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title: Numerical examples - Part 2
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# Numerical examples - Part 2
Unit 4, Lecture 40

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## Numerical examples on Heat Exchangers (Part 2)

- Course: Heat and Mass Transfer (DI05019071)
- Unit 4: Heat exchanger
- Lecture 40: Numerical examples (NTU Method & Cross-Flow)

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Welcome to Lecture 40. Today we continue our numerical analysis, focusing on the NTU method for predicting performance and using correction factors for complex flows.
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## Lecture Agenda

- 1. When to use LMTD vs. NTU Method
- 2. Recap of Effectiveness-NTU Formulas
- 3. Example 3: Performance Prediction (NTU Method)
- 4. Recap of Multipass Correction Factor (F)
- 5. Example 4: Shell and Tube Exchanger with 'F' Factor

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We will learn how to handle situations where you know the heat exchanger size but not how it will perform, and how to correct LMTD for non-ideal flow paths.
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## LMTD vs. Effectiveness-NTU Method

- LMTD Method is best for SIZING problems:
- - You know inlet and outlet temperatures.
- - You need to find the required Surface Area (A).
- 
- Effectiveness-NTU Method is best for RATING/PERFORMANCE problems:
- - You know the Surface Area (A) and inlet temperatures.
- - You need to predict outlet temperatures and Heat Transfer Rate (Q).

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If you try to use LMTD when outlet temperatures are unknown, you'll end up with iterative trial-and-error calculations. The NTU method avoids this.
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## Recap: Effectiveness-NTU Formulas

- Heat Capacity Rates: C_h = m_h * Cp_h, C_c = m_c * Cp_c
- Identify C_min (the smaller of C_h and C_c) and C_max.
- Capacity Ratio: c = C_min / C_max
- Maximum possible heat transfer: Q_max = C_min * (T_h,in - T_c,in)
- Number of Transfer Units (NTU) = (U * A) / C_min
- Effectiveness (ε) = Q_actual / Q_max
- ε is found from charts or formulas based on NTU, c, and flow arrangement.

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The core concept here is Q_max. The fluid with the smaller heat capacity rate dictates the maximum possible temperature change, which is limited by the inlet temperature difference.
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## Example 3: Problem Statement (NTU Method)

- A counter-flow heat exchanger has a surface area of 10 m² and U = 200 W/m².K.
- Hot fluid (Cp = 2.0 kJ/kg.K) enters at 120°C at 1.0 kg/s.
- Cold fluid (Cp = 4.18 kJ/kg.K) enters at 20°C at 0.5 kg/s.
- Find: The rate of heat transfer and the outlet temperatures.

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Notice we only have inlet temperatures. This is a classic rating problem. We will use the Effectiveness-NTU method to solve it directly.
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## Example 3: Capacity Rates and NTU

- Step 1: Calculate Capacity Rates
- C_h = 1.0 kg/s * 2000 J/kg.K = 2000 W/K
- C_c = 0.5 kg/s * 4180 J/kg.K = 2090 W/K
- Therefore, C_min = C_h = 2000 W/K.
- Capacity ratio (c) = C_min / C_max = 2000 / 2090 = 0.957
- 
- Step 2: Calculate NTU
- NTU = (U * A) / C_min = (200 * 10) / 2000 = 2000 / 2000 = 1.0

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Always convert specific heat to J/kg.K to match U (W/m2K). Here, the hot fluid is the minimum capacity fluid. Our NTU is exactly 1.
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## Example 3: Effectiveness and Q_actual

- Step 3: Find Effectiveness (ε) for Counter Flow
- Formula for Counter Flow: ε = [1 - exp(-NTU*(1-c))] / [1 - c*exp(-NTU*(1-c))]
- ε = [1 - exp(-1*(1-0.957))] / [1 - 0.957*exp(-1*(1-0.957))]
- ε = [1 - exp(-0.043)] / [1 - 0.957*exp(-0.043)]
- ε = [1 - 0.9579] / [1 - 0.957(0.9579)] = 0.0421 / (1 - 0.9167) = 0.505 (or 50.5%)
- 
- Step 4: Calculate Q_max and Q_actual
- Q_max = C_min * (T_h,in - T_c,in) = 2000 * (120 - 20) = 200,000 W = 200 kW
- Q_actual = ε * Q_max = 0.505 * 200 kW = 101 kW

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We use the specific formula for counter-flow. An effectiveness of 50.5% means the exchanger is transferring roughly half of the theoretically possible maximum heat.
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## Example 3: Outlet Temperatures

- Step 5: Calculate Outlet Temperatures
- From Q = C_h * (T_h,in - T_h,out)
- 101,000 = 2000 * (120 - T_h,out)
- 50.5 = 120 - T_h,out  =>  T_h,out = 69.5°C
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- From Q = C_c * (T_c,out - T_c,in)
- 101,000 = 2090 * (T_c,out - 20)
- 48.33 = T_c,out - 20  =>  T_c,out = 68.33°C

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Using basic energy balance with our actual Q, we find the outlet temperatures. Without the NTU method, finding these would have required guessing and checking.
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## Recap: Multipass Correction Factor (F)

- For Cross-Flow and Multipass Shell-and-Tube exchangers, the flow is neither pure parallel nor pure counter-flow.
- We calculate LMTD assuming pure counter-flow, then apply a correction factor (F).
- Q = U * A * F * ΔT_lm,CF
- F is obtained from charts based on parameters P and R:
- - P = (t2 - t1) / (T1 - t1)  (Effectiveness parameter)
- - R = (T1 - T2) / (t2 - t1)  (Capacity ratio parameter)

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In industrial shell and tube exchangers, fluid makes multiple passes. The flow is mixed. We account for this deviation from ideal counter-flow using the F factor.
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## Example 4: Problem Statement (Shell & Tube)

- A 1-shell pass, 2-tube pass heat exchanger is used to heat water in tubes from 20°C to 70°C.
- Hot oil in the shell cools from 140°C to 90°C.
- Total heat transfer rate Q = 150 kW.
- U = 400 W/m².K.
- Find the required surface area.

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Here we have all temperatures and Q, so it's a sizing problem. We use LMTD, but because it's a 1-2 shell and tube, we MUST use the correction factor.
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## Example 4: Solution

- Step 1: Calculate LMTD for Counter Flow
- ΔT1 = 140 - 70 = 70°C,  ΔT2 = 90 - 20 = 70°C
- Since ΔT1 = ΔT2, LMTD = 70°C.
- 
- Step 2: Find P and R parameters
- Let tube fluid be lower case (t), shell fluid be upper case (T).
- P = (70 - 20) / (140 - 20) = 50 / 120 = 0.417
- R = (140 - 90) / (70 - 20) = 50 / 50 = 1.0
- 
- Step 3: From chart for 1-2 exchanger, at P=0.417, R=1.0, F ≈ 0.85

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Because ΔT1 equals ΔT2, the LMTD is just that constant value, 70. We calculate P and R, look up the specific chart for a 1-2 exchanger, and find F is about 0.85.
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## Example 4: Final Area Calculation

- Step 4: Calculate Area
- Formula: Q = U * A * F * ΔT_lm
- 150,000 W = 400 W/m².K * A * 0.85 * 70 K
- 150,000 = A * (400 * 0.85 * 70)
- 150,000 = A * 23,800
- Area = 150,000 / 23,800 = 6.30 m²

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By applying the F factor, we correctly size the area to 6.3 square meters. If we ignored F, we would have undersized the heat exchanger, and it wouldn't meet the requirements.
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## Summary & Unit 4 Conclusion

- NTU Method is essential when outlet temperatures are unknown (rating/performance problems).
- Correction factor (F) is mandatory for multipass and cross-flow designs when using LMTD.
- This concludes Unit 4: Heat Exchangers. You should now be able to analyze and size various heat transfer equipment.

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This brings us to the end of Unit 4. You have learned the physics of heat exchangers, compact designs like heat pipes, and the mathematical tools (LMTD and NTU) to design them.
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