22-Mec-B2 Environmental Control in Buildings · May 2015

Question 3 of 8: Centrifugal fan power and the fan laws

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Professional Engineers of Ontario / EGBC annual examination, 07-Mec-B2 (now 22-Mec-B2) Environmental Control in Buildings, May 2015 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate on the cover of the first workbook which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for the refrigerant are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to submit a clear statement of any interpretation assumptions with the answer — that latitude is used explicitly below wherever the printed data are redundant or incomplete.

All eight problems are worked here. Every psychrometric state has been recomputed from the ASHRAE formulation for saturation vapour pressure rather than scaled off a chart, so the numbers below are tighter than a graphical solution would be; chart-quality agreement (about $\pm 0.2$ K in temperature and $\pm 0.0002$ kg/kg in humidity ratio) is all that an examiner expects, and a candidate reading the appended charts should reproduce every answer to within that band.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry), Ch. 6 (cooling loads), Ch. 10 (cooling towers), Ch. 15 (fans and duct design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 8 (energy estimating and degree-day methods), Ch. 12–13 (fluid flow, fans and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (fans and duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope), Ch. 30 (refrigerant properties).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression refrigeration and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality, and ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020 (Part 6, and Appendix C for design temperatures and degree-days), National Energy Code of Canada for Buildings 2020, Health Canada Residential Indoor Air Quality Guidelines, and Environment and Climate Change Canada Canadian Climate Normals.

Psychrometric relations used throughout. At barometric pressure $p$ with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation, the humidity ratio, specific enthalpy and humid volume of moist air are

$$W = 0.621945\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t), \qquad v = \frac{0.287042\,(t + 273.15)\,(1 + 1.6078\,W)}{p}$$

with $h$ in kJ per kg of dry air, $t$ in $\,{}^{\circ}$C and $p$ in kPa. In inch-pound units the enthalpy becomes $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air with $t$ in $\,{}^{\circ}$F. Wet-bulb temperatures are obtained by solving the adiabatic-saturation equation, not by eye. Mixing two air streams is exact in moisture and in enthalpy, so $W$ and $h$ of the mixture are the mass-weighted averages and the mixed dry bulb follows from them; weighting the dry bulb directly is the usual shortcut and differs here by about $0.01$ K.

Question 3: Centrifugal fan power and the fan laws (20 marks)

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. A centrifugal fan discharging into a circular duct, with the duty measured as a static pressure. The air is colder and the barometer lower than standard, so the density must be corrected before any pressure is converted to power.

Given data
Quantity	Symbol	Value
Volumetric flow rate	$Q$	20 000 cfm
Fan speed	$N_1$	2400 rpm
Duct diameter	$D$	32 in.
Fan static pressure	$p_s$	4.8 in. w.g.
Air temperature	$t$	40 °F
Barometric pressure	$p_b$	29.0 in. Hg
Fan total efficiency	$\eta$	70%
Increased fan speed	$N_2$	3200 rpm

Find. (a) the shaft power at 2400 rpm; (b) the shaft power at 3200 rpm with the same fan, density and duct system; and (c) an explanation of how air density affects flow rate, developed head and horsepower.

Fan and discharge duct. The duty against which the fan works is the fan total pressure — the measured static pressure plus the velocity pressure the fan must also generate to move the air along the duct at 3581 ft/min.

Approach. Convert the volumetric flow to a duct velocity, correct the air density for temperature and barometer, add the velocity pressure to the stated static pressure to get the fan total pressure, and apply the air-power relation and the fan efficiency. The speed change is then a straight application of the fan laws.

Duct area and air velocity. The duct is circular, so$$A = \frac{\pi D^2}{4} = \frac{\pi}{4}\left(\frac{32}{12}\right)^2 = 5.5851 \text{ ft}^2, \qquad V = \frac{Q}{A} = \frac{20\,000}{5.5851} = 3581 \text{ ft/min}$$ which is a normal main-duct velocity for an industrial system.
Air density at the stated conditions. From the ideal gas law written for air in these units,$$\rho = 1.325\,\frac{p_b}{T} = 1.325\,\frac{29.0}{459.67 + 40} = 0.07690 \text{ lbm/ft}^3$$ which is $2.5$ % above the standard $0.075$ lbm/ft$^3$ — the cold air more than offsets the low barometer.
Velocity pressure and fan total pressure. The velocity pressure in inches of water is$$p_v = \rho\left(\frac{V}{1096.7}\right)^2 = 0.07690\left(\frac{3581}{1096.7}\right)^2 = 0.820 \text{ in. w.g.}$$ The fan must generate the static pressure and the velocity pressure, so the fan total pressure is $p_t = p_s + p_v = 4.8 + 0.820 = 5.620$ in. w.g.
(a) Shaft power at 2400 rpm. Air power in these units is $Q\,p_t/6356$ horsepower, and the shaft power is that divided by the total efficiency:$$P_{air} = \frac{Q\,p_t}{6356} = \frac{20\,000 \times 5.620}{6356} = 17.68 \text{ hp}$$$$P_{shaft} = \frac{P_{air}}{\eta} = \frac{17.68}{0.70} = \boxed{25.26 \text{ hp}} = 18.84 \text{ kW}$$
(b) Shaft power at 3200 rpm. The fan size, the air density and the duct system are all unchanged, so the fan laws apply directly with the speed ratio $N_2/N_1 = 3200/2400 = 1.3333$:$$Q \propto N: \quad Q_2 = 20000 \times 1.3333 = 26667 \text{ cfm}$$$$p \propto N^2: \quad p_{t,2} = 5.620 \times 1.3333^2 = 9.991 \text{ in. w.g.}$$$$P \propto N^3: \quad P_2 = 25.26 \times 1.3333^3 = \boxed{59.88 \text{ hp}} = 44.65 \text{ kW}$$ A 33%% increase in speed costs 137%% more power — the single most important practical consequence of the cube law, and the reason variable-speed drives save so much energy at part load.
(c) The effect of air density. At a fixed rotational speed a fan is a volumetric machine: its blades sweep the same volume per revolution whatever the gas in them, so the volume flow rate is independent of density and stays at 20 000 cfm whether the air is cold, hot or at altitude. The pressure the fan develops, however, comes from accelerating and then decelerating a mass of gas, so developed head measured as a pressure is directly proportional to density: $p \propto \rho$ at constant speed. Since power is the product of a volume flow and a pressure, horsepower is also directly proportional to density, $P \propto \rho$. Two consequences matter in practice. First, if the same fan handled standard air rather than the $0.07690$ lbm/ft$^3$ air of this problem, the fan total pressure would fall to $5.481$ in. w.g. and the shaft power to $24.64$ hp — so a motor selected on a warm commissioning day can be overloaded on the first cold night, which is why fan motors are sized at the coldest density the system will see. Second, if instead the head is expressed as a column of the gas itself (feet of air rather than inches of water) it is independent of density, which is why fan curves are published at a stated standard density and corrected by a simple ratio.

Check: static versus total pressure. The question states the duty as a static pressure. The answers above use the fan total pressure, $p_t = p_s + p_v = 5.620$ in. w.g., with the total efficiency of 70%, which is the convention in ASHRAE Fundamentals Ch. 21 and in both recommended texts. Had the 70% been read as a static efficiency, the shaft power would be $Q\,p_s/(6356 \times 0.70) = 21.58$ hp and the 3200 rpm answer $51.15$ hp. Either convention is creditable provided it is stated; the total-pressure basis is used here and throughout.

Final results
Quantity	Symbol	Result
Duct cross-sectional area	$A$	5.5851 ft$^2$
Duct velocity	$V$	3581 ft/min
Air density	$\rho$	0.07690 lbm/ft$^3$
Velocity pressure	$p_v$	0.820 in. w.g.
Fan total pressure	$p_t$	5.620 in. w.g.
(a) Shaft power at 2400 rpm	$P_1$	25.26 hp (18.84 kW)
(b) Flow at 3200 rpm	$Q_2$	26667 cfm
(b) Total pressure at 3200 rpm	$p_{t,2}$	9.991 in. w.g.
(b) Shaft power at 3200 rpm	$P_2$	59.88 hp (44.65 kW)

Topic. Fan total pressure, air power and the fan laws.

Key relations. $A = \pi D^2/4$; $V = Q/A$; $\rho = 1.325\,p_b/T$ (lbm/ft$^3$, in. Hg, $^{\circ}$R); $p_v = \rho\,(V/1096.7)^2$; $p_t = p_s + p_v$; $P_{air} = Q\,p_t/6356$ hp; $P_{shaft} = P_{air}/\eta$. Fan laws at constant size and density: $Q \propto N$, $p \propto N^2$, $P \propto N^3$; at constant size and speed, $Q$ is independent of $\rho$ while $p \propto \rho$ and $P \propto \rho$.

Why this works. A fan adds energy to the air in two forms: it raises the pressure against which the ductwork resists, and it leaves the air moving. Only the sum of the two — the fan total pressure — represents the work actually done, which is why the velocity pressure belongs in the power calculation even though the gauge reads only the static component. The constant 6356 converts cfm times inches of water into horsepower. The fan laws then follow from geometric similarity: the volume swept scales with speed, the velocity head with speed squared, and the product of the two with speed cubed.

Common pitfall. Omitting the velocity pressure, which here understates the power by about 15%, or using the standard density 0.075 lbm/ft$^3$ instead of the 0.07690 lbm/ft$^3$ the stated temperature and barometer require, which understates it by a further 2%. A third trap in part (c) is asserting that a fan moves less air in dense cold gas; it moves the same volume and a greater mass.

Source. ASHRAE Handbook – Fundamentals (2021) Ch. 21 (fans, fan laws and duct design); McQuiston, Parker & Spitler, 6th ed., Ch. 12 (fluid flow, pumps and fans).

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