Heat Exchanger MCQ Quiz - Objective Question with Answer for Heat Exchanger - Download Free PDF
Last updated on Jun 11, 2025
Latest Heat Exchanger MCQ Objective Questions
Heat Exchanger Question 1:
Air at a temperature of 15°C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 800°C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 650°C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle, where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/s, the rate of heat transfer to the air in the heat exchanger, by taking the enthalpy of air as h = Cpt, where Cp is the specific heat equal to 1.005 KJ/Kg and t being the temperature, is
Answer (Detailed Solution Below)
Heat Exchanger Question 1 Detailed Solution
Concept:
The rate of heat transfer in a heat exchanger is calculated using enthalpy change when there is no change in kinetic or potential energy. Enthalpy for air is given by:
\( h = C_p \cdot T \)
Calculation:
- Initial temperature: \( T_1 = 15^\circ C = 288~K \)
- Final temperature after heat exchanger: \( T_2 = 800^\circ C = 1073~K \)
- Specific heat: \( C_p = 1.005~kJ/kg·K \)
- Mass flow rate: \( \dot{m} = 2~kg/s \)
Heat added in exchanger:
\( \dot{Q} = \dot{m} \times C_p \times (T_2 - T_1) \)
\( \dot{Q} = 2 \times 1.005 \times (1073 - 288) = 2 \times 1.005 \times 785 = 1580~kJ/s \)
Heat Exchanger Question 2:
In a double pipe counter-flow heat exchanger, 10,000 kg/hr of oil [] is cooled from 80 °C to 50 °C by water [] of flow rate 8000 kg/hr entering at 25 °C. What will be the outlet temperature of water?
Answer (Detailed Solution Below)
Heat Exchanger Question 2 Detailed Solution
Concept:
In a heat exchanger, heat lost by hot fluid = heat gained by cold fluid.
Given:
- Oil: \( m_o = 10000 \, \text{kg/hr}, \, C_{p,o} = 2.09 \, \text{kJ/kg·K}, \, T_{in,o} = 80^\circ C, \, T_{out,o} = 50^\circ C \)
- Water: \( m_w = 8000 \, \text{kg/hr}, \, C_{p,w} = 4.18 \, \text{kJ/kg·K}, \, T_{in,w} = 25^\circ C \)
Step 1: Heat lost by oil
\( Q = m_o \times C_{p,o} \times (T_{in} - T_{out}) = 10000 \times 2.09 \times 30 = 627000 \, \text{kJ/hr} \)
Step 2: Heat gained by water
\( Q = m_w \times C_{p,w} \times (T_{out} - 25) = 8000 \times 4.18 \times (T_{out} - 25) \)
\( 627000 = 33440 \times (T_{out} - 25) \Rightarrow T_{out} - 25 = \frac{627000}{33440} = 18.75 \)
\( \Rightarrow T_{out} = 25 + 18.75 = {43.75^\circ C} \)
Heat Exchanger Question 3:
In a parallel flow heat exchanger, if the heat capacity ratio is 1, then what is the maximum value of effectiveness?
Answer (Detailed Solution Below)
Heat Exchanger Question 3 Detailed Solution
Explanation:
Effectiveness of a Parallel Flow Heat Exchanger
Definition: In heat exchanger analysis, the effectiveness (ε) is a measure of how well a heat exchanger transfers heat relative to its maximum possible heat transfer. It is defined as the ratio of the actual heat transfer to the maximum possible heat transfer that could occur if the outlet temperature of one fluid reached the inlet temperature of the other fluid.
Mathematically, the effectiveness (ε) is expressed as:
ε = Qactual / Qmax
Where:
- Qactual: The actual heat transfer rate in the heat exchanger.
- Qmax: The maximum possible heat transfer rate, which occurs when one of the fluids undergoes the maximum possible temperature change.
For a parallel flow heat exchanger, the two fluids enter the heat exchanger at their respective inlet temperatures and flow in the same direction. The effectiveness of such a heat exchanger depends on the heat capacity ratio (Cr) and the number of transfer units (NTU).
Heat Capacity Ratio (Cr):
The heat capacity ratio is defined as:
Cr = Cmin / Cmax
Where:
- Cmin: The smaller heat capacity rate of the two fluids (C = ṁ × cp).
- Cmax: The larger heat capacity rate of the two fluids.
When the heat capacity ratio (Cr) is equal to 1, it implies that both fluids have the same heat capacity rate.
Effectiveness for Parallel Flow Heat Exchanger:
The effectiveness of a parallel flow heat exchanger is given by the following equation:
ε = [1 - exp(-NTU × (1 + Cr))] / [1 + Cr]
Where:
- NTU: Number of Transfer Units, a dimensionless parameter that represents the size and effectiveness of the heat exchanger.
- Cr: Heat capacity ratio.
Analysis for Cr = 1:
When Cr = 1, the equation for effectiveness simplifies to:
ε = [1 - exp(-2 × NTU)] / 2
The maximum value of effectiveness occurs when NTU approaches infinity (i.e., a very large heat exchanger). In this case:
exp(-2 × NTU) → 0
Therefore, the effectiveness becomes:
ε = [1 - 0] / 2 = 0.5
Conclusion: For a parallel flow heat exchanger with a heat capacity ratio (Cr) of 1, the maximum value of effectiveness is 0.5.
Heat Exchanger Question 4:
The heat exchanger effectiveness is defined as:
Answer (Detailed Solution Below)
Heat Exchanger Question 4 Detailed Solution
Explanation:
The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer.
\(\epsilon = \frac{{{\rm{Actual\;heat\;transfer}}}}{{{\rm{Maximum\;possible\;heat\;transfer}}}} = \frac{Q}{{{Q_{max}}}}\)
\(NTU = \frac{{UA}}{{{C_{min}}}}\)
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.
Capacity Ratio:
\(R = \frac{{{C_{min}}}}{{{C_{max}}}}\)
Effectiveness of a parallel flow heat exchanger:
\({\epsilon_{parallel}} = \frac{{1 - \exp \left[ { - NTU\left( {1 + R} \right)} \right]}}{{1 + R}}\)
Effectiveness of a counterflow heat exchanger:
\({\epsilon_{counter}} = \frac{{1 - \exp \left[ { - NTU\left( {1 - R} \right)} \right]}}{{1 - R\exp \left[ { - NTU\left( {1 - R} \right)} \right]}}\)
Therefore effectiveness is the function of both NTU and Capacity Ratio
Important Point:
For condensor and evaporator (R = 0)
In condenser and evaporators in which one fluid remains at constant temperature throughout the exchanger. Here Cmax = ∞ and thus \(R = \frac{{{C_{min}}}}{{{C_{max}}}} = 0\)
By using the above case we arrive at the following common expression for parallel flow as well as counter – flow heat exchangers.
\(\epsilon = 1 - {\rm{exp}}\left( { - NTU} \right)\)
Heat Exchanger Question 5:
Which of the following statement is incorrect in case of assumption made while deriving LMTD (Logarithmic Mean Temperature Difference) expression?
Answer (Detailed Solution Below)
Heat Exchanger Question 5 Detailed Solution
Explanation:
Logarithmic Mean Temperature Difference (LMTD)
- The Logarithmic Mean Temperature Difference (LMTD) is a logarithmic average of the temperature difference between the hot and cold fluids at each end of the heat exchanger.
- It is used to determine the temperature driving force for heat transfer in flow systems, most notably in heat exchangers.
- LMTD is used in the design and analysis of heat exchangers because it simplifies the heat transfer calculations.
- It assumes that the temperature difference between the hot and cold fluids changes logarithmically along the length of the heat exchanger.
- The LMTD method provides an accurate measure of the average temperature difference when the temperatures of the fluids vary linearly.
Advantages:
- Provides a more accurate representation of the temperature driving force for heat exchangers.
- Useful in the design and analysis of various types of heat exchangers, including shell and tube, plate, and finned tube heat exchangers.
Disadvantages:
- Assumes steady-state conditions, which may not be valid in all practical scenarios.
- Requires knowledge of the inlet and outlet temperatures of both hot and cold fluids, which may not always be available.
Applications: LMTD is widely used in the design, analysis, and performance evaluation of heat exchangers in various industries, including power generation, chemical processing, HVAC, and refrigeration.
Additional InformationLMTD calculation for Parallel flow heat exchanger:
The formula for calculation of LMTD (Log Mean Temperature Difference) for a parallel flow heat exchanger is given by-
\(LMTD_{parallel}=\frac{{{θ _1} - {θ _2}}}{{\ln \left( {\frac{{{θ _1}}}{{{θ _2}}}} \right)}}=\frac{{{θ _2} - {θ _1}}}{{\ln \left( {\frac{{{θ _2}}}{{{θ _1}}}} \right)}}\)
where θ1 = Th1 - Tc1 and θ2 = Th2 - Tc2
LMTD calculation for Counterflow heat exchanger:
\(LMTD_{counter}=\frac{{{θ _1} - {θ _2}}}{{\ln \left( {\frac{{{θ _1}}}{{{θ _2}}}} \right)}}=\frac{{{θ _2} - {θ _1}}}{{\ln \left( {\frac{{{θ _2}}}{{{θ _1}}}} \right)}}\)
where θ1 = Th1 - Tc2 and θ2 = Th2 - Tc1
Top Heat Exchanger MCQ Objective Questions
In a heat exchanger, it is observed that ΔT1 = ΔT2, where ΔT1 is the temperature difference between the two single phase fluid streams at one end and ΔT2 is the temperature difference at the other end. This heat exchanger is
Answer (Detailed Solution Below)
Heat Exchanger Question 6 Detailed Solution
Download Solution PDFExplanation:
In case of the counter-flow heat exchanger when the heat capacities of both the fluids are the same.
i.e. ṁhch = ṁccc
Q = ṁhch(Th1 – Th2) = ṁccc(Tc2 – Tc1)
⇒ (Th1 – Th2) = (Tc2 – Tc1)
⇒ (Th1 – Tc2) = (Th2 – Tc1)
⇒ ΔT1 = ΔT2
For parallel flow heat exchanger, ΔT1 will always be greater than ΔT2.
If one of the two fluids flowing through a heat exchanger of NTU = 2 remains at constant temperature throughout the exchanger length, the effectiveness of the heat exchanger will be
Answer (Detailed Solution Below)
Heat Exchanger Question 7 Detailed Solution
Download Solution PDFExplanation:
Heat Exchanger:
It is the steady flow adiabatic open system in which two flowing fluids exchange or transfer heat between them, without losing or gaining any heat from the ambient.
In the heat exchanger, the rate of enthalpy decrease of hot fluid = The rate of enthalpy increase of cold fluid.
Using the energy balance equation\(\dot{m}_h\times c_{ph}\times{(T_{hi}\;-\;T_{he})}=\dot{m}_c\times c_{pc}\times{(T_{ce}\;-\;T_{ci})}\)
ṁh, ṁc = Mass flow rate of hot and cold fluid.
cph, cpc = Specific heats of hot and cold fluid.
Thi, Tci = Hot and cold fluid inlet temperature, The,Tce = Hot and cold fluid exit temperature.
Capacity rate ratio (C):
\(C = \frac{{{{\left( {\dot m{c_p}} \right)}_{smaller}}}}{{{{\left( {\dot m{c_p}} \right)}_{larger}}}}\), (0 ≤ C ≤ 1)
C = 0, when one of the fluids undergoing phase change like in steam condenser or evaporator or boiler.
Number of transfer unit (NTU)
\(NTU = \frac{{UA}}{{{{\left( {\dot m{c_p}} \right)}_{smaller}}}}\), U = overall heat transfer coefficient, A = heat transfer area
The formula of effectiveness for:
Capacity rate ratio (C) |
Parallel Flow Heat Exchanger |
Counter Flow Heat Exchanger |
0 < C < 1 |
\({\epsilon_{parallel}} = \frac{{1 - {e^{ - \left( {1 + c} \right)NTU}}}}{{1 + c}}\) |
\({\epsilon_{counter}} = \frac{{1 - {e^{ - \left( {1 - c} \right)NTU}}}}{{1 - c{e^{ - \left( {1 - c} \right)NTU}}}}\) |
C = 1 |
\({\epsilon_{parallel}} = \frac{{1 - {e^{ - 2NTU}}}}{2}\) |
\(\epsilon_{counter} = \frac{{NTU}}{{\;1 + NTU}}\) |
C = 0 |
\({\epsilon_{parallel}} = 1 - {e^{ - NTU}}\) |
\({\epsilon_{counter}} = 1 - {e^{ - NTU}}\) |
Calculation:
Given:
NTU = 2 and C = 0 (since phase change)
\({\epsilon_{parallel}} = {\epsilon_{counter}} =1 - {e^{ - NTU}}= 1 - {e^{ - 2}}\)
In a double-pipe heat exchanger, the cold fluid is water with an inlet temperature 20°C and mass flow rate 20 kg/s and the hot fluid water inlet temperature 80°C and mass flow rate 10 kg/s. Assume that for water Cp = 4.2 kJ/kg°C, independent of temperature. What is the maximum temperature to which the cold fluid can be heated in a parallel flow and in a counter flow heat exchanger?
Answer (Detailed Solution Below)
Heat Exchanger Question 8 Detailed Solution
Download Solution PDFFor parallel flow Heat exchanger:
For maximum temperature, the outlet temperature of cold fluid must be equal to the outlet temperature of hot water.
Heat lost by hot water = Heat gained by cold water
\({m_n}{C_{ph}}\left( {{f_{{n_1}}} - {f_{{n_2}}}} \right) = {M_c}{C_{pc}}\left( {{t_{{c_1}}} - {t_{{c_2}}}} \right)\)
\(\because {t_{{n_2}}} = {t_{{c_2}}}\;\& \;{C_{ph}} = {C_{pc}}\)
\(\therefore \;Let\;{t_{{n_2}}} = {t_{{c_2}}} = t\)
\(\therefore 10 \times {C_{ph}} \cdot \left( {80 - t} \right) = 20 \times {C_{pc}} \times \left( {t - 20} \right)\)
80 - t = 2t - 40
⇒ 3t = 120
∴ t = 40°C
For counter flow Heat exchanger:
In Counter flow heat exchanger for maximum temperature, the outlet temperature of hot water is equal to the inlet temperature of cold water
\(\therefore {t_{{n_2}}} = {t_{{c_1}}} = 20^\circ C\)
∴ Heat lost by hot water = Heat gained by cold water
\({m_h}{C_{ph}} \times \left( {{t_{{n_1}}} - {t_{{n_2}}}} \right) = {m_c} \times {C_{pc}} \times \left( {{t_{{c_2}}} - {t_{{c_1}}}} \right)\)
\(10\;\left( {80 - 20} \right) = 20 \times \left( {{t_{{c_2}}} - 20^\circ } \right)\)
\(\frac{{60}}{2} = {t_{{c_2}}} - 20\)
\(\therefore {t_{{c_2}}} = 50^\circ C\)
A heat exchanger is used to heat cold water (CP = 4.18 kJ/kgK) entering at 12 ºC at a rate of 1.2 kg/s by hot air (CP = 1.0 kJ/kgK) at 90ºC at rate of 2.5 kg/s. The highest rate of heat transfer in the heat exchangers is:
Answer (Detailed Solution Below)
Heat Exchanger Question 9 Detailed Solution
Download Solution PDFConcept:
In a heat exchanger, the maximum possible heat transfer is given by:
Qmax = (ṁ × Cp)min × (ΔT)max
And, (ΔT)max = (Thi - Tci )
Thi = Hot air entering temperature, Tci = Cold water entering temperature
Calculation:
Given:
For water, ṁc = 1.2 kg/s, Cp for water = 4.18 kJ/kgK,
for air, ṁh = 2.5 kg/s, Cp = 1.2 kJ/kgK
ṁc × Cc = 1.2 × 4.18 = 5.06 kW/K
ṁh × Ch = 2.5 × 1.005 = 2.5125 kW/K
Cmin = ṁh × Ch
Maximum heat transfer rate = \({\dot{m}}_h~×~C_h~×~(T_{h_i} ~-~T_{c_i})\) = 2.5125 × (90 - 12) = 2.5125 × 78 = 195.975 kW
In counter flow heat exchangers between two fluids
Answer (Detailed Solution Below)
Heat Exchanger Question 10 Detailed Solution
Download Solution PDFExplanation:
In counter-flow heat exchangers, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions.
In counter-flow heat exchangers, the arrangement can be seen graphically.
In a counterflow heat exchanger:
- The fluid entering at one end is in its hottest state (for the hot fluid) or coldest state (for the cold fluid) compared to its state at the exit.
- This leads to a situation where one fluid enters the heat exchanger at its hottest while the opposite fluid enters at its coldest.
- The temperature gradients facilitate heat transfer from the hot to the cold fluid as they flow in opposite directions.
- By the time the fluids reach the exit, the initially hot fluid has transferred some of its heat to the initially cold fluid, cooling down in the process, while the initially cold fluid has warmed up.
- This leads to an effective heat exchange process, often making counterflow heat exchangers more efficient than parallel flow designs where both fluids move in the same direction.
So, at the inlet of a counterflow heat exchanger, we would not find both fluids in their hottest states or both in their coldest states. The design ensures that heat transfer occurs from the hot to the cold fluid effectively, leveraging the temperature difference along the entire length of the heat exchanger
Saturated vapor is condensed to saturated liquid in condenser. The heat capacity ratio is \({C_r} = \frac{{{C_{min}}}}{{{C_{max}}}}\). The effectiveness (ε) of the condenser is
Answer (Detailed Solution Below)
Heat Exchanger Question 11 Detailed Solution
Download Solution PDFConcept:
The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer.
\(\epsilon = \frac{{{\rm{Actual\;heat\;transfer}}}}{{{\rm{Maximum\;possible\;heat\;transfer}}}} = \frac{Q}{{{Q_{max}}}}\)
\(NTU = \frac{{UA}}{{{C_{min}}}}\)
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.
Capacity Ratio:
\(C_r = \frac{{{C_{min}}}}{{{C_{max}}}}\)
Effectiveness of a parallel flow heat exchanger:
\({\epsilon_{parallel}} = \frac{{1 - \exp \left[ { - NTU\left( {1 + C_r} \right)} \right]}}{{1 + C_r}}\)
Effectiveness of a counter-flow heat exchanger:
\({\epsilon_{counter}} = \frac{{1 - \exp \left[ { - NTU\left( {1 - C_r} \right)} \right]}}{{1 ~-~ C_r\exp \left[ { - NTU\left( {1 - C_r} \right)} \right]}}\)
For condenser and evaporator (Cr = 0)
In condenser and evaporators in which one fluid remains at a constant temperature throughout the exchanger. Here Cmax = ∞ and thus \(C_r = \frac{{{C_{min}}}}{{{C_{max}}}} = 0\)
By using the above case, we arrive at the following common expression for parallel flow as well as counter – flow heat exchangers.
\(\epsilon = 1 - {\rm{exp}}\left( { - NTU} \right)\)
Effectiveness (ε) and NTU relation for condenser may be written as
Answer (Detailed Solution Below)
Heat Exchanger Question 12 Detailed Solution
Download Solution PDFExplanation:
The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer.
\(ϵ = \frac{{{\rm{Actual\;heat\;transfer}}}}{{{\rm{Maximum\;possible\;heat\;transfer}}}} = \frac{Q}{{{Q_{max}}}}\)
\(NTU = \frac{{UA}}{{{C_{min}}}}\)
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.
Capacity Ratio:
\(C_r = \frac{{{C_{min}}}}{{{C_{max}}}}\)
Effectiveness of a parallel flow heat exchanger:
\({ϵ_{parallel}} = \frac{{1 - \exp \left[ { - NTU\left( {1 + C_r} \right)} \right]}}{{1 + C_r}}\)
Effectiveness of a counter-flow heat exchanger:
\({ϵ_{counter}} = \frac{{1 - \exp \left[ { - NTU\left( {1 - C_r} \right)} \right]}}{{1 ~-~ C_r\exp \left[ { - NTU\left( {1 - C_r} \right)} \right]}}\)
For condenser and evaporator (Cr = 0)
In condenser and evaporators in which one fluid remains at a constant temperature throughout the exchanger. Here Cmax = ∞ and thus \(C_r = \frac{{{C_{min}}}}{{{C_{max}}}} = 0\)
By using the above case, we arrive at the following common expression for parallel flow as well as counter – flow heat exchangers.
ϵ = 1 - exp(-NTU)
exp(-NTU) = 1 - ϵ
-NTU = ln (1 - ϵ)
NTU = -ln (1 - ϵ)
_______ is not the example of recuperator type heat exchanger.
Answer (Detailed Solution Below)
Heat Exchanger Question 13 Detailed Solution
Download Solution PDFExplanation:
Recuperator type heat exchanger:
- A recuperator is a specialized counter-flow energy recovery heat exchanger that is used to recover waste heat and is installed in the supply and exhaust air streams of an air handling system or in the exhaust gases of an industrial operation.
- In most cases, they are utilized to draw heat from the exhaust and transfer it to the combustion system's incoming air. By using waste energy in this way to heat the air, they are able to balance part of the fuel used and raise the system's overall energy efficiency.
- The recuperator is used to recover, or reclaim, this heat in order to reuse, or recycle, it in a variety of processes that employ combustion to generate heat.
- Recuperators are often used in Gas turbines, Automobile radiators, and condensers.
Which type of heat exchanger an automobile radiator is?
Answer (Detailed Solution Below)
Heat Exchanger Question 14 Detailed Solution
Download Solution PDFExplanation:
Cross-flow:
It is a cross-flow type. As the air flows perpendicular to the water flow in this type of configuration and the same mechanism is occurring in an automobile radiator.
Parallel flow:
When the fluids enter and leave at the same end as it is in concentric tube arrangement then the flow is known as parallel-flow.
Counterflow:
In the counterflow configuration, the fluids enter and leave at the opposite end.
Air can be best heated by steam in a heat exchanger of
Answer (Detailed Solution Below)
Heat Exchanger Question 15 Detailed Solution
Download Solution PDFExplanation:-
Double Pipe Type Heat Exchanger:
- It is the simplest form of a heat exchanger.
- It consists of a tube or pipe fixed concentrically inside a larger pipe or tube.
- It is used when flow rates of fluid are less and heat duty is small.
- For better heat transfer rate, fins are placed as it increases the heat transfer area.
∵ \(\varepsilon \; =\sqrt {\frac{{kp}}{{hA}}} \)
⇒ \(\varepsilon \; \propto \frac{1}{\sqrt{h}}\)
∵ hsteam > hair
∴ air should be heated with double pipe types with fin on the air-side.
Important Points
Plate Type Heat Exchanger:
- They are well suited for liquid-to-liquid heat transfer.
- The hot and cold fluid flow in alternate passages and thus cold fluid stream is surrounded by two hot fluid stream.
- Heat transfer capacity is enhanced by adding more plates in series.
Shell and Tube Type Heat Exchanger:
- It is also suitable for cooling a liquid by another liquid i.e liquid-to-liquid heat transfer.
- Tubes are relatively cheaper to fabricate than the shell body, therefore it is preferred to use corrosive or fouling fluid service on the tube side.
- The viscous fluid is generally kept in the shell side.