A new reservoir will use gravity to supply drinking water to a water treatment plant serving several surrounding towns, as shown in Figure 13.28. The required ﬂowrate is 5000 gal/min. The surface of the reservoir is 200 ft above the plain where the water treatment plant is located, and the supply

Question

A new reservoir will use gravity to supply drinking water to a water treatment plant serving several surrounding towns, as shown in Figure 13.28. The required ﬂowrate is 5000 gal/min. The surface of the reservoir is 200 ft above the plain where the water treatment plant is located, and the supply pipe is commercial steel, 3 ft in diameter. If the minimum pressure required at the water treatment plant is 50 psig, how far away can the reservoir be located with this size pipe? Assume that minor losses are negligible and that the water is at 50°F.

Accepted Answer

We are asked to determine the maximum distance between a water treatment plant and a reservoir given a set of piping constraints. Figure 13.28 serves as an appropriate sketch for this ﬂow situation. This is a type 2 problem in that the unknown is a length. 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]

To ﬁnd the velocity in the pipe, [latex]\bar V_2[/latex], we write [latex]\bar V_2[/latex] = Q/A and insert the data to obtain

[latex]\bar V_2=\frac{(5000\ gal/min)\left(\frac{1\ min}{60\ s} ight)\left(\frac{1\ \mathsf{ft}^3}{7.48\ gal} ight) }{π(3\ \mathsf{ft})^2/4}=1.58\ \mathsf{ft}/s[/latex]

Using properties for water at 50°F, we ﬁnd Re = [latex]\bar V_2D/ u[/latex] = [(1.58 ft/s)(3 ft)]/( 1.407 × 10⁻⁵ ft²/s) = 3.4 × 10⁵; thus the ﬂow is turbulent, as expected. From Figure 13.12, the relative roughness for D = 3 ft commercial steel pipe is e/D = 5 × 10⁻⁵. The friction factor from Figure 13.13, or calculated using the Colebrook or Chen equation, is found to be f = 0.0147.

To apply Eq. 13.29a, note that at point 1 on the reservoir surface, p₁ = 14.7 psia = 0 psig , [latex]\bar V^2_1[/latex] is negligible and z₁ = 200 ft. At point 2 we have p₂ = 50 psig, [latex]\bar V_2[/latex]= 1.58 ft/s, and z₂ = 0 ft. We will assume [latex]α_2[/latex] = 1 as usual. The major loss is h_L = f (L/D)( [latex]\bar V^2_2[/latex] /2), so after neglecting the minor losses, Eq. 13.29a becomes g(z₁ − z₂)−( p₂/ρ + [latex]\frac{1}{2} \alpha_2\bar V^2_2[/latex] ) = f (L/D)([latex]\bar V^2_2[/latex] /2). Solving for the length, we obtain

[latex]L=\frac{2D}{f\bar V^2_2}\left[g(z_1-z_2)-\left(\frac{p_2}{ ho} +\frac{1}{2}\alpha _2\bar V^2_2 ight) ight] [/latex]

Inserting the data yields:

[latex]L=\frac{2(3\ \mathsf{ft})}{(0.0147)(1.58\ \mathsf{ft}/s)^2 }[/latex]

[latex] imes \left[(32.2\ \mathsf{ft}/s^2)(200\ \mathsf{ft})-\frac{ (50\ lb_\mathsf{f}/in.^2)(144\ in.^2/\mathsf{ft}^2)}{1.94\ slugs/\mathsf{ft}^3 }-\frac{1}{2}(1.58\ \mathsf{ft}/s)^2 ight][/latex]

= 4.46 × 10⁵ ft

This is a distance of about 84 miles. Note that in this problem we have used gage pressure consistently throughout. In earlier problems we used absolute pressure throughout. Both approaches are correct, but we cannot mix gage and absolute pressure in the head loss equation.

Question 13.13: A new reservoir will use gravity to supply drinking water to...

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