Heat Transfer Operations Questions and Answers – Boilers – Boiling Curve Calculations

This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Boilers – Boiling Curve Calculations”.

1. The milk while boiling in an open vessel spills out because of the ___________ phenomenon.
a) Sub-cooled boiling
b) Super-cooled boiling
c) Sub-heated boiling
d) Super-heated boiling
View Answer

Answer: a
Explanation: This phenomenon of milk spill has been experienced by everyone at some point or the other, and to overcome this problem we reduce the gas flame. This is so because the milk experiences sub-cooled boiling when the bubbles pile up against one another before breaking which increases its volume considerably.

2. What is the relation between the vapour temperature and the liquid temperature on nucleation for a bubble inside the fluid?
a) Tv – Tsat = (2 σ/r – pg) Rv Tv/pv hfg
b) Tv – Tsat = (2 σ/r – pg) Rv Tv2/pv hfg
c) Tv – Tsat = (2 σ/r – pg) Rv Tv2pv hfg
d) Tv – Tsat = (2 σ – pg) Rv Tv2/pv hfg
View Answer

Answer: b
Explanation: The equation for the relation between the vapour temperature and the liquid temperature on nucleation for a bubble inside the fluid is given as-
Tv – Tsat = (2 σ/r – pg) Rv Tv2/pv hfg, which can be easily derived by sufficient energy balance.

3. Spherical bubbles of diameter 3 mm are observed in the bubble boiling phase of water at 1 atmosphere pressure. Assuming pure water vapour is present inside the bubbles, calculate the temperature of that vapour.
Given Surface tension = 73mN/m
R/pgfg = 0.000014
a) 119.387 ℃
b) 200.217 ℃
c) 300.217 ℃
d) 400.217 ℃
View Answer

Answer: a
Explanation: Tv – Tsat = (2 σ/r – pg) Rv Tv2/pv hfg is the required equation, here as we do not have any condensable gas present, pg=0, hence the equation becomes
Tv – Tsat = (2 σ/r) Rv Tv2/pv hfg representing as a quadratic we get, 0.00136 Tv2 – Tv + 100 = 0
Shift+Solve on Scientific calculator gives us 119.4℃

4. Spherical bubbles of diameter 3 mm are observed in the bubble boiling phase of water at 1 atmosphere pressure. Assuming pure wateRvapour is present inside the bubbles, calculate the temperature of that vapour.
Given Surface tension = 73mN/m
R/pgfg = 0.000014
What region of boiling is this ocurring if we consider the two answers we get?
a) Evaporation and Bubble growth
b) Stable film and Evaporation
c) Unstable film and Radiation
d) Nucleate and Stable film
View Answer

Answer: d
Explanation: Tv – Tsat = (2 σ/r – pg) Rv Tv2/pv hfg is the required equation, here as we do not have any condensable gas present, pg=0, hence the equation becomes
Tv – Tsat = (2 σ/r) Rv Tv2/pv hfgrepresenting as a quadratic we get, 0.00136 Tv2 – Tv+ 100 = 0
Shift+Solve on Scientific calculator gives us 119.4℃ and 615.91℃. Hence the latter is Nucleate while the former is Stable film.

5. Which one of the following statement is incorrect?
a) In sub-cool heating, the temperature of the heating surface is less than the normal boiling point of the liquid
b) Nucleate boiling is promoted on a smooth surface than a rough one
c) Film boiling region is usually avoided in commercial equipment
d) The point of transition from nucleate to film boiling is known as the burn-out point on the boiling curve
View Answer

Answer: b
Explanation: A rough surface is certainly better than a smooth surface in terms of heating a fluid as the surface area is more in the latter one.
advertisement
advertisement
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. Consider a situation when evaporation takes place at the liquid-vapour interface in which the heat transfer is mainly due to free convection then the Nusselt Number for the film follows which one of the following relation?
a) Nu = F1 Gr F2 Pr
b) Nu = 4 F1 Gr F2 Pr
c) Nu = 2 F1 Gr F2 Pr
d) Nu = 3 F1 Gr F2 Pr
View Answer

Answer: a
Explanation: When the evaporation takes place on the liq-vap interface, the relation that defines the film coefficient is Nu = F1 Gr F2 Pr, where F1 and F2 are geometry dependent variables and Gr, Pr are Grayshoff and Prandtl Numbers.

7. The heat flux Q/A depends on the specific heat of the liquid Cf as which one of the following given?
a) Q/A= δf hfg [(pf – pg) g/σ]0.55 [Cf dt/hfg p] 3
b) Q/A = δf hfg [(pf – pg) g/σ]0.6 [Cf dt/hfg p Csf]3
c) Q/A = δf hfg [(pf – pg) g/σ]0.5 [Cf dt/hfg p Csf]3
d) Q/A = δf hfg [(pf – pg) g/σ]0.5 [Cf dt/hfg p Csf]
View Answer

Answer: c
Explanation: The correct relation between the heat flux Q/A and the specific heat of the liquid Cf is represented as above. The derivation can be found in J.P. Holman book.
\(\frac{q}{A}=\mu_Lh_{fg}[\frac{g(\rho_L-\rho_v)}{\sigma}]^\frac{1}{2}[\frac{C_{pL}(T_s T_{sat})}{C_{sf}h_{fg}Pr_L^n}]^3\).
advertisement

8. The Nusselt number for nucleate boiling is correctly represented by which equation?
a) Nu=\(\frac{Q}{A}\) Db ⁄ (Ts-Tsat) KL
b) Nu=\(\frac{Q}{A}\) Db(Ts-Tsat) KL
c) Nu=\(\frac{Q}{A}\) Db (Ts-Tsat) ⁄ KL
d) Nu=\(\frac{Q}{A}\) Db KL ⁄ (Ts-Tsat)
View Answer

Answer: a
Explanation: When the boiling is nucleate boiling, the relation that defines the film coefficient is Nu = F1 Gr F2 Pr, where F1 and F2 are geometry dependent variables and Gr, Pr are Grayshoff and Prandtl Numbers. Here Gr=l3 βgΔT ⁄ V2, hence the final answer on substituting is Nu=\(\frac{Q}{A}\) Db ⁄ (Ts-Tsat) KL.

9. What is the term KL in the giver equation of Nusselt number for nucleate boiling heat transfer correlations?
Nub = \(\frac{(\frac{q}{A})D_b}{(T_s-T_{sat})k_L}\)
a) Thermal conductivity of the vapour
b) Thermal conductivity of the liquid
c) Thermal conductivity of the film
d) Thermal conductivity of the Bubble
View Answer

Answer: b
Explanation: The equation of the nusselt number for nucleate boiling is Nu=\(\frac{Q}{A}\) Db ⁄ (Ts-Tsat) KL, where q/A is the total heat flux, Db is the maximum bubble diameter as it leaves the surface, Ts-Tsat is the excess temperature, KL is the thermal conductivity of the liquid and Pr{L} is the Prandtl number of the liquid.
advertisement

10. Why does the heat flux decrease when the temperature increases way above the nucleate boiling point?
a) Steam bubbles no longer break away from the solid surface of the channel
b) Radiation takes over
c) Convection takes over
d) Fouling occurs
View Answer

Answer: a
Explanation: The heat flux decreases because at temperatures very much above natural boiling point, the bubbles no longer break away from the solid surface of the channel and form a film of vapour which is a bad conductor of heat, and hence the flux decreases.

11. Avoiding the Critical heat flux is an engineering problem in heat transfer applications, such as nuclear reactors, where fuel plates must not be allowed to overheat. To cope with this, what measure can be taken?
a) Decrease the pressure
b) Increase the pressure
c) Add more water
d) Add turbulence
View Answer

Answer: b
Explanation: Increasing the pressure increases the boiling point and hence shifts the boiling curve, hence now at this higher pressure CHF is avoided.

Sanfoundry Global Education & Learning Series – Heat Transfer Operations.

To practice all areas of Heat Transfer Operations, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.