A farmer needs to pump 50°F water to a spray irrigation system 100 ft away and 13 ft above the pump as shown in Figure 13.30. The irrigation system requires 30 gal/min at a minimum of 25 psig, and the farmer’s pump supplies 30 gal/min at no more than 50 psig. What is the minimum diameter of PVC

Question

A farmer needs to pump 50°F water to a spray irrigation system 100 ft away and 13 ft above the pump as shown in Figure 13.30. The irrigation system requires 30 gal/min at a minimum of 25 psig, and the farmer’s pump supplies 30 gal/min at no more than 50 psig. What is the minimum diameter of PVC pipe that will meet the requirements if minor losses are neglected?

Accepted Answer

We are asked to determine the minimum pipe diameter that will meet the requirements of a speciﬁed irrigation system. Figure 13.30 serves as an appropriate sketch for this ﬂow situation. Since the unknown is D, this is a type 4 problem. We will apply Eq. 13.29a

[latex]\left(\frac{p_1}{ ho}+\frac{1}{2}\alpha _1 \bar V^2_1+gz_1 ight)-\left(\frac{p_2}{ ho}+\frac{1}{2}\alpha _2 \bar V^2_2+gz_2 ight)=\sum{h_L}+\sum{h_M} [/latex]

with the ﬁrst point at the pump exit, and the second point at the inlet of the irrigation system. The elevations of these two points are known, and we also know [latex]\bar V =\bar V_1 =\bar V_2[/latex], and minor losses are negligible. We will assume turbulent ﬂow with [latex]\alpha[/latex] = 1. Under these conditions the equation reduces to (p₁ − p₂)/ρ + g(z₁ − z₂) = f (L/D)([latex]\bar V^2[/latex]/2). Since z₁ − z₂ = −H, we have:

[latex](p_1 − p_2)−ρgH= ρf\frac{L}{D}\frac{\bar V^2}{2}[/latex] (A)

Solving for the pressure at the pump we obtain

[latex]p_1 = p_2+ρgH+ ρf\frac{L}{D}\frac{\bar V^2}{2}[/latex] (B)

We see that the pump output pressure p₁, must at a minimum be greater than the ambient hydrostatic pressure to ρgH at the pump outlet plus the amount needed to overcome the frictional pressure drop in the pipe, ρf (L/D)([latex]\bar V^2[/latex]/2). In addition, the pump must provide the pressure p₂ required at the irrigation system. To ﬁnd the maximum frictional pressure drop that can be allowed, we evaluate the left-hand side of (A), using the known values of p₂ and p₁, to obtain

(p₁ − p₂) – ρgH = (50 − 25) lb_f/in.² −(1.94 slugs/ft³)(32.2 ft/s²)(13 ft) [latex]\left(\frac{1\ \mathrm{ft} ^2}{144\ in.^2} ight) [/latex]

= 19.4 psi

We must now select a Schedule 40 pipe size whose diameter D results in a frictional pressure drop less than or equal to the maximum allowable pressure drop of 19.4 psi. We can express this requirement by using (A) and the preceding calculation as ρf (L/D)([latex]\bar V^2[/latex]/2) ≤ 19.4 psi. We will select an initial value of [latex]D^{(1)}_{40}[/latex] , use the inside diameter corresponding to this pipe size and the speciﬁed ﬂowrate to calculate [latex]\bar V^{(1)}[/latex], Re⁽¹⁾, and f⁽¹⁾, calculate [ρf (L/D)([latex]\bar V^2[/latex]/2)]⁽¹⁾, then increase or decrease the [latex]D^{(2)}_{40}[/latex] value to the next larger or smaller pipe size as needed to satisfy the inequality. We will assume a relative roughness of zero for smooth plastic.

To illustrate the ﬁrst iteration, note that for an initial guess of [latex]\frac{3}{4}[/latex] in. pipe, D = 0 .06867 ft. Using property values of ρ = 1.94 slugs/ft³ and ν = 1.407 × 10⁻⁵ ft²/s for water at 50°F, we ﬁnd that the average velocity [latex]\bar V^{(1)}[/latex] is given by

[latex]\bar V^{(1)}=\frac{Q}{πD^2/4}=\frac{4(30\ gal/min) }{π(0.06867\ \mathrm{ft} )^2(7.48\ gal/\mathrm{ft} ^3)(60\ s/min)}=18.0\ \mathrm{ft} /s[/latex]

and the remaining parameters are found to be

[latex]Re^{(1)}=\frac{\bar V^{(1)}D^{(1)}}{ u }=\frac{(18.0\ \mathrm{ft} /s)(0.06867\ \mathrm{ft} )}{(1.407×10^{−5}\ \mathrm{ft} ^2/s)}=87,850[/latex]

f ⁽¹⁾ = 0.0182

[latex]ρf\frac{L}{D}\frac{\bar V^2}{2}=\frac{(1.94\ slugs/\mathrm{ft} ^3)(0.0182)\left(\frac{100\ \mathrm{ft} }{0.06867\ \mathrm{ft} } ight)\frac{1}{2}(18.0\ \mathrm{ft} /s)^2}{144\ in.^2/\mathrm{ft} ^2 }=57.8\ psi[/latex]

Using this procedure we can generate the data shown in the following table:

D₄₀ (in.)	D (ft)	[latex]\bar V_2[/latex](ft/s)	Re	f	[latex]ρf\frac{L}{D}\frac{\bar V^2}{2} \left(psi ight) [/latex]
[latex]\frac{3}{4}[/latex]	0.06867	18.0	8.8 × 10⁴	0.0182	57.8
1	0.08742	11.1	6.9 × 10⁴	0.0192	18.2
[latex]1\frac{1}{4}[/latex]	0.1150	6.43	5.3 × 10⁴	9.0205	4.98

The table indicates that a nominal 1 in. diameter pipe is satisfactory. If we assume that the pump pressure with this size pipe is 50 psig, then by rearranging (B), the pressure at the irrigation system is found to be

p₂ = p₁ − ρgH − [latex]ρf\frac{L}{D}\frac{\bar V^2}{2}[/latex] = (50 − 5.6 − 18.2) psig = 26.2 psig

which is 1.2 psig above the required pressure of 25 psig. Since there are many uncertainties in pipe ﬂow calculations and the predicted margin is small, it may be prudent to select the [latex]1\frac{1}{4}[/latex] in. pipe if the economics allow.

Question 13.15: A farmer needs to pump 50°F water to a spray irrigation syst...

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