Free PE Mech — HVAC practice problems

Ten real questions from our PE Mechanical: HVAC and Refrigeration bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

4 with figures10 knowledge areasFull worked solutions

1
medium
A. Sensible & total energy recovery
A sensible recovery wheel preheating the outdoor air for an office AHU draws $4000\,\text{cfm}$ of outdoor air at $T_{oa}=10\,°\text{F}$ and preheats it to $T_{supply}=52\,°\text{F}$ leaving the recovery device. Using $Q_s=1.08\,\text{cfm}\,\Delta T$ , the recovered SENSIBLE heat rate is most nearly:
1. 114000 Btu/h
2. 181000 Btu/h
3. 185000 Btu/h
4. 756000 Btu/h
Show the answer & worked solution
Answer: B.181000 Btu/h
$\Delta T=T_{supply}-T_{oa}=52-10=42\,°\text{F}$ ; $Q_s=1.08\,\text{cfm}\,\Delta T=1.08(4000)(42)=181440$ Btu/h. Distractors: used the LATENT coefficient $0.68$ in place of the sensible $1.08$ ( $114240$ ); used the TOTAL-enthalpy coefficient $4.5$ as if $\Delta T$ were a $\Delta h$ in Btu/lb ( $756000$ ); and used $1.10$ instead of $1.08$ ( $184800$ ).
2
medium
A. Load components & supply airflow
An office AHU cooling a perimeter zone must offset a room SENSIBLE load of $Q_s=60000\,\text{Btu/h}$ . The room is held at $T_{room}=75\,°\text{F}$ and the coil delivers supply air at $T_{supply}=55\,°\text{F}$ (see the AHU schematic). Using $Q_s=1.08\,\dot V\,\Delta T$ , the required supply-air flow is most nearly:
1. 667 cfm
2. 3240 cfm
3. 4410 cfm
4. 2780 cfm
Show the answer & worked solution
Answer: D.2780 cfm
$\Delta T=T_{room}-T_{supply}=75-55=20\,°\text{F}$ ; $\dot V=\dfrac{Q_s}{1.08\,\Delta T}=\dfrac{60000}{1.08(20)}=2778$ cfm. Distractors: used the LATENT coefficient $0.68$ in place of the sensible $1.08$ ( $4412$ ); used the total-enthalpy coefficient $4.5$ ( $667$ ); and MULTIPLIED by $1.08$ instead of dividing ( $1.08\,Q_s/\Delta T=3240$ ).
3
easy
A. LMTD, effectiveness-NTU & coil duty
A chilled-water cooling coil runs in pure COUNTERFLOW. The hot stream enters at $T_{h,in}=110$ °F and leaves at $T_{h,out}=75$ °F; the cold stream enters at $T_{c,in}=45$ °F and leaves at $T_{c,out}=60$ °F. The log-mean temperature difference is most nearly:
1. 40 °F
2. 34.1 °F
3. 50 °F
4. 39.2 °F
Show the answer & worked solution
Answer: D.39.2 °F
Counterflow terminal differences: $\Delta T_1=T_{h,in}-T_{c,out}=110-60=50$ °F and $\Delta T_2=T_{h,out}-T_{c,in}=75-45=30$ °F. $\text{LMTD}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}=\dfrac{50-30}{\ln(50/30)}=39.15$ °F. Distractors: used the ARITHMETIC mean $(\Delta T_1+\Delta T_2)/2=40$ °F (always larger than the log-mean); paired the ends as in PARALLEL flow, $T_{h,in}-T_{c,in}$ and $T_{h,out}-T_{c,out}$ ( $34.1$ °F); and took the LARGER terminal difference $50$ °F alone (no averaging).
4
easy
A. Vapor-compression cycle, COP & capacity
A walk-in cooler operates the R-134a cycle shown (states 1→2→3→4) with $h_1=104$ , $h_2=119$ , $h_3=41.6$ , and $h_4=41.6\,\text{Btu/lb}$ . The coefficient of performance for cooling is most nearly:
1. 0.806
2. 5.16
3. 4.16
4. 0.24
Show the answer & worked solution
Answer: C.4.16
$\text{COP}=\dfrac{h_1-h_4}{h_2-h_1}=\dfrac{104-41.6}{119-104}=\dfrac{62.4}{15}=4.16$ . Distractors: SWAPPED numerator and denominator, $\dfrac{h_2-h_1}{h_1-h_4}=0.24$ (work over effect); used the heating-COP form with the full drop $h_2-h_4$ in the numerator, $\dfrac{h_2-h_4}{h_2-h_1}=5.16$ (one greater than the cooling COP); and divided the effect by the condenser REJECTION instead of the work, $\dfrac{h_1-h_4}{h_2-h_3}=0.81$ .
5
hard
A. Cooling-tower range, approach & water balance
A water-cooled chiller serving an office chiller plant carries $300\,\text{tons}$ of cooling. The condenser rejects about $15000\,\text{Btu/h per ton}$ (refrigeration plus compressor work) to a tower run at a $10\,°\text{F}$ range. Using $Q=500\,\dot V\,(\text{range})$ , the condenser-water flow required is most nearly:
1. 417000 gpm
2. 900 gpm
3. 720 gpm
4. 7500 gpm
Show the answer & worked solution
Answer: B.900 gpm
Heat rejection $=300(15000)=4500000$ Btu/h; $\dot V=\dfrac{Q}{500\,(\text{range})}=\dfrac{4500000}{500(10)}=900\,\text{gpm}$ (≈ $3$ gpm/ton). Distractors: used $12{,}000$ Btu/h per ton (cooling only, omitted the compressor work, $720$ gpm); used $60$ instead of $500$ in $Q=500\,\dot V\,\Delta T$ — i.e. dropped the $8.33\,\text{lb/gal}$ water factor ( $7500$ gpm); and used the AIR coefficient $1.08$ instead of $500$ for water ( $416667$ ).
6
medium
A. Economics & electrical for HVAC
A 460-V chiller compressor motor draws a line current of 95 A at a line-to-line voltage of 460 V with a power factor of 0.88. The real (true) electrical power it draws is most nearly:
1. 115 kW
2. 38.5 kW
3. 75.7 kW
4. 66.6 kW
Show the answer & worked solution
Answer: D.66.6 kW
$P=\dfrac{\sqrt3\,V\,I\,\text{PF}}{1000}=\dfrac{1.732(460)(95)(0.88)}{1000}=66.6$ kW. Distractors: dropped the $\sqrt3$ (used the single-phase form $VI\,\text{PF}/1000=38.5$ ); dropped the power factor, giving the apparent power $\sqrt3\,VI/1000=75.7$ kVA (not real kW); and used $3$ instead of $\sqrt3$ , $3VI\,\text{PF}/1000=115.4$ (over by $\sqrt3$ ).
7
easy
A. Duct sizing, friction & fan pressure
A supply trunk run is $L=150\,\text{ft}$ long and is sized for a friction rate of $0.08\,\text{in. w.g. per }100\text{ ft}$ . The straight-duct friction loss over the run is most nearly:
1. 0.12 in. w.g.
2. 12 in. w.g.
3. 0.08 in. w.g.
4. 0.012 in. w.g.
Show the answer & worked solution
Answer: A.0.12 in. w.g.
$\Delta P_f=\text{rate}\times\dfrac{L}{100}=0.08\times\dfrac{150}{100}=0.12$ in. w.g. Distractors: multiplied by the full length, forgetting the friction rate is per 100 ft ( $0.08\times150=12$ , $100\times$ high); divided by $1000$ instead of $100$ ( $0.012$ ); and reported the friction RATE itself $0.08$ (ignored the run length).
8
easy
A. Combustion, boiler/furnace efficiency & fuel rate
On a natural-gas boiler, a tune-up technician measures an actual air–fuel ratio of $\text{AFR}_{actual}=17.64\,\text{lb air/lb fuel}$ against the stoichiometric value $\text{AFR}_{stoich}=14.7$ . The percent excess air is most nearly:
1. 2.94 %
2. 20 %
3. 83.3 %
4. 120 %
Show the answer & worked solution
Answer: B.20 %
Percent theoretical air $=\dfrac{\text{AFR}_{actual}}{\text{AFR}_{stoich}}\times100=\dfrac{17.64}{14.7}\times100=120\%$ , so percent EXCESS air $=120-100=20\%$ . Distractors: reported the percent THEORETICAL air $120\%$ (forgot to subtract the $100\%$ ); INVERTED the ratio to $\text{AFR}_{stoich}/\text{AFR}_{actual}\times100=83.3\%$ ; and used the absolute AFR DIFFERENCE $\text{AFR}_{actual}-\text{AFR}_{stoich}=2.94$ as if it were already a percentage.
9
easy
A. Valve/damper sizing, sensors & control response
A chilled-water terminal control valve passes $40\,\text{gpm}$ of water with a pressure drop of $\Delta p=4\,\text{psi}$ across the valve. The required valve flow coefficient $C_v$ is most nearly:
1. 3.16 (gpm/psi⁰.5)
2. 20 (gpm/psi⁰.5)
3. 10 (gpm/psi⁰.5)
4. 80 (gpm/psi⁰.5)
Show the answer & worked solution
Answer: B.20 (gpm/psi⁰.5)
$C_v=\dfrac{\text{gpm}}{\sqrt{\Delta p}}=\dfrac{40}{\sqrt{4}}=\dfrac{40}{2}=20$ . Distractors: divided by $\Delta p$ itself instead of $\sqrt{\Delta p}$ ( $=10$ ); MULTIPLIED by $\sqrt{\Delta p}$ instead of dividing ( $=80$ ); and square-rooted the flow as well, $\sqrt{\text{gpm}}/\sqrt{\Delta p}$ ( $=3.16$ ).
10
easy
A. Chillers, heat pumps & thermal storage
A performance test rates a water-cooled centrifugal chiller at 0.56 kW/ton at full load (1 ton of cooling = 3.516 kW). Its coefficient of performance for cooling is most nearly:
1. 1.97
2. 6.28
3. 1.79
4. 21.4
Show the answer & worked solution
Answer: B.6.28
$\text{COP}=\dfrac{3.516}{\text{kW/ton}}=\dfrac{3.516}{0.56}=6.28$ . Distractors: MULTIPLIED by the kW/ton instead of dividing, $3.516\times0.56=1.97$ ; computed the EER $12/(\text{kW/ton})=21.43$ and mislabeled it COP (off by the $3.412$ Btu/Wh factor); and dropped the $3.516$ , using $1/(\text{kW/ton})=1.79$ .

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