Learning Objectives:

• Classify the types of heat exchangers with suitable examples.
• Define and calculate Logarithmic Mean Temperature Difference (LMTD) for parallel and counter flow heat exchangers.

Solved Example: 34325

In a heat exchanger, if the temperature of the hot fluid decreases, what happens to the LMTD?

A. It increases

B. It decreases

C. It remains constant

D. It becomes irrelevant

Correct Answer: B

Solved Example: 34534

In a counterflow heat exchanger, the temperature difference is generally:

A. Greater than in a parallel flow heat exchanger

B. The same as in a parallel flow heat exchanger

C. Less than in a parallel flow heat exchanger

D. Insignificant

Correct Answer: A

Solved Example: 34534

The significance of using a logarithmic mean is that:

A. It reflects the non-linear nature of heat transfer

B. It simplifies the calculation process

C. It provides a constant value

D. It is easier to compute

Correct Answer: A

Solved Example: 345345

The units of LMTD are typically:

A. KJ/kg K

B. Celsius

C. KJ/kg

D. None of the above

Correct Answer: B

Solved Example: 435345

The log mean temperature difference is primarily used to:

A. Calculate the heat transfer rate

B. Determine the efficiency of a heat exchanger

C. Estimate thermal resistance

D. None of the above

Correct Answer: A

Solved Example: 435345

LMTD is defined as:

A. The average temperature difference between two fluids

B. The arithmetic mean of temperature differences

C. The logarithmic mean of temperature differences

D. None of the above

Correct Answer: C

Solved Example: 45345645

What parameters affect the LMTD in a heat exchanger?

A. Flow arrangement

B. Specific heat

C. Overall heat transfer coefficient

D. All of the above

Correct Answer: D

Solved Example: 84-1-02

The flow rates of hot and cold water streams running through a parallel flow heat exchanger are 0.2 Kg/sec and 0.5 Kg/sec respectively. The inlet temperatures on hot and cold sides are 75$^\circ$C and 20$^\circ$C respectively. The exit temperature of hot water is 45$^\circ$C. Calculate LMTD of heat exchanger.

A. 32$^\circ$C

B. 40.5$^\circ$C

C. 29.12$^\circ$C

D. 45.24$^\circ$C

\begin{align*} \mathrm{Heat\ Lost} &= \mathrm{Heat\ gained}\\ \dot{m_c} C_C (T_{c_2} - T_{c_1}) &=\dot{m_h} C_h (T_{h_1} - T_{h_2})\\ (0.5) (T_{c_2} - 20) &= (0.2) (75 - 45)\\ T_{c_2} &= 32 ^\circ C \end{align*} \begin{align*} \theta_1 &= 75 - 20 = 55 ^\circ C\\ \theta_2 &= 45 - 32 = 13 ^\circ C\\ \end{align*} \[LMTD = \dfrac{{{\theta _2} - {\theta _1}}}{{\ln \left( {\dfrac{{{\theta _2}}}{{{\theta _1}}}} \right)}} = \dfrac{{55 - 13}}{{\ln \left( {\dfrac{{55}}{{13}}} \right)}} = 29.11^\circ C\]

Correct Answer: C

Learning Objectives:

• To determine heat transfer rate from heat exchanger for parallel and counter flow arrangement.

\[\begin{aligned} \epsilon &= \dfrac{\dot{Q}} {\dot Q_{max}}\\ \epsilon &= \frac{C_H(T_{Hi}- T_{Ho})}{C_{min}(T_{Hi}- T_{Ci})}\\ \epsilon &= \frac{C_C(T_{Co}- T_{Ci})}{C_{min}(T_{Hi}- T_{Ci})}\end{aligned}\]

Solved Example: 84-2-02

A heat exchanger is used to heat cold water (C$_P$= 4.18 kJ/kgK) entering at 12 $^\circ$C at a rate of 1.2 kg/s by hot air (C$_P$= 1.0 kJ/kgK) at 90$^\circ$C at rate of 2.5 kg/s. The highest rate of heat transfer in the heat exchangers is:

A. 82 kW

B. 195 kW

C. 391 kW

D. None of these

\[Q_{\mathrm{max}} = (\dot{m} C_p)_{\mathrm{min}} \times (\Delta T)_\mathrm{{max}}\] For water: \[(\dot{m} C_p) = 1.2 \times 4.18 = 5.06\ \mathrm{kW/K}\] For Air: \[(\dot{m} C_p) = 2.5 \times 1.005 = 2.51\ \mathrm{kW/K}\] Overall: \[(\dot{m} C_p)_{\mathrm{min}} = \mathrm{min}(5.06, 2.51) = 2.51\ \mathrm{kW/K}\] Max heat transfer: \begin{align*} &=(\dot{m} C_p)_{\mathrm{min}} (T_{hi} - T_{ci})\\ &=2.51 (90 - 12)\\ &=195\ \mathrm{kW} \end{align*}

Correct Answer: B

Solved Example: 84-2-03

Water (C$_p$ = 4.18 kJ/kg.K) at 80$^\circ$C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/s. Air (C$_p$ = 1 kJ/kg.K) enters at 30$^\circ$C with a mass flow rate of 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in $^\circ$C) is

A. 40

B. 20

C. 10

D. 5

\begin{align*} \epsilon &= \dfrac{T_{h_1} - T_{h_2}}{T_{h_1}- T_{c_1}}\\ 0.8 &= \dfrac{80 - T_{h_2}}{80- 30}\\ T_{h_2} &= 40^\circ C \end{align*} \begin{align*} \mathrm{Heat\ Lost} &= \mathrm{Heat\ Gained}\\ \dot{m_h} c_h (T_{h_1} - T_{h_2}) &= \dot{m_c} c_c (T_{c_2} - T_{c_1})\\ (0.5) (4.18) (80 - 40) &= 1 \times 2.09 (T_{c_2} - 30)\\ T_{c_2} &= 40 ^\circ C \end{align*} \begin{align*} \theta_1 &= T_{h_1} - T_{c_2} = 80 - 70 = 10 ^\circ C\\ \theta_2 &= T_{h_2} - T_{c_1} = 40 - 30 = 10 ^\circ C \end{align*} \[LMTD = \dfrac{{{\theta _2} - {\theta _1}}}{{\ln \left( {\dfrac{{{\theta _2}}}{{{\theta _1}}}} \right)}} = \dfrac{{10 - 10}}{{\ln \left( {\dfrac{{10}}{{10}}} \right)}}\]

This is $\dfrac{0}{0}$ case, hence using L'Hospital's rule LMTD = 10 $^\circ$C.

Correct Answer: C

Learning Objectives:

• Use (NTU) Method is used to calculate the rate of heat transfer in heat exchangers (especially counter current exchangers) as compared to LMTD method.

The Number of Transfer Units (NTU) Method is used to calculate the rate of heat transfer in heat exchangers (especially counter current exchangers) when there is insufficient information to calculate the Log-Mean Temperature Difference (LMTD). In heat exchanger analysis, if the fluid inlet and outlet temperatures are specified or can be determined by simple energy balance, the LMTD method can be used; but when these temperatures are not available The NTU or The Effectiveness method is used.

Learning Objectives:

• Develop relations for effectiveness, and analyze heat exchangers when outlet temperatures are not known using the effectiveness-NTU method.

\[C_r = \dfrac{C_{min}}{C_{max}}\]

For Parallel Flow Concentric Tube:

For Counterflow Concentric Tube:

\[\epsilon = \dfrac{1 - e^{-NTU (1 - C_r)}}{1 - C_r e^{-NTU (1 - C_r)}}..............C_r <1\] \[\epsilon = \dfrac{NTU}{1+NTU}........... C_r = 1\] \[NTU = \dfrac{1}{C_r - 1 } ln\left( \dfrac{\epsilon - 1}{\epsilon C_r -1}\right)\] \[NTU = \dfrac{\epsilon}{1-\epsilon}.......... C_r = 1\]

Solved Example: 84-4-03

The concept of overall coefficient of heat transfer is used in heat transfer problems of:

A. Conduction

B. Convection

C. Radiation

D. Conduction and convection

Correct Answer: D

Solved Example: 84-4-06

A steam pipe is to be insulated by two insulating materials put over each other. For best results:

A. Better insulation should be put over pipe and inferior one over it

B. Inferior insulation should be put over pipe and better one over it

C. Both may be put in any order

D. Whether to put inferior OIL over pipe or the better one would depend on steam temperature

Correct Answer: A

Solved Example: 84-4-07

Joule sec is the unit of:

A. Universal gas constant

B. Kinematic viscosity

C. Thermal conductivity

D. Planck's constant

Correct Answer: D

Solved Example: 84-4-08

In regenerator type heat exchanger, heat transfer takes place by:

A. Direct mixing of hot and cold fluids

B. A complete separation between hot and cold fluids

C. Flow of hot and cold fluids alternately over a surface

D. Generation of heat again and again

Correct Answer: C

Solved Example: 84-4-09

A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30$^\circ$C and the entire stored 120 litres of water is initially cooled to 10$^\circ$C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is: (Specific heat of water is 4.2 kJ/kgK and the density of water is 1000 kg/m$^3$)

A. 1600

B. 2067

C. 2533

D. 3933

Rate of heat generated $\dfrac{dQ}{dt}$ = 3KW

Let at any time 't', temperature of cooler = T

Correct Answer: B

Solved Example: 84-4-10

The maximum effectiveness for a parallel flow heat exchanger is:

A. 70%

B. 100%

C. 50%

D. 20%

\[\epsilon = \dfrac{1 - exp(- (1 + R)NTU)}{1 + R}\] As R increases, the effectiveness decreases.
The max. value of R = 1. At this situation, \begin{align*} \epsilon &= \dfrac{1 - exp(- (1 + 1)NTU)}{1 + 1}\\ &= \dfrac{1 - exp(- 2NTU)}{2}\\ \epsilon_{max} &= \dfrac{1}{2} \end{align*}

Correct Answer: C