Free PE Mech — HVAC practice problems

Ten real questions from our PE Mechanical: HVAC and Refrigeration bank, drawn from 10 of its 14 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

4 with figures10 topicsFull worked solutions

1
medium
A. Sensible & total energy recovery
At a site $5{,}000\,\text{ft}$ above sea level, a sensible recovery wheel preheats the outdoor air for an office air handler. The wheel handles $4{,}000\,\text{cfm}$ of outdoor air entering at $20\,°\text{F}$ and an equal $4{,}000\,\text{cfm}$ of exhaust air entering at $72\,°\text{F}$ (both airflows at site conditions), and its sensible effectiveness at this balanced flow is $70\%$ . The recovered sensible heat rate (Btu/h) is most nearly:
Figure — drag or pinch to move around
1. 57,400 Btu/h
2. 134,000 Btu/h
3. 160,000 Btu/h
4. 185,000 Btu/h
Show the answer & worked solution
Answer: B.134,000 Btu/h
With balanced flow the maximum sensible transfer uses the span between the two entering streams (§9.2.7, p. 483): $t_3-t_1=72-20=52\,°\text{F}$ . At $5{,}000$ ft the standard-air sensible factor is $C_s=0.92$ (§9.1.7, p. 454; §9.1.8, p. 456), applied directly to the site airflow, so $q_{s,max}=C_s\,Q\,(t_3-t_1)=0.92(4{,}000)(52)=191{,}360$ Btu/h and $q_s=\varepsilon_s\,q_{s,max}=0.7\times191{,}360=133{,}952$ Btu/h. Distractors: used the sea-level sensible factor $1.10$ at $5{,}000$ ft ( $160{,}000$ ); applied $\varepsilon_s$ to the exhaust temperature alone, $0.7\times72\,°\text{F}$ , dropping the outdoor-air temperature from the span ( $185{,}000$ ); and applied $1-\varepsilon_s=0.3$ , the share not recovered ( $57{,}400$ ).
2
medium
A. LMTD, effectiveness-NTU & coil duty
In a heat-recovery plate heat exchanger, $40\,\text{gpm}$ of process water is cooled from $140$ °F to $65$ °F by $75\,\text{gpm}$ of domestic water entering at $50$ °F. The two streams flow in counterflow, both are water, and the overall heat-transfer coefficient is $U=500\,\text{Btu/(hr-ft}^2\text{-°F)}$ . The required heat-transfer area is most nearly:
Figure — drag or pinch to move around
1. 60 ft²
2. 92.3 ft²
3. 103 ft²
4. 200 ft²
Show the answer & worked solution
Answer: C.103 ft²
Duty from the process water (§9.1.7, standard water): $\dot Q=500\,(40)(140-65)=1{,}500{,}000$ Btu/hr. The domestic water absorbs the same duty, so $T_{Co}=T_{Ci}+\dfrac{\dot Q}{500\,(75)}=50+40=90$ °F. Counterflow terminal differences (§5.5.3): $T_{Hi}-T_{Co}=140-90=50$ °F and $T_{Ho}-T_{Ci}=65-50=15$ °F, so $\Delta T_{lm}=\dfrac{50-15}{\ln(50/15)}=29.07$ °F. With $F=1$ for counterflow (§5.5.1), $A=\dfrac{\dot Q}{U F\,\Delta T_{lm}}=\dfrac{1{,}500{,}000}{500\,(1)(29.07)}=103$ ft². Distractors: used the ARITHMETIC mean of the terminal differences, $(50+15)/2=32.5$ °F, in place of the log-mean ( $92.3$ ft²); assumed the domestic water warms as much as the process water cools, $T_{Co}=50+75=125$ °F, ignoring the different flows, which leaves $15$ °F at both ends ( $200$ ft²); and used the LARGER terminal difference, $50$ °F, as the driving difference ( $60$ ft²).
3
medium
A. Load components & supply airflow
An office AHU cooling a perimeter zone at a sea-level site must offset a total room cooling load of $75{,}000\,\text{Btu/h}$ with a sensible heat ratio of $0.80$ . The room is held at $75\,°\text{F}$ , and the supply air enters the room at $55\,°\text{F}$ (see the AHU schematic). The required supply-air flow (cfm) is most nearly:
Figure — drag or pinch to move around
1. 833 cfm
2. 2730 cfm
3. 3410 cfm
4. 4340 cfm
Show the answer & worked solution
Answer: B.2730 cfm
The supply air is sized on the room SENSIBLE load; the latent part is removed at the coil. With SHR $=$ sensible/total (§7.1), $q_s=\text{SHR}\,q_t=0.80(75{,}000)=60{,}000$ Btu/h. $\Delta t=t_{room}-t_{supply}=75-55=20\,°\text{F}$ , and the §9.1.7 sea-level sensible relation $q_s=1.10\,Q\,\Delta t$ gives $Q=\dfrac{q_s}{1.10\,\Delta t}=\dfrac{60{,}000}{1.10(20)}=2{,}727$ cfm. Distractors: paired the total load with the total-heat factor $4.5$ but a temperature difference; $4.5$ multiplies an enthalpy difference ( $833$ ); sized the air on the TOTAL load, $75{,}000/[1.10(20)]$ , though the latent part is removed at the coil and does not set the airflow ( $3{,}409$ ); and divided by the latent factor per grain, $4840/7000=0.691$ , instead of the sensible $1.10$ ( $4{,}339$ ).
4
medium
A. Vapor-compression cycle, COP & capacity
A food-plant cold room is served by an open-drive compressor operating 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 compressor's electric motor has an efficiency of $0.90$ . The cooling coefficient of performance based on the electrical input to the motor is most nearly:
Figure — drag or pinch to move around
1. 3.74
2. 4.16
3. 6.24
4. 12.8
Show the answer & worked solution
Answer: A.3.74
From the diagram's enthalpies (§4.5.2), the cycle COP is $\text{COP}_{ref}=\dfrac{h_1-h_4}{h_2-h_1}=\dfrac{104-41.6}{119-104}=\dfrac{62.4}{15}=4.16$ , cooling per unit of SHAFT work. The motor needs $W_{el}=W_{shaft}/\eta_m$ (§1.11.1), so on the electrical input $\text{COP}_{el}=\text{COP}_{ref}\,\eta_m=4.16(0.90)=3.74$ . With an open drive the motor loss leaves as heat to the machine room and never reaches the refrigerant, so $h_2$ is unchanged. Distractors: stopped at the cycle COP, $4.16$ , leaving out the motor loss that the electrical input has to cover; left $h_4$ out of the numerator, $\dfrac{h_1}{h_2-h_1}\,\eta_m=6.24$ ; and multiplied the electrical COP by $3.412$ Btu/h per W, $12.77$ , which turns a COP into an EER (§9.6.2), not a dimensionless COP.
5
hard
A. Cooling-tower range, approach & water balance
A water-cooled chiller serving an office building carries $300\,\text{tons}$ of refrigeration at a full-load efficiency of $0.60\,\text{kW/ton}$ . Its cooling tower sends condenser water to the chiller at $85\,°\text{F}$ and receives it back at $95\,°\text{F}$ , with a design wet-bulb of $78\,°\text{F}$ . The condenser-water flow (gpm) is most nearly:
1. 720 gpm
2. 843 gpm
3. 900 gpm
4. 1,200 gpm
Show the answer & worked solution
Answer: B.843 gpm
The tower rejects the refrigeration effect plus the compressor input. Compressor input $=300\times0.60=180\,\text{kW}$ , so $Q_o=300(12{,}000)+180(3{,}412)=3{,}600{,}000+614{,}160=4{,}214{,}160$ Btu/h (1 ton $=12{,}000$ Btu/h and 1 W $=3.412$ Btu/h, §1.1). Range $=95-85=10\,°\text{F}$ , and from $Q_o=500\,(\text{gpm})(T_{ent}-T_{lvg})$ (§9.3.9.2), $\text{gpm}=\dfrac{4{,}214{,}160}{500\times10}=843$ gpm (≈ $2.81$ gpm/ton). Distractors: used the refrigeration load alone, $300\times12{,}000$ Btu/h, leaving out the compressor heat the tower also rejects, $720$ gpm; used the 3 gpm/ton rule of thumb, which rests on the $15{,}000$ Btu/h cooling-tower ton ( $0.88$ kW/ton) and a $10\,°\text{F}$ range, $900$ gpm; and used the approach, $85-78=7\,°\text{F}$ , as the water $\Delta t$ instead of the range, $1{,}200$ gpm.
6
medium
A. Duct sizing, friction & fan pressure
A supply trunk run has $150\,\text{ft}$ of straight duct sized for a friction rate of $0.08\,\text{in. w.g. per }100\text{ ft}$ , plus fittings whose loss coefficients sum to $\Sigma C=2.2$ , all referenced to the duct velocity of $1{,}100\,\text{ft/min}$ . For standard air, the total pressure loss of the run (in. w.g.) is most nearly:
1. 0.12 in. w.g.
2. 0.195 in. w.g.
3. 0.246 in. w.g.
4. 0.286 in. w.g.
Show the answer & worked solution
Answer: D.0.286 in. w.g.
Straight duct: $0.08\times\dfrac{150}{100}=0.12$ in. w.g. Velocity pressure (§9.3.6.1): $p_v=(1{,}100/4{,}005)^2=0.0754$ in. w.g., so the fittings lose $\Sigma C\,p_v=2.2\times0.0754=0.166$ in. w.g. (§9.3.6.4). Section total (§9.3.6.6): $0.12+0.166=0.286$ in. w.g. Distractors: left out the fittings, reporting the straight-duct friction alone ( $0.12$ ); took the friction rate itself as the straight-duct loss, ignoring the $150$ -ft length ( $0.08+0.166=0.246$ ); and added one velocity pressure for the fittings instead of $\Sigma C\cdot p_v$ ( $0.12+0.0754=0.1954$ ).
7
medium
A. Combustion, boiler/furnace efficiency & fuel rate
A packaged boiler in a school heating plant fires $12\,\text{gal/h}$ of No. 2 fuel oil, and the combustion air supplied to its burner is measured at $1{,}590\,\text{lb/h}$ . The percent excess air is most nearly:
1. 18.3 %
2. 20 %
3. 25 %
4. 28.6 %
Show the answer & worked solution
Answer: C.25 %
The §10.3 table of approximate air requirements (p. 549) gives $106$ lb of air per gallon for No. 2 fuel oil, so the theoretical air is $106\times12=1{,}272$ lb/h. Percent excess air (§10.2, p. 548; §10.5.1, p. 553) $=\dfrac{\text{air supplied}-\text{theoretical air}}{\text{theoretical air}}\times100=\dfrac{1{,}590-1{,}272}{1{,}272}\times100=25\%$ . Distractors: read the No. 1 fuel oil row of the table ( $103$ lb of air per gallon) instead of the No. 2 fuel oil row, so the theoretical air came out $1{,}236$ lb/h ( $28.6\%$ ); read the No. 5 fuel oil row of the table ( $112$ lb of air per gallon) instead of the No. 2 fuel oil row, so the theoretical air came out $1{,}344$ lb/h ( $18.3\%$ ); and divided the excess air by the air supplied instead of the theoretical air, $318/1{,}590\times100=20\%$ .
8
medium
A. Valve/damper sizing, sensors & control response
A two-way control valve is being selected for an office air handler's chilled-water coil. At the design flow of $60\,\text{gpm}$ the coil's water-side pressure drop is $12\,\text{ft}$ of water, and the branch piping, fittings and balancing valve add another $4\,\text{ft}$ of water. The valve is to have an authority of $50\%$ at design flow. The required valve flow coefficient $C_v$ is most nearly:
1. 15 (gpm/√psi)
2. 22.8 (gpm/√psi)
3. 26.3 (gpm/√psi)
4. 32.2 (gpm/√psi)
Show the answer & worked solution
Answer: B.22.8 (gpm/√psi)
Authority (§11.3.2, p. 559) is the open-valve drop over the total branch drop, valve included: $\beta=\dfrac{\Delta p_{valve}}{\Delta p_{valve}+\Delta p_{coil+piping}}$ , so $\Delta p_{valve}=\dfrac{\beta}{1-\beta}\,\Delta p_{coil+piping}$ . The coil and piping drop $12+4=16$ ft of water $=16/2.31=6.93$ psi, so $\Delta p_{valve}=\dfrac{0.5}{0.5}\times6.93=6.93$ psi. Then (§11.3.8, p. 560; water, so no specific-gravity correction) $C_v=\dfrac{Q}{\sqrt{\Delta p}}=\dfrac{60}{\sqrt{6.93}}=22.8$ . Distractors: used the feet of water as if they were psi, $\Delta p_{valve}=16$ ( $C_v=15$ ); took the authority against the coil and piping alone, $\Delta p_{valve}=0.5\times6.93=3.46$ psi, leaving the valve out of the total ( $C_v=32.2$ ); and counted only the coil's $12$ ft and dropped the piping, fittings and balancing valve from the branch ( $\Delta p_{valve}=5.19$ psi, $C_v=26.3$ ).
9
medium
A. Chillers, heat pumps & thermal storage
During a field performance test, a water-cooled centrifugal chiller cools $420\,\text{gpm}$ of chilled water from $58\,°\text{F}$ to $44\,°\text{F}$ while drawing $150\,\text{kW}$ of electric power. Its coefficient of performance for cooling is most nearly:
1. 4.1
2. 5.74
3. 6.74
4. 19.6
Show the answer & worked solution
Answer: B.5.74
Cooling delivered (§9.1.7, p. 455): $q=500\,Q_w\,\Delta t=500(420)(58-44)=2{,}940{,}000$ Btu/h. Electric input in the same units ( $1$ W $=3.412$ Btu/h, §1.1, p. 2): $W=150\,\text{kW}\times3{,}412\,\text{Btu/h per kW}=511{,}800$ Btu/h. COP for cooling (§4.5.1, p. 290): $\text{COP}=\dfrac{Q_L}{W}=\dfrac{2{,}940{,}000}{511{,}800}=5.74$ . Distractors: took the load from the 2.4-gpm-per-ton rule of thumb, which assumes a $10\,°\text{F}$ range, instead of the measured $14\,°\text{F}$ ( $420/2.4=175$ tons, COP $4.1$ ); reported the heat-pump COP, adding the work to the heat removed, $(Q_L+W)/W=6.74$ ; and divided Btu/h by watts without the $3.412$ Btu/h-per-watt factor, $2{,}940{,}000/150{,}000=19.6$ , which is the EER, not the COP.
10
medium
A. Hydronic flow, head loss, steam & expansion
An office air handler's chilled-water coil cools $22{,}000\,\text{cfm}$ of standard air from $80\,°\text{F}$ dry bulb with an enthalpy of $31.5\,\text{Btu/lb}$ dry air to $55\,°\text{F}$ dry bulb with an enthalpy of $22.6\,\text{Btu/lb}$ dry air. Chilled water is supplied at $44\,°\text{F}$ and returns at $58\,°\text{F}$ . The required chilled-water flow (gpm) is most nearly:
1. 86.4 gpm
2. 126 gpm
3. 160 gpm
4. 176 gpm
Show the answer & worked solution
Answer: B.126 gpm
The coil both cools and dehumidifies, so the water carries the total load, which the air's enthalpy change captures. Air side (§9.1.7, p. 454): $q_t=4.5\,Q_s\,\Delta h=4.5(22{,}000)(31.5-22.6)=881{,}100$ Btu/h. Water side (§9.1.7, p. 455), with the water's own rise $\Delta t=58-44=14\,°\text{F}$ : $Q_w=\dfrac{q_w}{500\,\Delta t}=\dfrac{881{,}100}{500(14)}=126$ gpm. Distractors: sized the water for the sensible load only, $1.10(22{,}000)(80-55)=605{,}000$ Btu/h, missing the latent heat of the moisture the coil condenses ( $86.4$ gpm); used the leaving-air to supply-water difference ( $11\,°\text{F}$ ) as the water $\Delta t$ ( $160$ gpm); and applied the 2.4-gpm-per-ton rule of thumb, which assumes a $10\,°\text{F}$ rise, to the $73.4$ -ton load instead of the actual $14\,°\text{F}$ ( $176$ gpm).

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