In the early stages of the design of the chilled water line of Example 13.2, the friction factor was assumed to be f = 0.02 at a design ﬂowrate of 400 gal/min, with a water temperature of 40°F. What is the pressure drop predicted by using this value of f? If the line is installed so that the pipe

Question

In the early stages of the design of the chilled water line of Example 13.2, the friction factor was assumed to be f = 0.02 at a design ﬂowrate of 400 gal/min, with a water temperature of 40°F. What is the pressure drop predicted by using this value of f? If the line is installed so that the pipe exit is actually 3 ft above the inlet, what is the predicted pressure drop?

Accepted Answer

The chilled water line in Example 13.2 is horizontal, thus by applying Eq. 13.10 we have (p₁ − p₂)/ρ = f (L/D)( [latex]\bar V^2[/latex]/2).

[latex]\left(\frac{p_1}{\rho }+gz_1 \right)-\left(\frac{p_2}{\rho }+gz_2 \right)=f\frac{L}{D}\frac{\bar V^2}{2}[/latex] (13.10)

The volume ﬂowrate is given by Q = A [latex]\bar V[/latex], and for a design ﬂowrate of 400 gal/min, the average velocity is calculated as

[latex]\bar V=\frac{Q}{A}=\frac{(400\ gal/min)\left(\frac{1\ \mathsf{ft}^3}{7.48\ gal} \right)\left(\frac{1\ min}{60\ s} \right)}{\left(\frac{\pi (0.33\ \mathsf{ft})^2}{4} \right)}=10.4\ \mathsf{ft}/s[/latex]

Solving for the pressure drop we have

[latex]p_1 − p_2 = ρf\frac{L}{D}\frac{\bar V^2 }{2}= (1.94\ slugs/\mathsf{ft}^3)(0.02) \left(\frac{65\ \mathsf{ft}}{0.33\ \mathsf{ft}}\right)\left(\frac{1}{2} \right) (10.4\ \mathsf{ft}/s)^2 [/latex]

= 413 lb_f/ft² = 2.9 psia

We see that the predicted pressure drop for a horizontal pipe is about the same as the measured value of 3 psia.

If the pipe is inclined, applying Eq. 13.10 gives (p₁/ρ + gz₁) − (p₂/ρ + gz₂) = f (L/D)( [latex]\bar V^2[/latex]/2). Solving for the pressure difference we have p₁ − p₂ = ρ f (L/D)( [latex]\bar V^2[/latex]/2) + ρg(z₂ − z₁). The effect of friction is the same irrespective of the inclination as long as the ﬂowrate is the same, so from the preceding calculation we have

[latex]ρf\frac{L}{D}\frac{\bar V^2 }{2}[/latex] = 413 lb_f/ft² = 2.9 psia

Since the exit of the pipe is now 3 ft above the inlet, we have

ρg(z₂ − z₁) = 1.94 slugs/ft³(32.2 ft/s²)(3 ft ) = 187.4 lb_f/ft² = 1.3 psia

and the predicted pressure drop is

[latex]p_1 − p_2 = ρf\frac{L}{D}\frac{\bar V^2 }{2}[/latex] + ρg(z₂ − z₁) = 2.9 psia + 1.3 psia = 4.2 psia

The additional pressure required at the inlet is not because of additional friction but because of the elevation change.

Question 13.3: In the early stages of the design of the chilled water line ...

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