter corresponding to $ND_{R-2}$ as given in a table of pipe properties is in fact the real internal diameter $D_{R-2}$ you were looking for. This new data shall also give a different fluid velocity $v_{R-2}$, which should be the real velocity.

The calculations to estimate the internal diameter

Step 1. Let us determine the Reynolds number for the pipe section $1-R$ as follows. Let us first determine  the internal pipe cross section area and the fluid velocity for this pipe section.

$A_{1-R}=\dfrac{\pi D_{1-R}^2}{4}=2.16 \times 10^{-3}$ m$^2$

$v_{1-R}=\dfrac{2.3 \times 10^{-3} \text{m$^3$/s}}{2.16 \times 10^{-3} \text{m}^2}$

$v_{1-R}=1.06$ m/s

Then,

$N_{Re,1-R}=\dfrac{v_{1-R}D_{1-R}}{\nu}$

$N_{Re,1-R}=\dfrac{\left(1.06 \text{m/s} \right) \left( 5.25 \times 10^{-2} m \right)}{ 8.93\times 10^{-7} \text{m$^2$/s}}$

$\Rightarrow N_{Re,1-R}=6.23 \times 10^{4}$

Step 2. The process to determine $N_{Re,R-2}$ is quite similar to that in Step 1. However, some little algebra has to be done. First, the corresponding internal pipe cross section area for the $R-2$ pipe section can be written as follows,

$A_{R-2}=\dfrac{\pi D_{R-2}^2}{4}$        Eq. (05)

and in the same way the fluid velocity for this section of the pipe is expressed as,

$v_{R-2}=\dfrac{2.3 \times 10^{-3} \text{m$^3$/s}}{\dfrac{\pi D_{R-2}^2}{4}}$

$v_{R-2}=\dfrac{9.2 \times 10^{-3} }{\pi D_{R-2}^2}$        Eq. (06)

where the units have been drop to avoid confusion. Notice that symbolic handling of certain parameters is preferable to avoid future numerical errors. Then, the Reynolds number for this pipe section is,

$N_{Re,R-2}=\dfrac{9.2 \times 10^{-3} }{\pi D_{R-2}^2} \dfrac{ D_{R-2} }{ 8.93\times 10^{-7} \text{m$^2$/s}}$

$\Rightarrow N_{Re,R-2}=\dfrac{1.03 \times 10^{4} }{\pi D_{R-2}}$        Eq. (07)

Step 3. The head loss $h_{L,1-R}$ for pipe section $1-R$ can now be estimated from Eq. (02) as follows. However, the friction factor $f_{F,1-R}$ must be calculated first. Then, with the corresponding Colebrook equation we have,

$\dfrac{1}{\sqrt{f_{F,1-R}}}=-2\log_{10}\left( \dfrac{\epsilon}{3.7D_{1-R}} + \dfrac{2.51}{N_{Re,1-R}\sqrt{f_{F,1-R}}} \right)$

which after substitution of all known data becomes,

$\dfrac{1}{\sqrt{f_{F,1-R}}}=-2\log_{10}\left( 2.32 \times 10^{-4} + \dfrac{4.03 \times 10^{-5}}{\sqrt{f_{F,1-R}}} \right)$        Eq. (08)

Next, Eq. (08) must be solved numerically. Fortunately, we already have a simple algorithm and a worksheet to find $f_{F,1-R}$ in a short time. Follow this link to the post on Solving the  Colebrook equation. Therefore,

$f_{F,1-R}=2.29\times 10^{-2}$

Then,

$h_{L,1-R}=f_{F,1-R}\dfrac{L_{1-R}}{D_{1-R}}\dfrac{v_{1-R}^2}{2g}$

$h_{L,1-R}=2.29\times 10^{-2} \left(\dfrac{50\text{m}}{5.25 \times 10^{-2}\text{ m}}\right) \left[\dfrac{\left(1.06 \text{m/s} \right)^2}{2\left( 9.81 \text{ m/s}^2 \right)}\right]$

$\Rightarrow h_{L,1-R}=1.25$ m

Step 4. Again, this step requires little algebra. In this case, $F_{F,R-2}$ can not be calculated since the internal diameter $D_{R-2}$ remains unknown. Then, the Colebrook equation,

$\dfrac{1}{\sqrt{f_{F,R-2}}}=-2\log_{10}\left( \dfrac{\epsilon}{3.7D_{R-2}} + \dfrac{2.51}{N_{Re,R-2}\sqrt{f_{F,R-2}}} \right)$

will only be simplified to,

$\dfrac{1}{\sqrt{f_{F,R-2}}}=-2\log_{10}\left( \dfrac{1.21\times 10^{-5}}{D_{R-2}} + \dfrac{2.43 \times 10^{-4}\pi D_{R-2}} { \sqrt{f_{F,R-2}}} \right)$        Eq. (09)

A direct consequence of Eq. (09) is that $h_{L,R-2}$ can not be calculated either. However, we do could write it in terms of $D_{R-2}$ only. This is as follows,

$h_{L,R-2}=f_{F,R-2}\dfrac{L_{R-2}}{D_{R-2}}\dfrac{v_{R-2}^2}{2g} + 2\left(f_{F,R-2}\dfrac{L_{R-2}}{D_{R-2}}\right)_{Elb} \dfrac{v_{R-2}^2}{2g}$        Eq. (10)

where the last term in the above equation was added to account for the losses due to the two short radius elbows in the $R-2$ pipe section. The subscript $Elb$ indicates the resistance coefficient $K$ for these fittings:

$K=30f_{F,R-2,T}$        Eq. (11)

$f_{F,R-2,T}$ stands for the fully turbulent friction factor in the $R-2$ pipe section (see Flow of fluids through valves, fittings and pipe. Crane Co).  This can be easily determined from the Colebrook equation by taking the limit of $N_{Re,R-2}\rightarrow \infty$. This is,

$\dfrac{1}{\sqrt{f_{F,R-2,T}}}=-2\log_{10}\left( \dfrac{\epsilon}{3.7D_{R-2}} \right)$

$\Rightarrow f_{F,R-2,T}=\dfrac{0.25}{\left[\log_{10}\left(\epsilon/D_{R-2}\right)\right]^2}$        Eq. (12)

In this way, Eq. (10) can still be simplified,

$h_{L,R-2}=f_{F,R-2}\dfrac{70 \text{m}}{D_{R-2}} \left(\dfrac{9.2 \times 10^{-3} }{\pi D_{R-2}^2} \right)^2\dfrac{1}{2\left( 9.81 \text{ m/s}^2 \right)}$

$+ 2\dfrac{7.5}{\left[\log_{10}\left(\epsilon/D_{R-2}\right)\right]^2}\left(\dfrac{9.2 \times 10^{-3} }{\pi D_{R-2}^2} \right)^2\dfrac{1}{2\left( 9.81 \text{ m/s}^2 \right)}$

where Eq. (06) was used. A further arithmetic simplification of the previous equation gives,

$h_{L,R-2}=f_{F,R-2}\dfrac{3.02 \times 10^{-4}}{\pi^2 D_{R-2}^5} + \dfrac{6.45 \times 10^{-5}}{\left[\log_{10}\left(\epsilon/D_{R-2}\right)\right]^2}$        Eq. (13)

Thus, Eqns. (09,13) are to be solved simultaneously for $D_{R-2}$. If you look closely to Eqns. (09,13) you will notice that the only unknown is $D_{R-2}$ since $f_{F,R-2}$ can be determined from Eq. (09) once $D_{R-2}$ is given.

Step 5. The energy balance equation for the whole pipe can be written as,

$\dfrac{P_1}{\gamma} + z_1 - h_{L,1-R} - h_{L,R-2} + \dfrac{v_{1-R}^2}{2g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_{R-2}^2}{2g}$        Eq. (14)

Since we have already calculated the head losses $h_{L,1-R}$ and $h_{L,R-2}$ in the previous steps, it would be useful to isolate these parameters from Eq. (14). Then,

$h_{L,1-R} + h_{L,R-2} = \dfrac{P_1-P_2}{\gamma} - \left(z_2 - z_1\right) +  \dfrac{v_{1-R}^2}{2g} - \dfrac{v_{R-2}^2}{2g}$        Eq. (15)

Here, shall be considered $z_1=0$ m and $z_2=20$ m. Next, with the known parameters Eq. (15) can be simplified to,

$h_{L,1-R} + h_{L,R-2} = 40.04 - \dfrac{4.31 \times 10^{-6} }{\pi^2 D_{R-2}^4}$        Eq. (16)

The mechanical balance of the energy equation given in Eq. (16) should equal the general approach by the Darcy equation. Then, after substitution of $h_{L,1-R}$ and Eq. (13) into Eq, (16) we obtain,

$1.25 + f_{F,R-2}\dfrac{3.02 \times 10^{-4}}{\pi^2 D_{R-2}^5} + \dfrac{6.45 \times 10^{-5}}{\left[\log_{10}\left(\epsilon/D_{R-2}\right)\right]^2} = 40.04 - \dfrac{4.31 \times 10^{-6} }{\pi^2 D_{R-2}^4}$

which reduces to,

 $f_{F,R-2}\dfrac{3.02 \times 10^{-4}}{\pi^2 D_{R-2}^5} + \dfrac{6.45 \times 10^{-5}}{\left[\log_{10}\left(\epsilon/D_{R-2}\right)\right]^2} = 38.79 - \dfrac{4.31 \times 10^{-6} }{\pi^2 D_{R-2}^4}$       Eq. (17)   

Equations (09,17) must be solved numerically to determine the internal diameter $D_{R-2}$.

Step 6. The solution by trial-and-error was used here. The table below summarizes the calculations as follows,

• the first column corresponds to the iteration,
• the second column to the internal pipe diameter for section $D_{R-2}$. The row labeled 1 of this column is the guess for the internal diameter. Notice that the guess is equal to $D_{1-R}$, which seems to be a good starting point,
• the third column is the result of the numerical solution of Eq. (09) for $f_{F,R-2}$ based on the supplied $D_{R-2}$ at the same row,
• columns 4 and 5 correspond to the evaluation ogf the LHS (left hand side) and RHS (right hand side) of Eq. (17), respectively, and
• the last column is the % error as calculated between the LHS and RHS. If Eq. (17) is satisfied, then the internal diameter $D_{R-2}$ has been found.

		From Eq. (09)	Check on Eq. (17)
	$D_{R-2}$ (m)	$f_{F,R-2}$	LHS	RHS	Error %
1	5.25E-02	2.29E-02	1.76E+00	3.87E+01	2105.87%
2	5.75E-02	2.29E-02	1.12E+00	3.88E+01	3369.07%
3	4.75E-02	2.29E-02	2.89E+00	3.87E+01	1237.04%
4	4.25E-02	2.29E-02	5.06E+00	3.87E+01	663.93%
5	3.75E-02	2.31E-02	9.52E+00	3.86E+01	305.07%
6	3.25E-02	2.34E-02	1.97E+01	3.84E+01	94.85%
7	2.75E-02	2.38E-02	4.63E+01	3.80E+01	17.95%
8	2.25E-02	2.46E-02	1.30E+02	3.71E+01	71.57%

A short explanation of the iterative calculations

The whole calculation begins with the internal diameter $D_{R-2}$ guess. However, the solution can be found in one of two possible ways: increasing the diamter or decreasing it by a given step.

From rows 1 to 2, the diameter was increased by 0.05. In this case the error also increased as shown in the last column. This means that the unknown diameter is not larger than that of the pipe section $1-R$.

Then, the guess diameter was decreased by a constant step of 0.05 from row 3 through 8. As you can see in the last column the % error decreased with every step so that the unknown diameter should be smaller than that of pipe section $1-R$.

The process should be followed until the % error were  as small as desired or needed. In this situation, the % error suddenly increases at row 8. This means that the error would start increasing again and that the best solution we could get is that in row 7. This is,

$D_{R-2} = 2.75 \times 10^{-2}$ m

The behavior seen in the table above is not what you would wish but it seem to be natural since Eq. (17) is trascendental.

Step 7. $D_{R-2}=2.75 \times 10^{-2}$ m $=27.5$ mm is the internal pipe diameter of pipe section $R-2$ according to the pipe data for section $1-R$, fluid data and pressure gradient across the whole system. However, this is in fact the theoretical internal diameter.

Then, what is the real internal pipe diameter?

The real internal diameter is related to a real pipe being manufactured and available as a standard. Then we must look for pipe datasheets for Sch.- 40 commercial steel pipe (see Flow of fluids through valves, fittings and pipe. Crane Co, for example).

It is found that the pipe with $ND_{R-2}=1$ inch and internal diameter $D_{R-2}=26.6$ mm $2.66\times 10^{-2}$ m is the closest according to the geometrical features. Therefore, in practice what you would use is a $ND_{R-2}=1$ inch Sch. - 40 commercial steel pipe.

Step 8. If we are forced to use only market available pipes the fluid velocity should be different from the theoretical one. The flow rate $Q$ should remain the same but this shall not be the case of $v_{R-2}$.

In other words,

$v_{R-2}=\dfrac{9.2 \times 10^{-3} }{\pi D_{R-2}^2}$
          $=4.14$ m/s

Other parameters like $h_{L,R-2}$ and $f_{F,R-2}$ must be calculated again to get a real picture of the solution in hand.

This is the end of the problem.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• Some basic concepts on thermodynamics
• Who was Clausius?
• Processes at constant volume and pressure
• Composition variables in mixtures
• The first law of thermodynamics and some concepts
• Partial and total derivatives (for thermodynamics)
• About pilot lights
• Solving the Colebrook equation

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Ildebrando.