Question 7 of 8: Fan energy, and duct pressure loss

Question 7: Fan energy, and duct pressure loss (20 points: a 10, b 10)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Part (a) is a fan straddling a contraction from 16" to 10" round duct at 1,000 cfm; part (b) is a constant-section 12" square galvanized duct carrying 2,000 cfm through 50 ft of straight run and two smooth-radius elbows.

Find. For (a) the fan arrangement, the energy required expressed as a head in feet of air, and the horsepower; for (b) the total pressure loss from A to D.

Approach. Fan total pressure is the rise in total pressure across the machine, so the velocity pressures at the two different duct sizes must be added to the measured statics before subtracting; the head and power follow directly. For part (b), friction is computed on the circular equivalent diameter with the Colebrook relation, and the two elbows are charged as loss coefficients on the actual velocity pressure.

1. Velocity pressures at inlet and discharge. The two areas are $A_{\text{in}}=\tfrac{\pi}{4}(16/12)^{2}=1.3963$ ft² and $A_{\text{out}}=0.5454$ ft², so the velocities are 716 fpm and 1,833 fpm. For standard air,$$p_{v}=\left(\frac{V}{4005}\right)^{2}\ \text{in. wg}\ \Rightarrow\ p_{v,\text{in}}=0.03198,\qquad p_{v,\text{out}}=0.20958$$Halving the diameter would quadruple the velocity; going from 16" to 10" raises the velocity pressure by a factor of 6.6, and that is the whole point of the question.
2. Fan total pressure. Total pressure is static plus velocity pressure at each plane:$$p_{t,\text{in}}=-0.15+0.03198=-0.11802,\qquad p_{t,\text{out}}=0.30+0.20958=0.50958\ \text{in. wg}$$$$\boxed{\ \mathrm{FTP}=p_{t,\text{out}}-p_{t,\text{in}}=0.62760\ \text{in. wg}\ }$$For completeness the fan static pressure, which is what a catalogue selection usually quotes, is $\mathrm{FSP}=\mathrm{FTP}-p_{v,\text{out}}=0.41802$ in. wg — note that it is not the 0.45 in. wg difference of the two static readings.
3. Energy in feet of air. One inch of water gauge is 5.192 lbf/ft², and the air weighs 0.075 lbf/ft³, so the equivalent column of air is$$H=\frac{\mathrm{FTP}\times5.192}{\rho}=\frac{0.62760\times5.192}{0.075}$$$$\boxed{\ H=43.45\ \text{ft of air}\ }$$The number is large only because air is light: the same energy is 0.62760 in. wg, about 156 Pa.
4. Air power and shaft power. With the customary constant,$$\mathrm{AHP}=\frac{Q\times\mathrm{FTP}}{6356}=\frac{1000\times0.62760}{6356}$$$$\boxed{\ \mathrm{AHP}=0.09874\ \text{hp}\ }$$Assuming a total fan efficiency of 65 %, reasonable for a small forward-curved or backward-inclined centrifugal unit at this duty, the shaft power is $\mathrm{BHP}=0.09874/0.65=0.1519$ hp, or 0.1133 kW — and the motor would be selected at the next standard size above it. The head cross-checks against the air power: $\dot{m}gH=1000\times0.075\times43.45$ ft·lbf/min divided by 33,000 returns the same 0.09874 hp.

[Figure not reproduced: Part (b) — the duct run redrawn from the examination figure. Fifty feet of straight 12" square duct, a 90° elbow of 13" centreline radius at B and a 120° elbow of 24" centreline radius at C. See the official exam paper.]

Turning to part (b), the duct section is constant from A to D, so the velocity pressure is the same everywhere and the total-pressure loss and the static-pressure loss are identical — a convenience that does not survive into any run that changes size.

1. Velocity and velocity pressure. The duct area is $12\times12/144=1.00$ ft², so$$V=\frac{Q}{A}=\frac{2000}{1.00}=2,000\ \text{fpm},\qquad p_{v}=\left(\frac{2,000}{4005}\right)^{2}=0.24938\ \text{in. wg}$$Two thousand feet per minute is a normal medium-velocity duct speed; the answer scales with the square of it.
2. Circular equivalent diameter. Friction data are tabulated for round duct, so the rectangular section is converted to the round duct that would lose the same pressure per unit length at the same flow:$$D_{e}=1.30\,\frac{(ab)^{0.625}}{(a+b)^{0.25}}=1.30\,\frac{(144)^{0.625}}{(24)^{0.25}}=13.12\ \text{in.}$$Note that $D_{e}$ exceeds the 12" hydraulic diameter; it is a friction-equivalence device, not a geometric one, and it must be used with the flow rate rather than with the actual velocity.
3. Friction factor and friction rate. In the equivalent round duct the velocity is 2,131 fpm and the Reynolds number is 247,288. Galvanized duct of average construction has an absolute roughness of about 0.0003 ft, so Colebrook's relation$$\frac{1}{\sqrt{f}}=-2\log_{10}\left(\frac{\varepsilon}{3.7D_{e}}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$$gives $f=0.01714$, and the friction rate is$$\frac{\Delta p}{100\ \text{ft}}=f\,\frac{100}{D_{e}}\,p_{v,e}=0.44390\ \text{in. wg per 100 ft}$$which agrees with the published ASHRAE friction chart read at 2,000 cfm and 13" diameter — the chart is generated from this same relation.
4. Straight-duct friction. Over the $10+20+20=50$ ft of run,$$\Delta p_{f}=0.44390\times\frac{50}{100}=0.22195\ \text{in. wg}$$
5. Elbow losses. Local losses are charged as a multiple of the velocity pressure, $\Delta p=C\,p_{v}$, with $C$ read from the ASHRAE table for smooth-radius rectangular elbows without vanes at aspect ratio $H/W=1.0$. Elbow 1 has $r/W=13/12=1.083$, interpolating to $C_{1}=0.262$ for a 90° turn. Elbow 2 has $r/W=24/12=2.0$, for which $C_{90}=0.20$; the angle correction for a 120° turn is $K_{\theta}=1.20$, giving $C_{2}=0.240$. Hence$$\Delta p_{1}=0.262\times0.24938=0.06534,\qquad \Delta p_{2}=0.240\times0.24938=0.05985\ \text{in. wg}$$The wide-radius 120° elbow costs barely more than the tight 90° one, which is the practical lesson: bend radius matters more than bend angle.
6. Total loss from A to D.$$\Delta p_{A\to D}=\Delta p_{f}+\Delta p_{1}+\Delta p_{2}=0.22195+0.06534+0.05985$$$$\boxed{\ \Delta p_{A\to D}=0.34714\ \text{in. wg}\ (86.5\ \text{Pa})\ }$$Friction accounts for 0.22195 in. wg and the two fittings for 0.12519 in. wg, so a third of the loss in this short run is in the fittings — the usual proportion, and the reason fitting selection deserves as much care as duct sizing.