Heat Transfer practice questions

Answer: C) 0.6

• AIncorrect. That's 1 − ε.
• BIncorrect. That's inverted.
• CCorrect. ε = q/q_max.
• DIncorrect. Divide, don't subtract.

Answer: A) Film boiling sets in, and the wire temperature rises sharply

When the heat flux is controlled, going past the critical heat flux causes a vapor film to cover the surface and boiling to shift to film boiling. The surface temperature then jumps sharply, which can cause burnout.

• ACorrect. Past the critical heat flux, a vapor film covers the wire and insulates it. Because the heat flux is fixed, the wire temperature jumps until film boiling and radiation can carry that flux away, and the jump can burn out the wire.
• BIncorrect. Once the critical heat flux is passed, nucleate boiling can't carry the imposed flux. The wire temperature goes up, not down.
• CIncorrect. Natural convection without boiling happens at low excess temperatures, below the start of nucleate boiling, not above the critical heat flux.
• DIncorrect. The boiling heat-transfer coefficient peaks near the critical heat flux. Beyond that point, the vapor blanket makes it drop sharply.

Answer: B) 30 K

• AIncorrect. The formula's limit is the common value.
• BCorrect. When the two ends are equal, LMTD equals that difference.
• CIncorrect. That isn't the mean.
• DIncorrect. That halves it.

Answer: D) 800 W

• AIncorrect. Emissivity below 1 reduces emission.
• BIncorrect. Multiply by ε.
• CIncorrect. That's 1 − ε.
• DCorrect. q = εq_black.

Answer: C) Buoyancy forces to viscous forces

• AIncorrect. That's the Nusselt number.
• BIncorrect. That's the Weber number.
• CCorrect. It plays the role Reynolds number plays in forced flow.
• DIncorrect. That's the Prandtl number.

Answer: D) Cannot exceed the hot-stream outlet temperature

• AIncorrect. It's limited by the hot stream.
• BIncorrect. That violates the second law.
• CIncorrect. That isn't possible with finite area.
• DCorrect. Both streams approach a common temperature.

Answer: A) 400 W/(m²·K)

Fouling resistance adds in series with the clean resistance: 1/U = 0.002 + 0.0005 = 0.0025 m²·K/W, so U = 400 W/(m²·K).

• ACorrect. Resistances in series add: 1/U = 0.002 + 0.0005 = 0.0025 m²·K/W, so U = 400 W/(m²·K).
• BIncorrect. This is the clean coefficient, 1/0.002. Fouling adds resistance, so U must be lower than this.
• CIncorrect. This is 1/0.0005, the reciprocal of the fouling resistance alone. It leaves out the film and wall resistances.
• DIncorrect. This adds the fouling resistance twice: 1/(0.002 + 0.001). The 0.0005 m²·K/W given is already the total fouling resistance.

Answer: B) Emissivity equals absorptivity

• AIncorrect. That isn't the law.
• BCorrect. A good absorber is a good emitter.
• CIncorrect. That's a blackbody.
• DIncorrect. That isn't the law.

Answer: D) Halves the time constant

• AIncorrect. τ varies with 1/h.
• BIncorrect. Better convection speeds the response.
• CIncorrect. τ depends on h.
• DCorrect. τ is inversely proportional to h.

Answer: C) 32.5 °C

With no losses, the hot-side duty equals the cold-side duty: 2(2)(50) = 200 kW = 4(4)ΔT. So ΔT = 12.5 K and the cold outlet is 32.5 °C.

• AIncorrect. This is a 25-K rise, which scales the 50-K drop by only the mass-flow ratio or only the cp ratio. You need the full capacity rates, m·cp: 4 kW/K for the hot stream and 16 kW/K for the cold stream.
• BIncorrect. This assumes the cold stream rises by the same 50 K. That only happens when both streams have equal capacity rates, and here they are 4 and 16 kW/K.
• CCorrect. Q = 2 × 2 × 50 = 200 kW. The cold-stream rise is 200 kW ÷ (4 × 4) kW/K = 12.5 K, so the outlet is 20 + 12.5 = 32.5 °C.
• DIncorrect. This is a 6.25-K rise, which comes from leaving out the hot-stream mass flow rate (2 × 50 = 100 kW, then 100 ÷ 16). The correct duty is 200 kW.