# Heat Exchangers Phase Change Phenomenon - Mechanical Engineering (MCQ) questions and answers

1)   Which of the following is/are example/s of direct contact type heat exchanger?

a. jet condenser
b. desuperheater
c. cooling tower
d. all of the above
 Answer  Explanation ANSWER: all of the above Explanation: No explanation is available for this question!

2)   How is the effective surface area of finned tube surface calculated?
Where,
Afin = the area of tube surface on which fines are provided
Aunfinned = the area of tube surface on which fines are not provided
ηfin = fin efficiency

a. A = Aunfinned + Afin
b. A = ηfin (Aunfinned + Afin)
c. A = Aunfinned + (ηfin x Afin)
d. A = (ηfin x Afin)
 Answer  Explanation ANSWER: A = Aunfinned + (ηfin x Afin) Explanation: No explanation is available for this question!

3)   For the same heat transfer Q and same overall heat transfer coefficient Uo, surface area required for cross flow operation is always

a. less than LMTD for parallel flow
b. more than LMTD for parallel flow
c. same as LMTD for parallel flow
d. unpredictable
 Answer  Explanation ANSWER: less than LMTD for parallel flow Explanation: No explanation is available for this question!

4)   In parallel flow heat exchangers,

a. the exit temperature of hot fluid is always equal to the exit temperature of cold fluid
b. the exit temperature of hot fluid is always less than the exit temperature of cold fluid
c. the exit temperature of hot fluid is always more than the exit temperature of cold fluid
d. we cannot predict comparison between exit temperatures of hot fluid and cold fluid
 Answer  Explanation ANSWER: the exit temperature of hot fluid is always more than the exit temperature of cold fluid Explanation: No explanation is available for this question!

5)   For the same heat transfer Q and same overall heat transfer coefficient Uo, surface area required for parallel flow operation is always

a. less than LMTD for counter flow
b. more than LMTD for counter flow
c. same as LMTD for counter flow
d. unpredictable
 Answer  Explanation ANSWER: more than LMTD for counter flow Explanation: No explanation is available for this question!

6)   For the same inlet and exit temperatures of two fluids, the LMTD for counterflow is always

a. smaller than LMTD for parallel flow
b. greater than LMTD for parallel flow
c. same as LMTD for parallel flow
d. unpredictable
 Answer  Explanation ANSWER: greater than LMTD for parallel flow Explanation: No explanation is available for this question!

7)   Which of the following temperature difference is safer than other to consider in designing of heat exchangers?

a. Arithmetic Mean Temperature Difference (ΔTam)
b. Logarithmic Mean Temperature Difference (LMTD)
c. Both have nothing to do with safety
d. Other
 Answer  Explanation ANSWER: Logarithmic Mean Temperature Difference (LMTD) Explanation: No explanation is available for this question!

8)   How can the arithmetic mean temperature difference and LMTD of a same heat exchanger be compared?

a. the arithmetic mean temperature difference is less than LMTD of a same heat exchanger
b. the arithmetic mean temperature difference is more than LMTD of a same heat exchanger
c. the arithmetic mean temperature difference and LMTD of a same heat exchanger are equal
d. none of the above
 Answer  Explanation ANSWER: the arithmetic mean temperature difference is more than LMTD of a same heat exchanger Explanation: No explanation is available for this question!

9)   When is the arithmetic mean temperature difference of heat exchanger used instead of LMTD?

a. when the temperature profiles of two fluids of heat exchanger are sloping downward with curve
b. when the temperature profiles of two fluids of heat exchanger are sloping upward with curve
c. when the temperature profiles of two fluids of heat exchanger are straight
d. none of the above
 Answer  Explanation ANSWER: when the temperature profiles of two fluids of heat exchanger are straight Explanation: No explanation is available for this question!

10)   The arithmetic mean temperature difference for parallel flow heat exchanger is given as

a. ΔTam = (ΔTi– ΔTe)
b. ΔTam = (ΔTi+ ΔTe)
c. ΔTam = (ΔTi– ΔTe) / 2
d. ΔTam = (ΔTi+ ΔTe) / 2
 Answer  Explanation ANSWER: ΔTam = (ΔTi+ ΔTe) / 2 Explanation: No explanation is available for this question!

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