Heat Exchangers
LMTD
Learning Objectives:

Classify the types of heat exchangers with suitable examples.

Analyze heat exchanger performance by using the method of log mean temperature difference (LMTD) as well as NTU method and heat exchanger effectiveness for parallel and counter flow heat exchangers.

Apply fundamentals of heat transfer to understand the design of heat exchangers and to be able to specify the type and size of heat exchanger to satisfy the needs of a particular engineering process application.
The logarithmic mean temperature difference (also known as log mean temperature difference or simply by its initials LMTD) is used to determine the temperature driving force for heat transfer in flow systems, most notably in heat exchangers. The LMTD is a logarithmic average of the temperature difference between the hot and cold feeds at each end of the double pipe exchanger. The larger the LMTD, the more heat is transferred. The use of the LMTD arises straightforwardly from the analysis of a heat exchanger with constant flow rate and fluid thermal properties.
For counterflow in tubular heat exchangers, \[\Delta T_{lm} = \dfrac{(T_{Ho} T_{Ci})  (T_{Hi} T_{Co})}{ln\left ( \dfrac{T_{Ho} T_{Ci}}{T_{Hi} T_{Co}} \right )}\]
For parallel flow heat exchanger, \[\Delta T_{lm} = \dfrac{(T_{Ho} T_{Co})  (T_{Hi} T_{Ci})}{ln\left ( \dfrac{T_{Ho} T_{Co}}{T_{Hi} T_{Ci}} \right )}\]
Heat Exchanger Effectiveness
Learning Objectives:

Classification of heat exchangers according to flow.

To determine heat transfer rate from heat exchanger for parallel and counter flow arrangement.
\[\begin{aligned} \epsilon &= \dfrac{\dot{Q}} {\dot Q_{max}}\\ \epsilon &= \frac{C_H(T_{Hi} T_{Ho})}{C_{min}(T_{Hi} T_{Ci})}\\ \epsilon &= \frac{C_C(T_{Co} T_{Ci})}{C_{min}(T_{Hi} T_{Ci})}\end{aligned}\]
NTU
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 LogMean 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. \[NTU = \dfrac{UA}{C_{min}}\] where,
NTU = number of transfer units
U = the overall heat transfer coefficient
A = is the heat transfer area.
EffectivenessNTU Relation
Learning Objectives:

Develop relations for effectiveness, and analyze heat exchangers when outlet temperatures are not known using the effectivenessNTU method.
\[C_r = \dfrac{C_{min}}{C_{max}}\]
For Parallel Flow Concentric Tube:
\[\epsilon = \dfrac{1  e^{NTU (1 + C_r)}}{1 + C_r}\] \[NTU =  \dfrac{ln\left[ 1 \epsilon (1 + C_r)\right]}{1+ C_r}\]
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: 84403
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: 84404
In heat exchangers, degree of approach is defined as the difference between temperatures of: (TNTRB 2012 ME)
A. Cold water inlet and outlet
B. Hot medium inlet and outlet
C. Hot medium outlet and cold water inlet
D. Hot medium outlet and cold water outlet
Correct Answer: D
Solved Example: 84405
In counter flow heat exchangers:
A. Both the fluids at inlet (of heat exchanger where hot fluid enters) are in their coldest state
B. Both the fluids at inlet are in their hottest state
C. Both the fluids at exit are in their hottest state
D. One fluid is in hottest state and other in coldest state at inlet
Correct Answer: B
Solved Example: 84406
A steam pipe is to be insulated by two insulating materials put over each other. For best results: (ISRO VSSC Tech. Asst. Mech. 2015)
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: 84407
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: 84408
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: 84409
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
Rate of cooling : \[ms \dfrac{dT}{dt}= 3KW P\] \[\int_{10}^{30} dt = \dfrac{(3KW  P)}{ms} \int_0^3 dt\] \[30  10 = \dfrac{(3KW  P) \times 3 \times 3600}{120 \times 4.2 \times 10^3}\] \[3KW  P = \dfrac{20 \times 120 \times 42 }{3 \times 36} \dfrac{2800}{3}\] \[P = 3000  933 = 2067\ W\]Correct Answer: B