PE Mechanical Practice Exam — Free Questions with Worked Solutions

Question 1 of 10

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A wall has an overall heat-transfer coefficient U = 0.10 Btu/h·ft²·°F and an area of 2,000 ft². The indoor-to-outdoor temperature difference is 30°F. The conduction (transmission) heat loss through the wall is most nearly:

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HVAC & Refrigeration · question 1 of 10
A wall has an overall heat-transfer coefficient U = 0.10 Btu/h·ft²·°F and an area of 2,000 ft². The indoor-to-outdoor temperature difference is 30°F. The conduction (transmission) heat loss through the wall is most nearly:
1. A.200 Btu/h
2. B.6,000 Btu/h
3. C.60,000 Btu/h
4. D.600,000 Btu/h
Answer: B — 6,000 Btu/h
Transmission load is q = U·A·ΔT (Handbook §5.2.1):
q = 0.10 × 2,000 × 30 = 6,000 Btu/h
Why the other options are wrong
A.
Omitted the temperature difference (computed U·A only).
C.
Omitted the U-factor (computed A·ΔT only).
D.
Used the R-value (R = 1/U = 10) in place of U: 10 × 2,000 × 30 = 600,000 Btu/h. The equation uses U, not R.
Handbook references
§5.2.1 — Composite Plane Wall (starts p. 305)
Look these up in the official NCEES PE Mechanical Reference Handbook ↗
HVAC & Refrigeration · question 2 of 10
Standard air flows at 2,500 cfm through a round duct 14 in in diameter. The air velocity is most nearly:
1. A.585 fpm
2. B.1,840 fpm
3. C.2,340 fpm
4. D.9,350 fpm
Answer: C — 2,340 fpm
Velocity = flow / area, with A = πd²/4 (d in feet):
d = 14/12 = 1.167 ft, A = π(1.167)²/4 = 1.07 ft²
V = 2,500/1.07 = 2,340 fpm
Why the other options are wrong
A.
Used A = πd² (omitted the /4): 2,500/4.28 = 585.
B.
Used A = d² (treated the round duct as square): 2,500/1.36 = 1,840.
D.
Used the radius in πd²/4 (a quarter of the real area): 2,500/0.27 = 9,350.
Handbook references
§9.3.6 — Duct Design — velocity and area (starts p. 489)
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HVAC & Refrigeration · question 3 of 10
A water-cooled centrifugal chiller has a full-load COP of 5.5. Its full-load efficiency expressed in kW per ton is most nearly:
1. A.0.64
2. B.1.6
3. C.2.2
4. D.5.5
Answer: A — 0.64
One ton is 12,000 Btu/hr = 3.517 kW of cooling. kW/ton = electrical input per ton of output = 3.517/COP = 3.517/5.5 = 0.64 kW/ton. (Good centrifugal machines run ~0.5–0.65 kW/ton at full load — a quick sanity range worth memorizing.)
Why the other options are wrong
B.
Inverted the conversion: 5.5/3.517 = 1.56.
C.
Took 12/5.5 = 2.2 — mixed Btu-based tons with kW without the 3,412 conversion.
D.
Quoted the COP itself.
Handbook references
§4.5.1 — Basic Cycles — COP definitions and 1 ton = 12,000 Btu/hr = 3,516 W (starts p. 289)
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HVAC & Refrigeration · question 4 of 10
Return air at 75°F db / 62.5°F wb (h = 28.1 Btu/lb) mixes with outdoor air at 95°F db / 78°F wb (h = 41.4 Btu/lb) in the ratio 75% return to 25% outdoor. The mixed-air enthalpy (Btu/lb) is most nearly: ____
Enter your response in the blank.
Answer: 31.4 Btu/lb (accepts 30.8–32 Btu/lb)
h_mix = 0.75 × 28.1 + 0.25 × 41.4 = 31.4 Btu/lb. Enthalpy mixes linearly on a mass basis; the ventilation load a system sees is this mixed-air enthalpy relative to the coil leaving condition.
Handbook references
§9.2.3 — Adiabatic Mixing of Two Moist Airstreams (starts p. 481)
§7.3 — Psychrometric Charts (humidity-ratio reads) (starts p. 389)
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HVAC & Refrigeration · question 5 of 10
From the economizer mixing temperatures shown — outdoor air 50°F, return air 74°F, mixed air 66°F — the outdoor-air fraction is most nearly:
1. A.33%
2. B.50%
3. C.67%
4. D.89%
Answer: A — 33%
The mixed temperature is the flow-weighted blend: T_mix = X_OA·T_OA + (1−X_OA)·T_RA.
Solve: X_OA = (T_mix − T_RA)/(T_OA − T_RA) = (66 − 74)/(50 − 74) = 8/24 = 33%
Why the other options are wrong
B.
Assumed an equal 50/50 blend rather than computing from the temperatures.
C.
Computed the return-air fraction (66 − 50)/(74 − 50) = 16/24 = 67% — the complement of the outdoor-air fraction.
D.
Took the ratio of the absolute temperatures, 66/74 = 89%, instead of the ratio of temperature differences.
Handbook references
§9.3.2 — Air-Handling Unit Mixed-Air Plenums (starts p. 487)
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HVAC & Refrigeration · question 6 of 10
Which of the following statements of the pump/fan affinity laws are correct? Select the three that apply.
1. A.Flow varies directly with speed (Q ∝ N)
2. B.Head (or pressure) varies with the square of speed (H ∝ N²)
3. C.Power varies with the cube of speed (P ∝ N³)
4. D.Head varies directly with speed (H ∝ N)
5. E.Flow varies with the square of speed (Q ∝ N²)
Answers: A — Flow varies directly with speed (Q ∝ N); B — Head (or pressure) varies with the square of speed (H ∝ N²); C — Power varies with the cube of speed (P ∝ N³)
The affinity laws: flow ∝ N (A), head ∝ N² (B), power ∝ N³ (C).
Why the other options are wrong
D.
Head goes with the SQUARE of speed, not directly.
E.
Flow goes DIRECTLY with speed, not the square.
Handbook references
§9.3.8.3 — Fan Affinity Laws (starts p. 522)
§3.7.6 — Pump Affinity Laws (starts p. 259)
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HVAC & Refrigeration · question 7 of 10
An energy retrofit costs $28,000 and saves $6,260/yr over 10 years, for which (P/A, 8%, 10) = 6.7101. The retrofit's savings-to-investment ratio (SIR) is most nearly:
1. A.0.67
2. B.1.0
3. C.1.5
4. D.2.2
Answer: C — 1.5
SIR = present value of savings ÷ investment = (6,260 × 6.7101)/28,000 = 42,005/28,000 = 1.50. An SIR above 1.0 means the discounted savings more than repay the investment (equivalent to NPV > 0); federal and institutional energy programs typically fund projects by descending SIR.
Why the other options are wrong
A.
Inverted the ratio: 28,000/42,005 = 0.67.
B.
Assumed break-even — skipped the arithmetic.
D.
Used UNdiscounted savings: 62,600/28,000 = 2.24 — ignores the time value the SIR exists to capture.
Handbook references
§1.6.2 — Economic Factor Tables (i = 8.00%) (starts p. 50)
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HVAC & Refrigeration · question 8 of 10
The cooling tower performance curve shown plots cold-water (leaving) temperature against ambient wet-bulb for the design range and flow. At a 68°F wet-bulb, the tower delivers cold water at most nearly:
1. A.68°F
2. B.73°F
3. C.77°F
4. D.83°F
Answer: C — 77°F
Enter the curve at 68°F wet-bulb and read up to the design-range line: 77°F leaving water — a 9°F approach at this condition. Tower curves are read exactly this way at off-design wet-bulbs; note the approach widens as wet-bulb falls, which is what limits water-side economizer hours.
Why the other options are wrong
A.
Water at the wet-bulb itself would need an infinite tower — zero approach is unreachable.
B.
73°F reads the curve at a 63°F wet-bulb — an axis misread one gridline left.
D.
83°F reads the design point (78°F wb) rather than the asked 68°F condition.
Handbook references
§9.3.9 — Cooling Towers and Fluid Coolers — range and approach (starts p. 524)
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HVAC & Refrigeration · question 9 of 10
The variable-flow chilled-water riser shown serves coils on six floors; the critical (top-floor) coil is 60 ft above the pump. At design flow the critical coil has a pressure drop of 12 ft of water, its two-way control valve is sized for 4 psi, and the branch piping to and from the riser loses 8 ft. The chiller evaporator's pressure drop is 18 ft. Using 0.433 psi per ft of water, the minimum differential-pressure setpoint (psi) measured between the supply and return mains at the critical branch takeoff is most nearly:
1. A.9
2. B.13
3. C.20
4. D.39
Answer: B — 13
The DP sensor at the critical takeoff must guarantee just enough differential to push design flow through everything DOWNSTREAM of that point: branch piping + coil + control valve.
$\Delta P = (12 + 8)\ \text{ft} \times 0.433 + 4\ \text{psi} = 8.66 + 4 = 12.7 \approx \mathbf{13\ psi}$
\Delta P = (12 + 8)\ \text{ft} \times 0.433 + 4\ \text{psi} = 8.66 + 4 = 12.7 \approx \mathbf{13\ psi}
The 60-ft elevation is irrelevant — this is a closed loop, so static height cancels between supply and return. The evaporator drop is upstream of the sensor and is overcome by the pump, not the setpoint.
Why the other options are wrong
A.
Forgot the control valve: $20\ \text{ft} \times 0.433 = 8.66 \rightarrow 9\ \text{psi}$ 20\ \text{ft} \times 0.433 = 8.66 \rightarrow 9\ \text{psi} — the valve must still see its design drop at full flow.
C.
Added the chiller evaporator ( $18 \times 0.433 = 7.8\ \text{psi}$ 18 \times 0.433 = 7.8\ \text{psi}): $12.7 + 7.8 = 20.5$ 12.7 + 7.8 = 20.5 — the evaporator is upstream of the sensor.
D.
Added the 60-ft elevation ( $26\ \text{psi}$ 26\ \text{psi}): static height cancels in a closed loop.
HVAC & Refrigeration · question 10 of 10
An open-plan office of 12,000 ft² houses 90 occupants (245 Btu/h sensible and 155 Btu/h latent each), with lighting at 0.9 W/ft² and plug loads at 0.5 W/ft². The envelope has a UA of 850 Btu/h·°F at a 20°F outdoor-to-indoor design difference. The 900 cfm of ventilation air arrives 20°F above room temperature and is ducted to the air handler, where it mixes with return air ahead of the cooling coil. The space sensible cooling load (Btu/h) is most nearly:
1. A.96,000
2. B.110,000
3. C.116,000
4. D.130,000
Answer: A — 96,000
The question asks for the SPACE sensible load, so two data items are out of scope: the occupants' latent heat (it is not sensible) and the outdoor-air load (the ventilation air is mixed and cooled at the air handler before it reaches the room, so it is a coil load, not a room load).
1. People, sensible only:
$q_{\text{people}} = 90 \times 245 = 22{,}050\ \text{Btu/h}$
q_{\text{people}} = 90 \times 245 = 22{,}050\ \text{Btu/h}
2. Lighting. The Handbook's lighting heat-gain relation $q_{\text{el}} = 3.412\,W F_{\text{ul}} F_{\text{sa}}$ q_{\text{el}} = 3.412\,W F_{\text{ul}} F_{\text{sa}}, with both factors taken as 1.0 and the installed wattage $W$ = 0.9 W/ft² × 12,000 ft²:
$q_{\text{light}} = 12{,}000 \times 0.9 \times 3.412 = 36{,}850\ \text{Btu/h}$
q_{\text{light}} = 12{,}000 \times 0.9 \times 3.412 = 36{,}850\ \text{Btu/h}
$q_{\text{el}} = 3.412\,W F_{\text{ul}} F_{\text{sa}}$ q_{\text{el}} = 3.412\,W F_{\text{ul}} F_{\text{sa}} · §9.1.3 Electric Lighting (starts p. 447)
3. Plug loads, the same watts-to-Btu/h conversion:
$q_{\text{plug}} = 12{,}000 \times 0.5 \times 3.412 = 20{,}472\ \text{Btu/h}$
q_{\text{plug}} = 12{,}000 \times 0.5 \times 3.412 = 20{,}472\ \text{Btu/h}
4. Envelope conduction:
$q_{\text{env}} = UA\,\Delta T = 850 \times 20 = 17{,}000\ \text{Btu/h}$
q_{\text{env}} = UA\,\Delta T = 850 \times 20 = 17{,}000\ \text{Btu/h}
$\dot{Q} = UA(T_{\text{inside}} - T_{\text{outside}})$ \dot{Q} = UA(T_{\text{inside}} - T_{\text{outside}}) · §5.2.1 Composite Plane Wall (starts p. 305)
5. Sum:
$q_{\text{space,s}} = 22{,}050 + 36{,}850 + 20{,}472 + 17{,}000 = 96{,}372 \approx \mathbf{96{,}000\ Btu/h}$
q_{\text{space,s}} = 22{,}050 + 36{,}850 + 20{,}472 + 17{,}000 = 96{,}372 \approx \mathbf{96{,}000\ Btu/h}
Adding the occupants' latent heat (13,950 Btu/h) or the outdoor-air sensible load (19,800 Btu/h) would charge the room with heat it never sees as sensible gain.
Why the other options are wrong
B.
Added the occupants' LATENT heat ( $90 \times 155 = 13{,}950$ 90 \times 155 = 13{,}950) into a sensible total: $110{,}322$ 110{,}322 Btu/h.
C.
Added the outdoor-air sensible load ( $1.10 \times 900 \times 20 = 19{,}800$ 1.10 \times 900 \times 20 = 19{,}800), which is removed at the coil, not in the space: $116{,}172$ 116{,}172 Btu/h.
D.
Added both the occupant latent heat and the outdoor-air load: $130{,}122$ 130{,}122 Btu/h.

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Machine Design & Materials · question 1 of 10
A round rod 1.25 in in diameter carries an axial tensile load of 15,000 lb. The normal stress is most nearly:
1. A.3,060 psi
2. B.9,600 psi
3. C.12,200 psi
4. D.48,900 psi
Answer: C — 12,200 psi
σ = P/A, with A = πd²/4 = π(1.25)²/4 = 1.227 in².
σ = 15,000/1.227 = 12,200 psi
Why the other options are wrong
A.
Used A = πd² (omitted the /4): 15,000/4.91 = 3,060 psi.
B.
Used A = d² (treated the section as a square): 15,000/1.5625 = 9,600 psi.
D.
Used the radius in πd²/4 (a quarter of the real area): 15,000/0.307 = 48,900 psi.
Handbook references
§2.10.5 — Uniaxial Loading — σ = P/A (starts p. 143)
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Machine Design & Materials · question 2 of 10
An electric motor draws 20 kW of electrical power and operates at 88% efficiency. What is the mechanical (shaft) power output of the motor?
1. A.2.4 kW
2. B.17.6 kW
3. C.20 kW
4. D.22.7 kW
Answer: B — 17.6 kW
Motor efficiency is mechanical output over electrical input: η = P_out / P_in. So:
P_out = η × P_in = 0.88 × 20 = 17.6 kW
The output is always less than the input (losses), so multiply by η (< 1).
Why the other options are wrong
A.
Used (1 − η) = 0.12, computing the LOSSES (2.4 kW) rather than the output power.
C.
Reported the input power (20 kW), ignoring the efficiency. Some power is always lost to heat, friction, etc.
D.
Divided by efficiency (20/0.88 = 22.7 kW). Output cannot exceed input for a motor — dividing applies when going from output back to required input.
Handbook references
§2.7.5 — Power and Efficiency (starts p. 128)
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Machine Design & Materials · question 3 of 10
A helical compression spring deflects 0.50 in under a 150-lb axial load. Its spring rate is most nearly:
1. A.0.003 lb/in
2. B.75 lb/in
3. C.300 lb/in
4. D.600 lb/in
Answer: C — 300 lb/in
Spring rate k = F/δ = 150/0.50 = 300 lb/in
Why the other options are wrong
A.
Inverted the ratio (δ/F = 0.50/150 = 0.003).
B.
Multiplied F × δ (150 × 0.50 = 75) instead of dividing.
D.
Used δ/2 in the denominator (150/0.25 = 600).
Handbook references
§2.16.1.2 — Mechanical Springs — rate (starts p. 177)
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Machine Design & Materials · question 4 of 10
A solid round shaft 1.75 in in diameter carries a torque of 8,000 in-lb. Express the maximum torsional shear stress in psi. Enter your response in the blank.
Answer: 7,600 psi (accepts 7,400–7,800 psi)
For a solid shaft: τ = 16T/(πd³) = 16 × 8,000/(π × 1.75³) = 128,000/16.84 = 7,600 psi
Handbook references
§2.11.1 — Torsion — τ = Tr/J for circular solid or thick-walled shafts (starts p. 147)
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Machine Design & Materials · question 5 of 10
A clevis joint's pin carries a 12,000-lbf load in DOUBLE shear (two shear planes). With an allowable shear stress of 15,000 psi, the minimum pin diameter is most nearly:
1. A.0.40 in
2. B.0.63 in
3. C.0.71 in
4. D.1.01 in
Answer: C — 0.71 in
Each of the two shear planes carries half the load, so the required area PER PLANE is
A = (12,000/2)/15,000 = 0.40 in²
Invert the circle:
d = √(4A/π) = √(4 × 0.40/π) = √0.509 = 0.714 ≈ 0.71 in
Why the other options are wrong
A.
Reported the required AREA (0.40 in²) as though it were the diameter.
B.
Dropped the 4/π in inverting the area: d = √0.40 = 0.63 in.
D.
Sized for SINGLE shear (full 12,000 lbf on one plane): A = 0.80 in² → d = 1.01 in, ignoring the clevis's second plane.
Handbook references
§2.18.1.1 — Bolted and Riveted Joints Loaded in Shear (starts p. 205)
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Machine Design & Materials · question 6 of 10
Which of the following statements about fits and tolerances are correct? Select the three that apply.
1. A.A clearance fit always leaves space between the mating parts
2. B.In an interference fit, the shaft is larger than the hole before assembly
3. C.Tolerance is the total permissible variation of a dimension
4. D.A transition fit always guarantees clearance
5. E.Specifying tighter tolerances generally lowers manufacturing cost
Answers: A — A clearance fit always leaves space between the mating parts; B — In an interference fit, the shaft is larger than the hole before assembly; C — Tolerance is the total permissible variation of a dimension
Clearance fit → always a gap (A); interference fit → shaft bigger than hole (B); tolerance = the allowed dimensional band (C).
Why the other options are wrong
D.
A transition fit may result in either slight clearance OR slight interference — it does not guarantee clearance.
E.
Tighter tolerances are harder to hold and RAISE manufacturing cost.
Handbook references
§2.2 — Cylindrical Fits and Tolerances (starts p. 91)
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Machine Design & Materials · question 7 of 10
For the simply supported beam shown, an 800-lb load acts 5 ft from A and a 1,200-lb load acts 15 ft from A (20-ft span). The MAXIMUM support reaction is most nearly:
1. A.900 lb
2. B.1,000 lb
3. C.1,100 lb
4. D.2,000 lb
Answer: C — 1,100 lb
Sum moments about A to get R_B: R_B·20 = 800(5) + 1,200(15) = 4,000 + 18,000 = 22,000 → R_B = 1,100 lb. Then R_A = (800+1,200) − 1,100 = 900 lb. The larger reaction is R_B = 1,100 lb.
Why the other options are wrong
A.
That's R_A (900 lb) — the smaller reaction.
B.
Split the total load evenly (2,000/2 = 1,000); the loads are unequal and off-center.
D.
Reported the total applied load (2,000 lb), not a single reaction.
Handbook references
§2.11.6 — Beams — reactions (starts p. 151)
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Machine Design & Materials · question 8 of 10
A compressed-air system retrofit at a furniture factory costs $46,000 installed. It is expected to save $1,150 per MONTH in electrical energy and $8,000 per YEAR in maintenance. Ignoring the time value of money, the simple payback period is most nearly:
1. A.2.1 yr
2. B.3.3 yr
3. C.5.0 yr
4. D.5.8 yr
Answer: A — 2.1 yr
1. Put both savings on the same annual basis: energy $1{,}150 \times 12 = 13{,}800$ 1{,}150 \times 12 = 13{,}800 $/yr; maintenance 8,000 $/yr. Total = 21,800 $/yr.
2. Simple payback:
$\frac{46{,}000}{21{,}800} = 2.11 \approx \mathbf{2.1\ yr}$
\frac{46{,}000}{21{,}800} = 2.11 \approx \mathbf{2.1\ yr}
The economic-analysis definitions give the simple payback period, which here has no salvage value to subtract · §1.6.1 Nomenclature and Definitions (starts p. 48)
Why the other options are wrong
B.
Counted the energy saving only: $46{,}000/13{,}800 = 3.3$ 46{,}000/13{,}800 = 3.3 yr, leaving the $8{,}000\ \text{USD}$ 8{,}000\ \text{USD} maintenance saving on the table.
C.
Added the monthly and yearly figures without converting: $46{,}000/(1{,}150 + 8{,}000) = 5.0$ 46{,}000/(1{,}150 + 8{,}000) = 5.0 yr — mixed time bases.
D.
Counted the maintenance saving only: $46{,}000/8{,}000 = 5.8$ 46{,}000/8{,}000 = 5.8 yr, ignoring the larger energy stream.
Handbook references
§1.6 — Economic Analysis — simple payback (starts p. 48)
§1.6.1 — Nomenclature and Definitions — Simple Payback Period expression (starts p. 48)
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Machine Design & Materials · question 9 of 10
A clearance hole in a bracket for a fixed-fastener joint is dimensioned 0.500–0.514 in diameter, with a position tolerance of 0.010 in diameter at maximum material condition (MMC) referenced to datums A, B and C. Its true position lies 0.75 in from the nearest part edge. To check the thinnest wall left at that edge, the designer needs the hole's resultant condition: the worst-case boundary the hole surface can reach toward the surrounding material, over every size and position the drawing allows. The resultant condition is most nearly:
1. A.0.490 in
2. B.0.524 in
3. C.0.528 in
4. D.0.538 in
Answer: D — 0.538 in
The hole is an internal feature of size whose position is toleranced at MMC, so bonus tolerance becomes available as the hole departs from MMC. The boundary the stem describes reaches furthest into the surrounding material when the hole is at its largest size (LMC) and is displaced by all the positional tolerance available at that size.
1. Size limits. For a hole, MMC is the smallest size, 0.500 in, and LMC the largest, 0.514 in.
the GD&T definitions explain MMC and LMC as the conditions holding the most and the least material in a feature, with the smallest hole as the MMC case · §1.7.4 Geometric Dimensioning and Tolerancing (GD&T) (starts p. 66)
2. Bonus tolerance grows with departure from MMC and is greatest at LMC:
$\text{bonus} = \text{LMC} - \text{MMC} = 0.514 - 0.500 = 0.014\ \text{in}$
\text{bonus} = \text{LMC} - \text{MMC} = 0.514 - 0.500 = 0.014\ \text{in}
the position row of the GD&T chart pictures a hole's tolerance zone growing from its MMC value to a larger one at LMC · §1.7.4 Geometric Dimensioning and Tolerancing (GD&T) (starts p. 66)
$\text{total positional tolerance at LMC} = 0.010 + 0.014 = 0.024\ \text{in}$
\text{total positional tolerance at LMC} = 0.010 + 0.014 = 0.024\ \text{in}
3. Resultant condition. The Handbook prints only the virtual condition, the inner boundary at MMC (MMC minus the geometric tolerance for an internal feature). The resultant condition is its counterpart at the opposite size limit: LMC plus the total positional tolerance available there.
$\text{RC} = 0.514 + 0.024 = \mathbf{0.538\ in}$
\text{RC} = 0.514 + 0.024 = \mathbf{0.538\ in}
the GD&T definitions give virtual condition in words as the boundary that sets clearance between mating parts, built from the MMC and the related geometric tolerance · §1.7.4 Geometric Dimensioning and Tolerancing (GD&T) (starts p. 66)
4. Use it. The thinnest wall to the edge is the edge distance less the boundary's radius: 0.75 − 0.538/2 = 0.481 in.
The virtual condition, by contrast, is $\text{MMC} - \text{tolerance} = 0.500 - 0.010 = 0.490$ \text{MMC} - \text{tolerance} = 0.500 - 0.010 = 0.490 in: the mating boundary a fastener must clear, not the boundary that sets the wall.
Why the other options are wrong
A.
Computed the virtual condition (inner boundary) instead of the resultant condition: MMC $-$ geometric tolerance $= 0.500 - 0.010 = 0.490\ \text{in.}$ = 0.500 - 0.010 = 0.490\ \text{in.} That boundary is used to check mating clearance for a pin, not the outer material boundary.
B.
Started from LMC but forgot the bonus tolerance, adding only the specified geometric tolerance: $0.514 + 0.010 = 0.524\ \text{in.}$ 0.514 + 0.010 = 0.524\ \text{in.} At MMC modifier the departure from MMC yields additional (bonus) positional tolerance that must be included.
C.
Included the bonus but dropped the specified $0.010\text{-in.}$ 0.010\text{-in.} geometric tolerance: LMC + bonus $= 0.514 + 0.014 = 0.528\ \text{in.}$ = 0.514 + 0.014 = 0.528\ \text{in.} The resultant condition adds BOTH the specified tolerance and the bonus to LMC.
Handbook references
§1.7.4 — Geometric Dimensioning and Tolerancing (GD&T) — Virtual Condition (starts p. 66)
§1.7 — Interpretation of Technical Drawings — MMC/LMC modifiers and positional tolerance zones (starts p. 59)
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Machine Design & Materials · question 10 of 10
A turbine rotor hub is shrunk onto a solid steel shaft. The shaft is 4.000 in in diameter; the hub is 8.000 in in outside diameter and 3.00 in long. Before assembly the shaft diameter exceeds the hub bore diameter by 0.0040 in. Both parts are steel with E = 30,000,000 psi, and the coefficient of friction at the interface is 0.15. The torque the joint can transmit by friction alone is most nearly:
1. A.127,000 lbf·in
2. B.170,000 lbf·in
3. C.254,000 lbf·in
4. D.339,000 lbf·in
Answer: A — 127,000 lbf·in
1. The 0.0040 in is measured on diameters; the interference each mating RADIUS must absorb is $\delta = 0.0040/2 = 0.0020$ \delta = 0.0040/2 = 0.0020 in.
The interference-fit section defines the interference in its contact-pressure equation as the radial interference · §2.11.3 Interference-Fit Stresses (starts p. 149)
2. Contact pressure for a hub of the same material shrunk on a SOLID shaft ( $r = 2.000$ r = 2.000 in, $r_o = 4.000$ r_o = 4.000 in):
$\begin{aligned} p &= \frac{E\delta}{r} \cdot \frac{r_o^2 - r^2}{2r_o^2} = \frac{30{,}000{,}000 \times 0.0020}{2.000} \times \frac{16.0 - 4.0}{2 \times 16.0} \\ &= 30{,}000 \times 0.375 = 11{,}250\ \text{psi} \end{aligned}$
\begin{aligned} p &= \frac{E\delta}{r} \cdot \frac{r_o^2 - r^2}{2r_o^2} = \frac{30{,}000{,}000 \times 0.0020}{2.000} \times \frac{16.0 - 4.0}{2 \times 16.0} \\ &= 30{,}000 \times 0.375 = 11{,}250\ \text{psi} \end{aligned}
$p = \dfrac{E\delta}{2bc^2}\left(c^2 - b^2\right)$ p = \dfrac{E\delta}{2bc^2}\left(c^2 - b^2\right) · §2.11.3 Interference-Fit Stresses (starts p. 149)
3. Friction acts over the whole bore area $2\pi rL$ 2\pi rL at a moment arm $r$ :
$T = f \cdot p \cdot (2\pi rL) \cdot r = 2\pi(0.15)(11{,}250)(2.000^2)(3.00) = \mathbf{127{,}000\ lbf·in}$
T = f \cdot p \cdot (2\pi rL) \cdot r = 2\pi(0.15)(11{,}250)(2.000^2)(3.00) = \mathbf{127{,}000\ lbf·in}
$T = 2\pi b^2 \mu p l$ T = 2\pi b^2 \mu p l · §2.11.3 Interference-Fit Stresses (starts p. 149)
(about 10,600 lbf·ft). Sanity check: the resulting hoop stress at the hub bore is $p(r_o^2 + r^2)/(r_o^2 - r^2) = 18{,}750$ p(r_o^2 + r^2)/(r_o^2 - r^2) = 18{,}750 psi, comfortably below the yield of a steel hub, so the fit is realistic.
$\sigma_{ot} = p\,\dfrac{c^2 + b^2}{c^2 - b^2}$ \sigma_{ot} = p\,\dfrac{c^2 + b^2}{c^2 - b^2} · §2.11.3 Interference-Fit Stresses (starts p. 149)
Why the other options are wrong
B.
Used $p = E\delta/(2r)$ p = E\delta/(2r), the limiting form for a shaft in an infinitely thick hub: $p = 15{,}000$ p = 15{,}000 psi and $T = 170{,}000$ T = 170{,}000 lbf·in. That limit corresponds to a geometry factor of 0.500; an 8.000-in hub on a 4.000-in shaft gives 0.375, so the finite hub OD must be carried through.
C.
Substituted the 0.0040-in DIAMETRAL interference where the radial interference belongs: $p$ doubles to $22{,}500$ 22{,}500 psi and $T = 254{,}000$ T = 254{,}000 lbf·in. Each radius takes half the diametral interference.
D.
Dropped the hub-geometry factor entirely, using $p = E\delta/r = 30{,}000$ p = E\delta/r = 30{,}000 psi and $T = 339{,}000$ T = 339{,}000 lbf·in. The factor $(r_o^2 - r^2)/(2r_o^2)$ (r_o^2 - r^2)/(2r_o^2) never exceeds 0.500 no matter how thick the hub, so omitting it overstates the pressure by at least a factor of two.
Handbook references
§2.11.3 — Interference-Fit Stresses (starts p. 149)
§2.19.1 — Cylindrical Pressure Vessel (starts p. 219)
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Thermal & Fluid Systems · question 1 of 10
Water flows at 0.05 m³/s through a pipe of inside diameter 0.10 m. The average flow velocity is most nearly:
1. A.1.6 m/s
2. B.5.0 m/s
3. C.6.4 m/s
4. D.25 m/s
Answer: C — 6.4 m/s
Velocity is flow divided by cross-sectional area (continuity, Handbook §3.3.1):
A = (π/4)D² = (π/4)(0.10)² = 7.85×10⁻³ m²
v = Q/A = 0.05/7.85×10⁻³ = 6.4 m/s
Why the other options are wrong
A.
Used A = πD² (forgot the ÷4): 0.05/0.0314 = 1.6 m/s.
B.
Used A = D² (dropped π/4 entirely): 0.05/0.01 = 5.0 m/s.
D.
Used the radius (0.05 m) in place of the diameter when computing the area: 25 m/s.
Handbook references
§3.3.1 — Continuity — Q = A·v (starts p. 223)
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Thermal & Fluid Systems · question 2 of 10
How much heat is required to raise 4 kg of water from 25°C to 75°C? Take the specific heat of water as 4.18 kJ/kg·K.
1. A.209 kJ
2. B.418 kJ
3. C.836 kJ
4. D.1,254 kJ
Answer: C — 836 kJ
Q = m·c·ΔT = 4 × 4.18 × (75 − 25) = 836 kJ
Why the other options are wrong
A.
Omitted the mass (4.18 × 50 = 209).
B.
Used the initial temperature 25 as ΔT (4 × 4.18 × 25 = 418).
D.
Used the final temperature 75 as ΔT (4 × 4.18 × 75 = 1,254).
Handbook references
§1.2.10 — Thermal Properties (starts p. 28)
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Thermal & Fluid Systems · question 3 of 10
A power cycle takes in 1,000 kJ of heat per cycle and produces 350 kJ of net work. The heat it rejects per cycle is most nearly:
1. A.350 kJ
2. B.650 kJ
3. C.1,000 kJ
4. D.1,350 kJ
Answer: B — 650 kJ
By the first law on the cycle, heat in equals work out plus heat rejected:
Q_out = Q_in − W_net = 1,000 − 350 = 650 kJ
(The thermal efficiency here is 350/1,000 = 35%.)
Why the other options are wrong
A.
Reported the net work (350 kJ) as the heat rejected.
C.
Reported the heat input (Q_in = 1,000 kJ).
D.
Added the work to the heat input (1,000 + 350) instead of subtracting.
Handbook references
§4.5.1 — Basic Cycles (starts p. 289)
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Thermal & Fluid Systems · question 4 of 10
A heat engine rejects heat to a sink at 300 K. For the engine to reach a thermal efficiency of 45%, the source must be at least at some minimum temperature. Express that minimum source temperature in kelvin. Enter your response in the blank.
Answer: 545 K (accepts 535–556 K)
No engine can beat the Carnot efficiency between the same two reservoirs, so the required efficiency sets a floor on the source temperature.
η = 1 − T_L/T_H → T_H = T_L/(1 − η)
T_H = 300/(1 − 0.45) = 300/0.55 = 545 K
Handbook references
§4.5.1 — Basic Cycles — Carnot efficiency (starts p. 289)
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Thermal & Fluid Systems · question 5 of 10
In a counterflow heat exchanger the hot fluid goes 300°F → 120°F and the cold fluid goes 80°F → 200°F. The log-mean temperature difference is most nearly:
1. A.40°F
2. B.65°F
3. C.70°F
4. D.100°F
Answer: B — 65°F
Counterflow end differences: ΔT₁ = 300 − 200 = 100°F, ΔT₂ = 120 − 80 = 40°F.
LMTD = (100 − 40)/ln(100/40) = 60/0.916 = 65°F
Why the other options are wrong
A.
Used only the small-end difference ΔT₂ = 40°F.
C.
Used the arithmetic mean (100 + 40)/2 = 70°F — the LMTD is lower.
D.
Used only the large-end difference ΔT₁ = 100°F.
Handbook references
§5.5.3 — Log Mean Temperature Difference (starts p. 313)
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Thermal & Fluid Systems · question 6 of 10
Which of the following statements about heat exchangers are correct? Select the three that apply.
1. A.For the same end temperatures, counterflow gives a higher LMTD than parallel flow
2. B.Plate heat exchangers provide a large heat-transfer area in a compact volume
3. C.A correction factor F (< 1) is applied to the LMTD for multipass or crossflow units
4. D.In parallel flow, the hot stream can be cooled below the cold-stream outlet temperature
5. E.Fouling increases the overall heat-transfer coefficient U
Answers: A — For the same end temperatures, counterflow gives a higher LMTD than parallel flow; B — Plate heat exchangers provide a large heat-transfer area in a compact volume; C — A correction factor F (< 1) is applied to the LMTD for multipass or crossflow units
Counterflow is thermodynamically more effective (higher LMTD) (A); plate units are compact/high-area (B); multipass/crossflow needs an F-correction on the LMTD (C).
Why the other options are wrong
D.
In parallel flow the outlet temperatures only approach each other — the hot stream cannot go below the cold outlet.
E.
Fouling adds resistance and LOWERS U.
Handbook references
§5.5 — Heat Exchangers — configurations and LMTD (starts p. 312)
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Thermal & Fluid Systems · question 7 of 10
A centrifugal pump draws 60°F water from the open supply tank shown, with the suction lift and the suction-line friction loss marked on the figure. The site barometer reads 30.0 in. Hg. The net positive suction head available at the pump inlet is most nearly:
1. A.22.4 ft
2. B.23.0 ft
3. C.25.4 ft
4. D.38.4 ft
Answer: A — 22.4 ft
Net positive suction head available is the absolute head at the pump inlet in excess of the head the liquid needs to stay liquid, with every term in feet of the water being pumped. The Handbook's relation is
$\text{NPSH}_A = h_p + h_z - h_{\text{vpa}} - h_f$
\text{NPSH}_A = h_p + h_z - h_{\text{vpa}} - h_f
where $h_p$ h_p is the atmospheric head on the supply surface and $h_z$ h_z is the elevation of that surface above the pump inlet, negative when the surface is below the inlet.
Atmospheric head. The Handbook's unit table gives 1 in of mercury = 0.491 psi, so the barometer stands at $30.0 \times 0.491 = 14.73$ 30.0 \times 0.491 = 14.73 psia. A pressure becomes a head of liquid through $h = P/\gamma$ h = P/\gamma; with P in psi, multiply by 144 in²/ft². The water table gives $\gamma = 62.37$ \gamma = 62.37 lbf/ft³ at 60°F:
$h_p = \frac{14.73 \times 144}{62.37} = 34.01\ \text{ft}$
h_p = \frac{14.73 \times 144}{62.37} = 34.01\ \text{ft}
Vapor-pressure head. The same 60°F row gives a vapor pressure of 0.26 psia, the absolute pressure below which the water would flash. In feet of the same water:
$h_{\text{vpa}} = \frac{0.26 \times 144}{62.37} = 0.60\ \text{ft}$
h_{\text{vpa}} = \frac{0.26 \times 144}{62.37} = 0.60\ \text{ft}
Elevation and friction. The supply surface is 8 ft below the pump inlet (a suction lift), so $h_z = -8$ h_z = -8 ft. The suction friction loss, $h_f = 3$ h_f = 3 ft, is head spent in the suction line before the water reaches the inlet.
$\text{NPSH}_A = 34.01 + (-8) - 0.60 - 3 = \mathbf{22.4\ ft}$
\text{NPSH}_A = 34.01 + (-8) - 0.60 - 3 = \mathbf{22.4\ ft}
Why the other options are wrong
B.
Ignored the vapor-pressure head → 34.01 − 8 − 3 = 23.01 ≈ 23.0 ft.
C.
Ignored the suction friction loss → 34.01 − 8 − 0.60 = 25.41 ≈ 25.4 ft.
D.
Took the suction lift as a positive elevation term, as for a flooded suction → 34.01 + 8 − 3 − 0.60 = 38.41 ≈ 38.4 ft.
Handbook references
§3.7.4 — Centrifugal Pump Characteristics — net positive suction head available (starts p. 255)
§1.1.1 — Units — the printed conversion 1 in of mercury = 0.491 psi (starts p. 2)
§1.2.9 — Properties of Water at Atmospheric Pressure — I-P table, 60°F row (γ = 62.37 lbf/ft³, vapor pressure 0.26 psi) (starts p. 26)
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Thermal & Fluid Systems · question 8 of 10
An instrument-air system installation has seven activities, with durations in weeks and predecessors as follows: A Layout and permits, 2, no predecessor; B Compressor pad, 3, after A; C Header piping, 8, after A; H Procure compressor, 9, no predecessor; D Set compressor, 4, after B and H; E Dryer and receiver, 3, after B; F Tie-in and test, 3, after C, D and E. The critical path and the resulting project duration are:
1. A.A-B-E-F, 11 weeks
2. B.A-B-D-F, 12 weeks
3. C.A-C-F, 13 weeks
4. D.H-D-F, 16 weeks
Answer: D — H-D-F, 16 weeks
Two activities have no predecessor, A and H, so the network has two starting points and not every path begins at A.
Paths and lengths:
A-B-D-F = 2 + 3 + 4 + 3 = 12 wk
A-B-E-F = 2 + 3 + 3 + 3 = 11 wk
A-C-F = 2 + 8 + 3 = 13 wk
H-D-F = 9 + 4 + 3 = 16 wk
The critical path is the longest: H-D-F at 16 weeks. Set compressor (D) waits on both B and H, and H finishes at week 9 against B's week 5, so H governs D's start.
Why the other options are wrong
A.
A-B-E-F, 11 weeks — picked the shortest path through the network. The critical path is the longest-duration path, not the shortest.
B.
A-B-D-F, 12 weeks — traced Set compressor (D) back through Compressor pad (B) only, overlooking that D also waits on Procure compressor (H), which finishes four weeks later than B.
C.
A-C-F, 13 weeks — took the longest path that starts at A, overlooking that H is a second activity with no predecessor and is never reached from A.
Handbook references
§1.8 — Project Planning and Scheduling — critical path (starts p. 70)
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Thermal & Fluid Systems · question 9 of 10
A 2.0-in-diameter water jet discharges freely from a nozzle at 60 ft/s and strikes a stationary curved vane which turns the jet through 135° from its original direction. The vane surface is smooth, so the jet leaves at the same speed at which it arrives, and the jet is surrounded by atmosphere along its whole length. Take the density of water as 1.94 slug/ft³. The magnitude of the resultant force the jet exerts on the vane is most nearly:
1. A.108 lbf
2. B.152 lbf
3. C.260 lbf
4. D.282 lbf
Answer: D — 282 lbf
Because the jet is free, the pressure is atmospheric at both ends of the control volume, so the only forces are the momentum flux terms and the reaction from the vane. Momentum is a vector: the turn produces force components along both axes and they must be combined.
Jet area and flow rate.
A = π(2.0/12)²/4 = 0.02182 ft²
Q = Av = 0.02182(60) = 1.309 ft³/s
Qρv = 1.309(1.94)(60) = 152.4 lbf ← the momentum flux of the jet
Components on the vane. With the incoming jet along +x and a turning angle α = 135°, the leaving velocity components are v cos 135° = −0.7071v and v sin 135° = +0.7071v:
F_x = Qρ(v₁ − v₂ cos α) = 152.4(1 − cos 135°) = 152.4(1.7071) = 260.1 lbf
F_y = Qρ·v₂ sin α = 152.4(0.7071) = 107.7 lbf
Resultant.
F = √(260.1² + 107.7²) = 282 lbf
The closed form is a useful check: for a smooth fixed vane the resultant is always F = 2Qρv·sin(α/2) = 2(152.4)sin 67.5° = 281.5 lbf, and it bisects the angle between the entering and leaving jets. Note that the x component alone, 260 lbf, is 92% of the resultant — close enough to look right, which is exactly why the vector step cannot be skipped.
Why the other options are wrong
A.
Reported only the transverse component, F_y = Qρ·v sin 135° = 152.4(0.7071) = 108 lbf. That is the force perpendicular to the incoming jet, not the resultant; the vane also has to arrest and reverse part of the jet's original momentum, which is the larger of the two components.
B.
Assumed the vane brings the jet to rest, so that F = Qρv = 1.309(1.94)(60) = 152 lbf. That is the flat-plate-normal-to-the-jet result. A vane that turns the flow past 90° sends momentum back the way it came, so the change in momentum — and the force — exceeds the incoming momentum flux.
C.
Reported only the axial component, F_x = Qρ·v(1 − cos 135°) = 152.4(1.7071) = 260 lbf, and stopped without combining it with F_y = 108 lbf. The turn is not symmetric about the x axis, so a transverse force of 108 lbf is genuinely present and the resultant is 282 lbf, not 260 lbf.
Handbook references
§3.3.1 — Continuity Equation — Q = Av (starts p. 223)
§3.5 — Impulse-Momentum Principle — ΣF = ΣQ₂ρ₂v₂ − ΣQ₁ρ₁v₁ (starts p. 249)
§3.5.3.1 — Fixed Blade — −F_x = Qρ(v₂ cos α − v₁), F_y = Qρ(v₂ sin α − 0) (starts p. 250)
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Thermal & Fluid Systems · question 10 of 10
A combined-cycle plant burns fuel in its gas turbine at a rate of 375,000,000 Btu/h. Air enters the gas-turbine compressor at 60°F, and the machine delivers 33% of the fuel energy as shaft work. Its exhaust leaves at 1,010°F at 1,000,000 lbm/h with a mean specific heat of 0.265 Btu/lbm·°F, then passes through a heat-recovery steam generator that discharges the gas to the stack at 290°F. The steam bottoming cycle turns 30% of the heat it takes from that generator into shaft work. The overall thermal efficiency of the plant is most nearly:
1. A.48%
2. B.53%
3. C.58%
4. D.63%
Answer: A — 48%
The two cycles are coupled through one quantity only: the heat the exhaust actually gives up inside the recovery steam generator. That is less than the gas turbine rejects, because the stack is still hot.
Gas-turbine work.
Ẇ_GT = 0.33(375,000,000) = 123,750,000 Btu/h
Exhaust capacity rate.
ṁ·c_p = 1,000,000(0.265) = 265,000 Btu/h·°F
Heat recovered. The generator can only take the exhaust from 1,010°F down to the stack temperature it is designed for:
Q̇_HRSG = 265,000(1,010 − 290) = 190,800,000 Btu/h
Bottoming-cycle work.
Ẇ_ST = 0.30(190,800,000) = 57,240,000 Btu/h
Overall efficiency. Both machines are charged against the one fuel stream:
η_C = (Ẇ_GT + Ẇ_ST)/Q̇_in = (123,750,000 + 57,240,000)/375,000,000 = 180,990,000/375,000,000
η_C = 0.483, i.e. 48%
The gas turbine rejects 375,000,000 − 123,750,000 = 251,250,000 Btu/h, but only 190,800,000 Btu/h of it is above the stack temperature and therefore available to the steam cycle. Roughly 61,000,000 Btu/h leaves up the stack and the bottoming cycle never sees it — which is why the combined efficiency is not the 53% an ideal-recovery estimate would give.
Why the other options are wrong
B.
Cooled the exhaust all the way to the 60°F compressor inlet instead of to the 290°F stack: Q̇ = 265,000(1,010 − 60) = 251,750,000 Btu/h, Ẇ_ST = 75,525,000, and η = (123,750,000 + 75,525,000)/375,000,000 = 53%. That assumes the recovery generator extracts every Btu the gas turbine rejects, which no finite surface can do — the stack temperature in the problem is what fixes the recoverable share.
C.
Computed the two work outputs correctly but charged the plant only for the fuel energy that did not leave up the stack: 375,000,000 − 265,000(290 − 60) = 314,050,000 Btu/h, giving 180,990,000/314,050,000 = 58%. Thermal efficiency is referenced to the energy supplied, and stack loss is a loss of the plant, not a reduction in what was purchased.
D.
Added the two cycle efficiencies, 33% + 30% = 63%. The efficiencies are referenced to different heat inputs — the bottoming cycle's 30% applies only to the 190,800,000 Btu/h it receives, not to the 375,000,000 Btu/h of fuel.
Handbook references
§4.5.4 — Turbines — Combined Cycle; η_C = Ẇ_OUT/Q̇_IN (starts p. 294)
§4.3.3.1 — Special Cases of the Steady-Flow Energy Equation — single-mass-stream heat transfer, h_i + q = h_e (starts p. 286)
§4.5.1 — Basic Cycles — heat-engine thermal efficiency η = W/Q_H (starts p. 289)
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HVAC & Refrigeration · question 1 of 10

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HVAC & Refrigeration · question 3 of 10

HVAC & Refrigeration · question 4 of 10

HVAC & Refrigeration · question 5 of 10

HVAC & Refrigeration · question 6 of 10

HVAC & Refrigeration · question 7 of 10

HVAC & Refrigeration · question 8 of 10

HVAC & Refrigeration · question 9 of 10

HVAC & Refrigeration · question 10 of 10

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Machine Design & Materials · question 1 of 10

Machine Design & Materials · question 2 of 10

Machine Design & Materials · question 3 of 10

Machine Design & Materials · question 4 of 10

Machine Design & Materials · question 5 of 10

Machine Design & Materials · question 6 of 10

Machine Design & Materials · question 7 of 10

Machine Design & Materials · question 8 of 10

Machine Design & Materials · question 9 of 10

Machine Design & Materials · question 10 of 10

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Thermal & Fluid Systems · question 1 of 10

Thermal & Fluid Systems · question 2 of 10

Thermal & Fluid Systems · question 3 of 10

Thermal & Fluid Systems · question 4 of 10

Thermal & Fluid Systems · question 5 of 10

Thermal & Fluid Systems · question 6 of 10

Thermal & Fluid Systems · question 7 of 10

Thermal & Fluid Systems · question 8 of 10

Thermal & Fluid Systems · question 9 of 10

Thermal & Fluid Systems · question 10 of 10