22-Mec-B2 Environmental Control in Buildings · May 2014
Question 3 of 8: Fan laws, duty point and throttling
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Professional Engineers of Ontario / Engineers Canada annual examination, 07-Mec-B2 Environmental Control in Buildings, May 2014. Three hours, open book, non-communicating calculator permitted. Eight problems of 20 points each; candidates answer five. All eight are solved here, because the set is a study resource. Psychrometric charts (SI and I-P) and a DuPont HFC-134a pressure-enthalpy diagram are attached to the paper.
Reference texts.
ASHRAE, Handbook — Fundamentals (2021): Ch. 1 Psychrometrics, Ch. 14 Climatic Design Information, Ch. 16 Ventilation and Infiltration, Ch. 18 Nonresidential Cooling and Heating Load Calculations, Ch. 21 Duct Design, Ch. 25–27 Heat, Air and Moisture Control.
F. C. McQuiston, J. D. Parker and J. D. Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley.
E. G. Pita, Air Conditioning Principles and Systems, 4th ed., Prentice Hall.
W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann (fan laws, spray chambers).
Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 9th ed., McGraw-Hill (vapour-compression cycles).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality; ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
National Research Council Canada, National Building Code of Canada and National Energy Code of Canada for Buildings.
Question 3: Fan laws, duty point and throttling (20 marks)
Given. A fan tested at 1000 r/min, to be run at 1500 r/min against a duct and filter in series.
Fan characteristic measured at 1000 r/min
Volume flow rate (m³/s)
0.5
1.0
1.5
2.0
2.5
3.0
Pressure (mm water)
46
52
53
48
37
19
Power consumption (kW)
0.80
1.10
1.40
1.70
2.05
2.60
The duct resistance is 38.1 mm of water at 1.42 m³/s and the filter resistance is 12.7 mm of water at the same flow; both obey a square law.
Find. At 1500 r/min with the filter fitted: the delivered volume flow, the power absorbed, the fan efficiency at that duty, and the extra series resistance needed to cut the flow by 0.47 m³/s.
Approach. Convert the measured characteristic to 1500 r/min with the fan laws, express the duct-plus-filter resistance as a square law, and intersect the two curves. Efficiency follows from air power over shaft power, and the throttling resistance is the vertical gap between the fan curve and the system curve at the reduced flow.
Scale the characteristic with the fan laws. For one fan at two speeds the volume varies as the speed, the pressure as its square and the power as its cube. With $N_2/N_1 = 1500/1000 = 1.5$: $$\dot V_2 = 1.5\,\dot V_1, \qquad \Delta p_2 = 1.5^{2}\,\Delta p_1 = 2.25\,\Delta p_1, \qquad P_2 = 1.5^{3}\,P_1 = 3.375\,P_1$$
Fan characteristic scaled to 1500 r/min
Volume flow rate (m³/s)
0.75
1.50
2.25
3.00
3.75
4.50
Pressure (mm water)
103.5
117.0
119.3
108.0
83.3
42.8
Power (kW)
2.70
3.71
4.73
5.74
6.92
8.78
Build the system characteristic. Resistances in series add at a common flow, so at 1.42 m³/s the combined loss is $38.1 + 12.7 = 50.8$ mm water, and for turbulent flow $\Delta p = k\,\dot V^{2}$ with $$k = \frac{50.8}{1.42^{2}} = 25.19\ \text{mm water per }(\text{m}^{3}/\text{s})^{2}$$
(a) Locate the operating point. The fan delivers what the system will accept, so the duty point is where the two curves cross: $$25.19\,\dot V^{2} = \Delta p_{fan}(\dot V) \;\Rightarrow\; \boxed{\dot V = 2.18\ \text{m}^{3}/\text{s}}, \qquad \Delta p = 119\ \text{mm water}$$reading the scaled characteristic through its tabulated points. In SI pressure units $\Delta p = 119.2 \times 9.807 = 1169$ Pa.
(b) Power absorbed. Reading the scaled power curve at the same flow, $$\boxed{P = 4.62\ \text{kW}}$$a little above the 4.73 kW tabulated at 2.25 m³/s, as expected.
(c) Fan total efficiency. The useful output is the air power, $$P_{air} = \dot V\,\Delta p = 2.18 \times 1169 = 2543\ \text{W} = 2.54\ \text{kW}$$$$\eta = \frac{P_{air}}{P_{shaft}} = \frac{2.54}{4.62} = \boxed{0.55}$$Fifty-five per cent is respectable for a fan of this class and confirms that the duty point sits close to the peak-efficiency region of the curve.
(d) Throttling resistance. Reducing the flow by 0.47 m³/s gives a new duty of $2.18 - 0.47 = 1.71$ m³/s. On the fan curve that flow corresponds to 118.0 mm water, while the existing duct and filter absorb only $$\Delta p_{sys} = 25.19\,(1.71)^{2} = 73.3\ \text{mm water}$$The damper or orifice placed in series must therefore dissipate the difference, $$\Delta p_{extra} = 118.0 - 73.3 = \boxed{44.7\ \text{mm water}}$$equivalent to a resistance coefficient $k_{extra} = 44.7/1.71^{2} = 15.4$ mm water per (m³/s)². The absorbed power falls only to 3.99 kW, so throttling wastes about 0.6 kW — the standard argument for speed control instead of damper control.
Figure 3.1 — Fan characteristics at 1000 and 1500 r/min with the duct-plus-filter system curve. The duty point is 2.18 m³/s at 119 mm water; throttling to 1.71 m³/s requires a damper absorbing 45 mm water, shown as the vertical gap between the fan and system curves.
