Heat Exchanger Analysis MCQ Quiz in తెలుగు - Objective Question with Answer for Heat Exchanger Analysis - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 16, 2025
Latest Heat Exchanger Analysis MCQ Objective Questions
Top Heat Exchanger Analysis MCQ Objective Questions
Heat Exchanger Analysis Question 1:
In a concentric double-pipe heat exchanger where one of the fluids undergoes phase change
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 1 Detailed Solution
Concept:
When one of the fluid in Heat Exchanger is undergoing a change of phase like,
1) The profile for Steam Generator is,
2) The profile for Evaporator is,
In both the cases, the temperature of one fluid remains constant.
So temperature differences interchange at inlet and outlet if changed direction but they do not change in magnitude.
Hence, both counter flow and parallel flow will have same LMTD.
LMTDcounter = LMTDparallel
- For a concentric double pipe heat exchanger where one of the fluids undergo a phase change the direction of flow of the two fluids is of no consequences.
- Because the (LMTD) logarithmic mean temperature difference of the two fluids in both cases i.e. parallel and counter flow are the same so that the heat transfer remains the same in both the cases.
Heat Exchanger Analysis Question 2:
A counter flow heat exchanger uses at steam 120°C. The oil enters the tube at 20°C with mass flow rate of 0.2 kg/s. Tubes are thin wall with diameter of 1.4 cm. and 35 m length & the overall heat transfer coefficient is 310 W/m2K. Take specific heat of oil 2.13 kJ/kg K & enthalpy of vaporization of water at 120°C is 831.235 kJ/kg. Find mass of steam condensed in gram per sec in the process (correct upto 2 decimal places)
Answer (Detailed Solution Below) 34.48 - 34.60
Heat Exchanger Analysis Question 2 Detailed Solution
Concept:
For phase change effectiveness is
ϵ = 1 - e-NTU
Calculation:
Given:
Thi = 120 °C, Tci = 20 °C, U = 310 W/m2K, D = 1.4 cm, L = 35 m, hfg = 831.235 kJ/kg
Oil mass flow rate = 0.2 kg/s, Specific heat of oil = 2.13 kJ/kg K
For phase change effectiveness is
ϵ = 1 - e-NTU
ϵ = 1 - e-1.12 = 0.673
Qactual = 28.703 mJ
Q = m × hfg
m = 34.53 gram/sec
Heat Exchanger Analysis Question 3:
The athematic mean temperature difference for counter-flow heat exchanger when hot water side temperatures are 90°C and 40°C, while cold side temperatures are 20° and 40°C
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 3 Detailed Solution
Concept:
For a heat exchanger with inlet temperature difference ΔT1 and outlet temperature difference ΔT2 ,
Counter-flow heat exchanger:
where θ1 = Th1 - Tc2, and θ2 = Th2 - Tc1
Calculation:
Given:
Th1 = 90°C, Th2 = 40°C, Tc1 = 20°, Tc2 = 40°C.
θ1 = Th1 - Tc2 = 90 - 40 ⇒ 50°C
θ2 = Th2 - Tc1 = 40 - 20 ⇒ 20°C
Additional Information Parallel flow heat exchanger:
where θ1 = Th1 - Tc1, and θ2 = Th2 - Tc2
Heat Exchanger Analysis Question 4:
A cross-flow type air heater has an area of 50 m2. The overall transfer coefficient is 100 W/m2 K; and heat capacity of the stream, be it hot or cold, is 1000 W/K. What is the NTU?
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 4 Detailed Solution
Concept:
NTU is a measure of the effectiveness of the heat exchanger.
The NTU is a measure of the heat transfer size of the exchanger; larger the value of NTU, the closer the heat exchanger approaches its thermodynamic limit.
Cmin is the minimum heat capacity between the hot and cold fluid
Calculation:
Given:
A = 50 m2, U = 100 W/m2 K,
Cmax = Cmin = 1000 W/K
Heat Exchanger Analysis Question 5:
In a certain exchanger, both the fluids have identical mass flow rate and specific heat product. The hot fluid enters at 76°C and leaves 47°C and cold fluid enters at 26°C and leaves at 55°C. The effectiveness of heat exchanger is
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 5 Detailed Solution
Concept:
Heat exchanger effectiveness:
Calculation:
Heat Exchanger Analysis Question 6:
In a counter flow heat exchanger, if the hot and cold fluids enter at T1 and T2, respectively, whereas cold fluid leaves at T3 and hot fluid leaves at T4, then LMTD is given by:
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 6 Detailed Solution
Concept:
For counterflow heat exchanger:
Log mean temperature difference is given by:
Where ΔTi = Thi – Tce
ΔTe = The - Tci
Thi & Tci = Temperature at which hot or and cold water enters respectively
The & Tce = Temperature at which hot air & cold water exits respectively
In the case of the balanced counter flow heat exchanger,
ṁhch = ṁccc = C
ṁh and ṁc are mass flow rates of hot and cold fluid, and ch and cc specific heat capacity of hot and cold fluid.
LMTD for counter-flow heat exchanger, ΔTi = ΔTe
Explanation:
Given:
Tci = T2, Tce = T4, Thi = T1, The = T4
ΔTi = T1 – T3
ΔTe = T4 – T2
Heat Exchanger Analysis Question 7:
Saturated steam at 100°C is condensing on shell side of a shell and tube heat exchanger. The cooling water enters the tube at 30°C and leaves at 70°C. Calculate arithmetic mean temperature difference in counter flow arrangement.
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 7 Detailed Solution
Concept:
- For the phase change heat exchanger, the LMTD of parallel flow and counterflow will be the same. So taking counter-flow heat exchanger:
- Arithmetic mean temperature difference is given by:
where,
Calculation:
Given:
⇒
Heat Exchanger Analysis Question 8:
Two fluid heat exchanger has inlet and outlet temperature of 65°C and 40°C for the hot fluid and 15°C and 30°C for the cold fluid. Find out whether it it counter flow or parallel flow and also calculate the effectiveness of heat exchanger.
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 8 Detailed Solution
Concept:
The heat exchanger effectiveness (ϵ) is defined as the ratio of actual heat transfer to the maximum possible heat transfer.
Qact = ṁhCph (Th1 – Th2) = ṁcĊpc (Tc2 – Tc1)
Fluid Capacity Rate, C : Ch = ṁhCph, Cc = ṁc Cpc
Qmax = Cmin(Th1 – Tc1)
Q = Ch(Th1 – Th2) = Cc(Tc2 – Tc1)
Q = Ch × 25 = Cc × 15
∴ Ch is less than Cc
The formula used will be:
Heat Exchanger Analysis Question 9:
The effectiveness of a counter-flow heat exchanger has been estimated as 0.25. Hot gases enter at 200°C and leave at 75°C. Cooling air enters at 40°C. The temperature of the air leaving the unit will be:
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 9 Detailed Solution
Concept:
Heat exchanger effectiveness:
Calculation:
Given:
ϵ = 0.25, Th1 = 200°C, Th2 = 75°C, Tc1 = 40°C
Heat capacities of gases and air are not given, so for calculating cold fluid outlet temperature, we have to assume the heat capacity of cold air is minimum.
Because then only, the effectiveness formula includes cold air outlet temperature.
Tce - 40 = 40
Tce = 80°C
Heat Exchanger Analysis Question 10:
The overall heat transfer coefficient for a shell and tube heat exchange for clean surfaces is Uo = 400 W/m2K. The fouling factor after one year of operation is found to be
Answer (Detailed Solution Below)
Heat Exchanger Analysis Question 10 Detailed Solution
Explanation:
Overall heat transfer coefficient (OHTC):
- OHTC is defined as the quantity of rate of heat transfer takes place due to any mode per unit area for unit temperature difference.
- It doesn’t have any physical meaning, it’s an experimentally determined coefficient value.
- The concept of OHTC is valid for a steady-state.
- In the case of a heat exchanger, there is a combined mode of heat transfer or multiple resistances present therefore we use,
where R Total is total resistance and U = OHTC
- Unit of ‘U’ is W/m2K
- U indicates the quantity of rate of heat transfer takes place due to any mode per unit area for the unit temperature difference.
Fouling factor indicates the resistance offered due to chemical deposits and scaling.
So,
Calculation:
Given:
Uo = 400 W/m2K,