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Examples of Pipeline Calculation and Selection Problems with Solutions

Problem 1. Minimum pipeline diameter determination

Statement of problem: A petrochemical plant transfers paraxylene С6Н4(СН3)2 at Т=30 °С and  a throughput capacity Q=20 m3/h through a steel pipe section with a length L=30 m. P-xylene density and viscosity is ρ=858 kg/m3 and μ=0.6 Cp, respectively. Assume absolute roughness ε for steel as 50 μm.

Input data: Q=20 m3/h; L=30 m; ρ=858 kg/m3; μ=0.6 cP; ε=50 mcm; Δp=0.01 mPa; ΔH=1.188 m.

Problem: Determine the minimum pipe diameter, at which the differential pressure across this section will not exceed Δp=0.01 mPa (ΔH=1.188 m of P-xylene column).

Solution: Since the flow velocity v and the pipe diameter d are unknown, it’s impossible to calculate either the Reynolds number Re, or the relative roughness ɛ/d. It is necessary to take the value of the coefficient of friction λ and calculate the corresponding value of d, using the energy loss equation and the continuity equation. Then, based on the d value, the Reynolds number Re and the relative roughness ɛ/d will be calculated. Next, using the Moody diagram, a new value of f will be obtained. Thus, using the method of successive iterations, the required value of the diameter d will be determined.

Using the continuity equation v=Q/F and the flow area formula F=(π·d²)/4,  we will transform the Darcy-Weisbach equation as follows:

∆H = λ · L/d · v²/(2·g) = λ · L/d · Q²/(2·g·F²) = λ · [(L·Q²) / (2·d·g·[(π·d²)/4]²)] = (8·L·Q²)/(g·π²) · λ/d5 = (8·30·(20/3600)²)/(9.81·3.14²) · λ/d5 = 7.658·10-5 · λ/d5

Then, we will express the diameter:

d = 5(7.658·10-5·λ)/∆H = 5(7.658·10-5·λ)/10000 = 0.0238·5λ

Now, let's express the Reynolds number through the diameter d:

Re = (ρ·v·d)/μ = (4·ρ·Q)/(π·μ·d) = (4·858·20)/(3.14·3600·0.6·10-3·d) = 10120/d

Proceeds in a similar way with a relative roughness:

ε/d = 0.00005/d

For the first stage of iteration, it’s necessary to select the value of the friction coefficient. Take the average value λ = 0.03. Next, let’s sequentially compute d, Re and ε/d:

d = 0.0238·5(λ) = 0,0118 m

Re = 10120/d = 857627

ε/d = 0.00005/d = 0.00424

Knowing these values, let’s proceed with a reverse operation to determine, using the Moody diagram, the value of the coefficient of friction λ, which will be 0.017. Further, let’s find d, Re and ε/d for the new λ value:

d = 0.,0238·5λ = 0,0105 m

Re = 10120/d = 963809

ε/d = 0.00005/d = 0.00476

Again, from the Moody diagram, we will receive a clarified value of λ (=0.0172). The value obtained differs from the previously chosen one only by [(0.0172-0.017)/0.0172]·100=1.16%, therefore, a new stage of iteration is not required and the values, which we ​​found earlier, are correct. Thus, the minimum diameter of the pipe is 0.0105 m.

Problem 2. Selecting Optimum Economic Solutions Based on Initial Data

Statement of problem: Two pipelines of a different diameter were proposed for technological process implementation. The first option assumes the use of larger-diameter pipes, which implies large capital expenditures (Cк1 = 200000 rubles), however, annual costs will be less and will amount to Се1 = 30000 rubles. For the second option, pipes of a smaller diameter were chosen, which reduces capital expenditures to Cк2 = 160000 rubles, but increases the annual maintenance costs to Се2 = 36000 rubles. Both options are designed for n = 10 years of operation.

Input data: Cк1 = 200000 RUB; Се1 = 30000 RUB; Cк2 = 160000 RUB; Се2 = 35000 RUB; n=10 years.

Problem: Determine the most cost-effective solution.

Solution: Obviously, the second option is more advantageous due to reduced capital expenditures, but in the first case there is an advantage due to lower current costs. Let’s use the formula for determining the payback period of additional capital expenditures due to savings in the maintenance:

no = (Cк1к2)/(Сe2e1) = (200000-160000)/(35000-30000) = 8 years

It follows that at a service life of up to 8 years, the second option will be more cost-effective due to lower capital expenditures, but the l total costs of the both projects will be equaled at the 8th year of operation, and ever since the first option will be more advantageous.

Since the pipeline it expected to be operated for 10 years, the advantage should be given to the first option.

Problem 3. Pipeline Optimum Diameter Selection and Calculation

Statement of problem: Two process trains are designed to handle non-viscous fluids at a flow rate of Q1 = 20 m3/h and Q2 = 30 m3/h. In order to simplify pipeline installation and maintenance, it was decided to use pipes of the same diameter for both trains.

Input data: Q1 = 20 m3/h; Q2 = 30 m3/h.

Problem: Determine the pipe diameter d, which is suitable for the given problem.

Solution: Since there are no additional requirements for the pipeline, the main criterion of compliance will be the pipeline ability to transfer fluid at the specified flow rates. Let’s use the tabular data of optimum velocities for non-viscous fluids in a pressure pipeline. This range will be 1.5 - 3 m/s.

It follows that it is possible to determine the optimum diameter ranges, corresponding to the optimum velocity rates, and to establish ​​the intersection area. Pipe diameters from this area will obviously meet the applicability requirements for the above listed flow cases.

Let's determine a range of optimum diameters for the case Q1 = 20 m3/h, using the flow calculation formula, having expressed the pipe diameter from it:

Q = [(π·d²)/4] · v


d = √(4·Q)/(π·v)

Let’s substitute the minimum and maximum value ​​of the optimum velocity:

d1min = √(4·20)/(3600·3.14·1.5) = 0.069 m

d1max = √(4·20)/(3600·3.14·3) = 0.049 m

Thus, pipes with a diameter of 49 to 69 mm are suitable for the train with a flow rate of 20 m3/h.

Let’s determine the range of optimum diameters for the case Q2 = 30 m3/h:

d2min = √(4·30)/(3600·3.14·1.5) = 0.084 m

d2max = √(4·30)/(3600·3.14·3) = 0.059 m

Finally we will find that the range of optimum diameters is 49-69 mm and 59-84 mm for the first and second cases, respectively. The intersection of these two ranges will provide a set of target values. Thus, pipes with a diameter from 59 to 69 mm can be used for both trains.

Problem 4. Determination of Water Flow Regime in Pipes

Statement of problem: Water at a rate of 90 m3/h flows through a pipeline, 0.2 m in diameter. Water temperature is t = 20°C; at this temperature, dynamic viscosity is 1·10-3 Pa·s and density is 998 kg/m3.

Input data: d = 0.2 m; Q = 90 m3/h; μ = 1·10-3; ρ = 998 kg/m3.

Problem: Identify water flow regime in the pipe.

Solution: The flow regime can be determined by the Reynolds criterion values (Re); to calculate this criterion, it’s necessary to preliminary determine the water flow velocity in the pipe (v). The value v can be calculated from the flow equation for round pipes:

Q = v·(π·d²)/4


v = Q·4/(π·d²) = [90/3600] · [4/(3.14·0.2²)] = 0.8 m/s

Using the flow velocity determined, let’s calculate the Reynolds criterion for it:

Re = (ρ·v·d)/μ = (998·0.8·0.2) / (1·10-3) = 159680

The critical Reynolds criterion value Reкр for round pipes is 2300. The criterion value obtained is greater than the critical value (159680 > 2300), hence, a turbulent flow regime is in the pipe.

Problem 5. Reynolds Criterion Value Determination

Statement of problem: Water flows in a rectangular inclined chute with a width w = 500 mm and a height h = 300 mm, not reaching a = 50 mm to the upper edge of the chute. The water flow rate is Q = 200 m3/h. In calculations, the density of water to assume as ρ = 1000 kg/m3 and dynamic viscosity μ = 1·10-3 Pa·s.

Input data: w = 500 mm; h = 300 mm; l = 5000 mm; a = 50 mm; Q = 200 m3/h; ρ = 1000 kg/m3; μ = 1·10-3 Pa·s.

Problem: Determine the Reynolds criterion value.

Solution: Since in this case the fluid flows through a rectangular chute instead of a round pipe, an equivalent diameter of the channel should be found for subsequent calculations. In general, it is calculated by the formula:

dэ = (4·Fж)/Pc

Fж – fluid flow cross-section area;
Pс – wetted perimeter.

Obviously that the fluid flow width coincides with the width channel w, whilst the fluid flow height is h-a mm. In this case we will receive:

Pc = w+2·(h-a) = 0.5+2·(0.3-0.05) = 1 m

Fж = w·(h-a) = 0.5·(0.3-0.05) = 0.125 m2

Now it’s possible to determine the equivalent diameter of the fluid flow:

dэ = (4·Fж)/Pc = (4·0.125)/1 = 0.5 m

Then, let’s use the formula for the flow, expressed in terms of the flow velocity and its cross-section area, to find the flow velocity:

Q = v·Fж m/s

v = Q/Fж = 200/(3600·0.125) = 0.45

Using the values we ​​found earlier, it becomes possible to use the formula for calculating the Reynolds criterion:

Re = (ρ·v·dэ)/μ = (1000·0.45·0.5) / (1·10-3) = 225000

Problem 6. Pipeline Pressure Head Loss Calculation and Determination

Statement of problem: A pump delivers water through a round pipe, whose configuration is shown in the figure, to the end consumer. Water flow rate is Q = 7 m3/h. The diameter of the pipe is d = 50 mm, and the absolute roughness Δ = 0.2 mm. In the calculations, assume the density of water as ρ = 1000 kg/m3 and the dynamic viscosity as μ = 1·10-3 Pa·s.

Input data: Q = 7 m3/h; d = 120 mm; Δ = 0.2 mm; ρ = 1000 kg/m3; μ = 1·10-3 Pa·s.

Problem: Calculate the pressure head loss in the pipeline (Hоп).

Solution: First, let’s find the flow velocity in the pipeline; for this purpose, use the fluid flow formula:

v = (4·Q) / (π·d²) = [(4·7)/(3.14·0.05²)] · 1/3600 = 1 m/s

The velocity calculated allows us to determine the Reynolds criterion value for the given flow:

Re = (w·d·ρ)/μ = (1·0.05·1000) / (1·10-3) = 50000

The total pressure head loss consists of a friction loss during fluid flow through the pipe (Hт) and a pressure head loss in local resistance spots (Hмс).

The friction loss can be calculated by the following formula:

Hт = [(λ·l)/dэ] · [v²/(2·g)]

λ – coefficient of friction;
L – total pipeline length;
[v²/(2·g)] – flow velocity head.

Let’s find the flow velocity head value:

v²/(2·g) = 1²/(2·9.81) = 0.051 m

To determine the coefficient of friction, a correct calculation formula, which depends on the Reynolds criterion value, should be chosen. For this purpose, we find a relative roughness of the pipe by the formula:

e = Δ/d = 0.2/50 = 0.004

Next, we calculate two additional values:

10/e = 10/0.004 = 2500

The previously found Reynolds criterion value falls within the interval 10/e < Re < 560/e, therefore, the following calculation formula must be used:

λ = 0.11·(e+68/Re)0.25 = 0.11·(0.004+68/50000)0.25 = 0.03

Now it’s possible to determine the friction pressure head loss value:

HT = [(λ·l)/d] · [v²/(2·g)] = [(0.03·30)/0.05] · 0.051 = 0.918 m

The total pressure head loss in local resistance spots is formed by the head losses in each of the local resistance spots (in this problem, this includes two turns and one normal valve). The total pressure head loss can be calculated by the formula:


where ζ – local resistance coefficient.

Since the tabulated values ​​of the pressure head coefficient do not include the value for 50mm-diameter pipes, an approximate calculation method should be used to determine these values. The coefficient of resistance (ζ) for a normal valve is 4.9 for a 40mm-diameter pipe and 4 for an 80mm-diameter pipe, respectively. Simply imagine that intermediate values ​​between these values lie on a straight line, i.e. their variation is described by the formula ζ = a·d+b, where a and b are the coefficients of the straight line equation. Let’s compose and solve the system of equations:


4,9 = a·40+b
4 = a·80+b



a = -0,0225
b = 5,8

Problem 7. Entire Pipeline Hydraulic Resistance Change Determination

Statement of problem: During a repair of the main pipeline, which has an inner diameter d1 = 0.5 m and is used to transfer water at a velocity v1 = 2 m/s, it became clear that a pipe section with a length L = 25 m had to be replaced. Since a pipe of the same diameter was not available in stock to replace the failed line section, a pipe with an inner diameter d2 = 0.45 m was installed. Absolute roughness is Δ1 = 0.45 mm for the 0.5m-diameter pipe and Δ2 = 0.2 mm for the 0.45m-diameter pipe. In the calculations, assume the density of water as ρ = 1000 kg/m3 and dynamic viscosity as μ = 1·10-3 Pa·s.

Input data: d1 = 0.5 m; d2 = 0.45 m; L = 25 m; v1 = 2 m/s; Δ1 = 0.45 mm; Δ2 = 0.2 mm; ρ = 1000 kg/m3; μ = 1·10-3 Pa·s.

Problem: Determine the change in the hydraulic resistance for the entire pipeline.

Solution: Since the remaining part of the pipeline was not changed, its hydraulic resistance also did not change after repair; thus, to solve the problem, it will be sufficient to compare the hydraulic resistance of the replaced and replacement pipe sections.

Let’s calculate the hydraulic resistance of the replaced pipe section (H1). Since the pipe does not have any sources of local resistance, it will sufficient to find the friction loss (Hт1):

Hт1 = [(λ1·l)/d1] · [(v1²)/(2·g)]

λ1 – coefficient of hydraulic resistance of the replaced section;
g – acceleration of gravity.

To find λ, it’s necessary to preliminary determine the relative roughness (e1) of the pipe and the Reynolds criterion (Re1):

e1 = Δ1/d1 = 0.45/500 = 0.0009

Re1 = (v1·d1·ρ)/μ = (2·0.5·1000)/(1·10-3) = 1000000

Let’s choose the calculation formula for λ1:

10/e1 = 10/0.0009 = 11111

560/e1 = 560/0.0009 = 622222

Since the calculated value Re1 > 560/e1, then the following formula should be used to find λ1:

λ1 = 0.11·e10.25 = 0.11·0.00090.25 = 0.019

Now it’s possible to find the pressure head drop in the replaced pipe section:

H1 = Hт1 = (λ1·l)/d1 ·[(v1²)/(2·g)] = (0.019·25)/0.5·2²/(2·9.81) = 0.194 m

Let’s calculate the hydraulic resistance of the replacement pipe section (H2). In this case, in addition to the friction pressure head drop (Hт2), there is a pressure head drop due to local resistance spots (Hмc2), i.e. a sharp pipeline narrowing at the replaced section inlet and a sharp expansion at the replaced section outlet.

First, we determine the friction pressure head drop in the replacement pipe section. Since the diameter became smaller and the flow rate remained the same, it’s necessary to find the new flow velocity v2. The sought-for value can be found from the equality of the flows calculated for the replaced and replacement sections:

v1·(π·d1²)/4 = v2·(π·d2²)/4


v2 = v1·(d1/d2)² = 2·(500/450)² = 2.47 m/s

The Reynolds criterion for the water flow in the replacement pipe section:

Re2 = (v2·d2·ρ)/μ = (2.47·0.45·1000)/(1·10-3) = 1111500

Now, let’s find the relative roughness for a 450mm-diameter pipe section and choose the formula for calculating the coefficient of friction:

e2 = Δ2/d2 = 0.2/450 = 0.00044

10/e2 = 10/0.00044 = 22727

560/e2 = 560/0.00044 = 1272727

The obtained value Re2 lies in the interval between 10/e1 and 560/e1 (22 727 < 1 111 500 < 1 272 727), so the following formula will be used to calculate λ2:

λ2 = 0.11·(e2+68/Re2)0.25 = 0.11·(0.00044+68/1111500)0.25 = 0.0165

Wherefrom it becomes possible to calculate the friction loss in the replacement pipe section:

Hт2 = [(λ2·l)/d2] · [(v2²)/(2·g)] = [(0.0165·25)/0.45] · [2.47²/(2·9.81)] = 0.285 m

The pressure head loss in the local resistance spots will be consisted of the losses at the replaced section inlet (sharp channel narrowing) and outlet (sharp channel expansion). Let’s find the ratio of the replacement pipe section area to the original pipe section area:

F2/F1 = (d2²)/(d1²) = (0.45/0.5)² = 0.81

From the tabular values, we select the local resistance coefficients: ζpc = 0.1 for a sharp narrowing and ζρρ = 0.04 for a sharp expansion. Using this data, let’s calculate the total pressure head loss in the local resistance spots:

Hмс2 = ∑ζмс · [v²/(2·g)] = [ζрс·(v1²)/(2·g)] + [ζрр·(v2²)/(2·g)] = [0.1·2²/(2·9.81)] + [0.04·2.47²/(2·9.81)] = 0.032 m

From the above it follows that the total pressure head drop in the replacement section is as follows:

H2 = Hт2+Hмс2 = 0.285+0.032 = 0.317 m

Knowing the pressure head loss in the replaced pipe section and in the replacement pipe section, we can determine the change in the loss:

∆H = 0.317-0.194 = 0.123 m

We will find that after the replacement of the pipeline section, its total pressure head loss has increased by 0.123 m.