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Heat exchangers are vital components in many mechanical engineering systems, enabling efficient transfer of thermal energy between two fluids at different temperatures without mixing them. Whether in power plants, refrigeration units, chemical industries, or automotive radiators, understanding how to analyze and design heat exchangers is essential for engineers.
Two primary methods are used to evaluate heat exchanger performance: the Log Mean Temperature Difference (LMTD) method and the Number of Transfer Units (NTU) - effectiveness method. Both approaches help calculate the rate of heat transfer and predict outlet temperatures, but they differ in assumptions and applicability.
This section will build your understanding from fundamental heat transfer principles to detailed analysis methods, supported by practical examples and tips tailored for competitive exams and real-world engineering challenges.
A heat exchanger is a device designed to transfer heat between two or more fluids at different temperatures without mixing them. The fluids may be liquids or gases, and the heat transfer occurs through a solid barrier or direct contact.
There are three common flow arrangements in heat exchangers:
Heat transfer in heat exchangers occurs mainly through conduction and convection. The hot fluid transfers heat to the exchanger surface by convection, heat conducts through the wall, and then convection transfers heat to the cold fluid.
Understanding these mechanisms helps in selecting materials, designing surfaces, and predicting performance.
The Log Mean Temperature Difference (LMTD) is a way to find an average temperature difference between the hot and cold fluids along the length of a heat exchanger. Since the temperature difference varies along the exchanger, simply using the arithmetic mean would be inaccurate.
Consider a heat exchanger where the temperature difference at one end is \( \Delta T_1 \) and at the other end is \( \Delta T_2 \). The LMTD is defined as:
This formula assumes steady-state operation, no heat loss to the surroundings, and constant fluid properties.
The LMTD represents the effective temperature difference that drives heat transfer across the exchanger. It is crucial for calculating the heat transfer rate using the formula:
When outlet temperatures are unknown, or fluid properties vary significantly, LMTD method becomes difficult to apply directly. This leads us to the NTU-effectiveness method.
The NTU-effectiveness method is a powerful alternative to LMTD analysis, especially when outlet temperatures are unknown or difficult to measure.
Effectiveness (\( \varepsilon \)) is defined as the ratio of actual heat transfer rate \( Q \) to the maximum possible heat transfer rate \( Q_{max} \):
The maximum possible heat transfer occurs when the fluid with the minimum heat capacity rate is heated or cooled to the inlet temperature of the other fluid:
The heat capacity rate \( C \) of a fluid stream is the product of its mass flow rate \( \dot{m} \) and specific heat capacity at constant pressure \( c_p \):
The Number of Transfer Units (NTU) is a dimensionless parameter representing the size of the heat exchanger relative to the fluid capacity:
The effectiveness depends on NTU, the heat capacity ratio \( C_r = \frac{C_{min}}{C_{max}} \), and the flow arrangement. Different correlations or charts exist for parallel flow, counterflow, and crossflow heat exchangers.
Using NTU and effectiveness, one can find the actual heat transfer rate without knowing outlet temperatures initially. This is especially useful in design and simulation.
Hot water enters a counterflow heat exchanger at 90°C and leaves at 60°C. Cold water enters at 30°C and leaves at 50°C. The overall heat transfer coefficient \( U \) is 500 W/m²·K, and the heat transfer area \( A \) is 5 m². Calculate the heat transfer rate \( Q \) using the LMTD method.
Step 1: Identify temperature differences at each end:
\( \Delta T_1 = T_{h,in} - T_{c,out} = 90 - 50 = 40^\circ C \)
\( \Delta T_2 = T_{h,out} - T_{c,in} = 60 - 30 = 30^\circ C \)
Step 2: Calculate LMTD:
\[ \Delta T_{lm} = \frac{40 - 30}{\ln\left(\frac{40}{30}\right)} = \frac{10}{\ln(1.333)} = \frac{10}{0.28768} = 34.78^\circ C \]
Step 3: Calculate heat transfer rate:
\[ Q = U A \Delta T_{lm} = 500 \times 5 \times 34.78 = 86,950 \text{ W} = 86.95 \text{ kW} \]
Answer: The heat transfer rate is approximately 86.95 kW.
A heat exchanger has hot water entering at 90°C with a mass flow rate of 2 kg/s and specific heat \( c_p = 4200 \) J/kg·K. Cold water enters at 30°C with mass flow rate 3 kg/s and same \( c_p \). The overall heat transfer coefficient is 400 W/m²·K, and the heat transfer area is 4 m². Calculate the heat exchanger effectiveness and heat transfer rate.
Step 1: Calculate heat capacity rates:
\( C_h = \dot{m}_h c_p = 2 \times 4200 = 8400 \text{ W/K} \)
\( C_c = \dot{m}_c c_p = 3 \times 4200 = 12600 \text{ W/K} \)
Step 2: Identify \( C_{min} \) and \( C_{max} \):
\( C_{min} = 8400 \text{ W/K} \), \( C_{max} = 12600 \text{ W/K} \)
Step 3: Calculate NTU:
\[ NTU = \frac{U A}{C_{min}} = \frac{400 \times 4}{8400} = \frac{1600}{8400} = 0.1905 \]
Step 4: Calculate capacity rate ratio \( C_r \):
\[ C_r = \frac{C_{min}}{C_{max}} = \frac{8400}{12600} = 0.6667 \]
Step 5: For a counterflow heat exchanger, effectiveness is given by:
\[ \varepsilon = \frac{1 - \exp[-NTU (1 - C_r)]}{1 - C_r \exp[-NTU (1 - C_r)]} \]
Calculate exponent:
\( -NTU (1 - C_r) = -0.1905 \times (1 - 0.6667) = -0.1905 \times 0.3333 = -0.0635 \)
Calculate numerator:
\( 1 - e^{-0.0635} = 1 - 0.9385 = 0.0615 \)
Calculate denominator:
\( 1 - 0.6667 \times 0.9385 = 1 - 0.6256 = 0.3744 \)
Effectiveness:
\( \varepsilon = \frac{0.0615}{0.3744} = 0.1643 \)
Step 6: Calculate maximum heat transfer rate:
\[ Q_{max} = C_{min} (T_{h,in} - T_{c,in}) = 8400 \times (90 - 30) = 8400 \times 60 = 504,000 \text{ W} \]
Step 7: Calculate actual heat transfer rate:
\[ Q = \varepsilon Q_{max} = 0.1643 \times 504,000 = 82,800 \text{ W} = 82.8 \text{ kW} \]
Answer: The heat exchanger effectiveness is 0.1643, and the heat transfer rate is approximately 82.8 kW.
A counterflow heat exchanger has hot fluid entering at 150°C and leaving at 100°C, and cold fluid entering at 30°C with unknown outlet temperature. The mass flow rates and specific heats are:
The overall heat transfer coefficient \( U = 600 \) W/m²·K and heat transfer area \( A = 6 \) m². Calculate the heat transfer rate using both LMTD and NTU methods and compare results.
Step 1: Calculate heat capacity rates:
\( C_h = 1.5 \times 4200 = 6300 \text{ W/K} \)
\( C_c = 2.0 \times 4200 = 8400 \text{ W/K} \)
Step 2: Find outlet temperature of cold fluid using energy balance (for LMTD):
Heat lost by hot fluid:
\( Q = C_h (T_{h,in} - T_{h,out}) = 6300 \times (150 - 100) = 315,000 \text{ W} \)
Assuming no losses, heat gained by cold fluid:
\( Q = C_c (T_{c,out} - T_{c,in}) \Rightarrow T_{c,out} = \frac{Q}{C_c} + T_{c,in} = \frac{315,000}{8400} + 30 = 37.5 + 30 = 67.5^\circ C \)
Step 3: Calculate temperature differences for LMTD:
\( \Delta T_1 = T_{h,in} - T_{c,out} = 150 - 67.5 = 82.5^\circ C \)
\( \Delta T_2 = T_{h,out} - T_{c,in} = 100 - 30 = 70^\circ C \)
Step 4: Calculate LMTD:
\[ \Delta T_{lm} = \frac{82.5 - 70}{\ln\left(\frac{82.5}{70}\right)} = \frac{12.5}{\ln(1.1786)} = \frac{12.5}{0.1643} = 76.07^\circ C \]
Step 5: Calculate heat transfer rate using LMTD:
\[ Q = U A \Delta T_{lm} = 600 \times 6 \times 76.07 = 273,852 \text{ W} = 273.85 \text{ kW} \]
Step 6: Calculate NTU and effectiveness:
\( C_{min} = 6300 \), \( C_{max} = 8400 \)
\[ NTU = \frac{U A}{C_{min}} = \frac{600 \times 6}{6300} = \frac{3600}{6300} = 0.5714 \]
\[ C_r = \frac{6300}{8400} = 0.75 \]
Effectiveness for counterflow:
\[ \varepsilon = \frac{1 - \exp[-NTU (1 - C_r)]}{1 - C_r \exp[-NTU (1 - C_r)]} \]
Calculate exponent:
\( -NTU (1 - C_r) = -0.5714 \times (1 - 0.75) = -0.5714 \times 0.25 = -0.1429 \)
Numerator:
\( 1 - e^{-0.1429} = 1 - 0.867 = 0.133 \)
Denominator:
\( 1 - 0.75 \times 0.867 = 1 - 0.650 = 0.35 \)
Effectiveness:
\( \varepsilon = \frac{0.133}{0.35} = 0.38 \)
Step 7: Calculate maximum heat transfer rate:
\[ Q_{max} = C_{min} (T_{h,in} - T_{c,in}) = 6300 \times (150 - 30) = 6300 \times 120 = 756,000 \text{ W} \]
Step 8: Calculate actual heat transfer rate:
\[ Q = \varepsilon Q_{max} = 0.38 \times 756,000 = 287,280 \text{ W} = 287.28 \text{ kW} \]
Comparison: LMTD method gave 273.85 kW, NTU method gave 287.28 kW. The small difference arises due to assumptions and rounding.
Answer: Both methods give comparable heat transfer rates, validating their use. NTU method is preferred when outlet temperatures are unknown.
A heat exchanger with \( U = 500 \) W/m²·K and \( A = 4 \) m² handles fluids with \( C_h = 6000 \) W/K and \( C_c = 4000 \) W/K. Calculate and compare the effectiveness for parallel flow and counterflow arrangements using NTU method.
Step 1: Calculate \( C_{min} \) and \( C_{max} \):
\( C_{min} = 4000 \), \( C_{max} = 6000 \)
Step 2: Calculate NTU:
\[ NTU = \frac{U A}{C_{min}} = \frac{500 \times 4}{4000} = \frac{2000}{4000} = 0.5 \]
Step 3: Calculate capacity rate ratio:
\[ C_r = \frac{4000}{6000} = 0.6667 \]
Step 4: Effectiveness for parallel flow:
\[ \varepsilon_{parallel} = \frac{1 - \exp[-NTU (1 + C_r)]}{1 + C_r} = \frac{1 - \exp[-0.5 \times (1 + 0.6667)]}{1 + 0.6667} \]
Calculate exponent:
\( -0.5 \times 1.6667 = -0.8333 \)
Numerator:
\( 1 - e^{-0.8333} = 1 - 0.4346 = 0.5654 \)
Denominator:
\( 1 + 0.6667 = 1.6667 \)
Effectiveness:
\( \varepsilon_{parallel} = \frac{0.5654}{1.6667} = 0.339 \)
Step 5: Effectiveness for counterflow:
\[ \varepsilon_{counter} = \frac{1 - \exp[-NTU (1 - C_r)]}{1 - C_r \exp[-NTU (1 - C_r)]} = \frac{1 - \exp[-0.5 \times (1 - 0.6667)]}{1 - 0.6667 \exp[-0.5 \times (1 - 0.6667)]} \]
Calculate exponent:
\( -0.5 \times 0.3333 = -0.1667 \)
Numerator:
\( 1 - e^{-0.1667} = 1 - 0.8465 = 0.1535 \)
Denominator:
\( 1 - 0.6667 \times 0.8465 = 1 - 0.5643 = 0.4357 \)
Effectiveness:
\( \varepsilon_{counter} = \frac{0.1535}{0.4357} = 0.352 \)
Answer: Counterflow arrangement has slightly higher effectiveness (0.352) than parallel flow (0.339), indicating better heat transfer performance.
A shell-and-tube heat exchanger has a heat transfer area of 10 m². The unit cost of manufacturing is Rs.12,000 per m². Estimate the total cost of the heat exchanger in INR.
Step 1: Multiply area by unit cost:
\[ \text{Cost} = 10 \times 12,000 = 120,000 \text{ INR} \]
Answer: The estimated cost of the heat exchanger is Rs.120,000.
When to use: When using NTU-effectiveness method to avoid incorrect effectiveness values.
When to use: When problem does not explicitly specify flow type but counterflow is common in practice.
When to use: To quickly solve NTU-effectiveness problems during competitive exams.
When to use: To avoid errors in temperature difference and logarithmic calculations.
When to use: Always, to maintain dimensional consistency and avoid calculation errors.
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