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title: Numerical examples - Part 1
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# Numerical examples - Part 1
Unit 4, Lecture 39

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## Numerical examples on Heat Exchangers (Part 1)

- Course: Heat and Mass Transfer (DI05019071)
- Unit 4: Heat exchanger
- Lecture 39: Numerical examples (LMTD Method)

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Welcome to Lecture 39. Today we will focus purely on problem-solving, applying the LMTD method to parallel and counter-flow heat exchangers.
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## Lecture Agenda

- 1. Problem Solving Strategy
- 2. Recap of Key Formulas (Energy Balance & LMTD)
- 3. Example 1: Parallel Flow Heat Exchanger
- 4. Example 2: Counter Flow Heat Exchanger
- 5. Comparison and Discussion

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We will review the fundamental equations, outline a strategy for tackling these problems, and then work through two comprehensive examples step-by-step.
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## Problem Solving Strategy

- Step 1: Identify the flow arrangement (Parallel, Counter, Cross, etc.).
- Step 2: Note down given values (mass flow rates, specific heats, inlet/outlet temperatures, U).
- Step 3: Perform an overall energy balance to find any missing temperatures: Q = mh*Cph*ΔTh = mc*Cpc*ΔTc.
- Step 4: Calculate temperature differences at both ends (ΔT1, ΔT2).
- Step 5: Calculate LMTD (ΔTm).
- Step 6: Use Q = U * A * ΔTm to find the unknown (Area, U, or Q).

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Always start by drawing the temperature profile. It helps visually confirm your ΔT1 and ΔT2 values and prevents simple arithmetic mistakes.
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## Recap of Key Formulas

- Energy Balance: Q = m_h * C_ph * (T_h,in - T_h,out) = m_c * C_pc * (T_c,out - T_c,in)
- Heat Transfer Rate: Q = U * A * ΔT_lm
- LMTD Formula: ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
- For Parallel Flow:
- - ΔT1 = T_h,in - T_c,in
- - ΔT2 = T_h,out - T_c,out
- For Counter Flow:
- - ΔT1 = T_h,in - T_c,out
- - ΔT2 = T_h,out - T_c,in

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Pay close attention to how ΔT1 and ΔT2 are defined based on the flow direction. This is where most errors occur.
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## Example 1: Problem Statement (Parallel Flow)

- Water (Cp = 4.18 kJ/kg.K) enters a parallel flow heat exchanger at 15°C at a rate of 0.5 kg/s.
- It is heated by hot oil (Cp = 2.1 kJ/kg.K) entering at 100°C at a rate of 0.8 kg/s.
- The overall heat transfer coefficient is 350 W/m².K.
- The outlet temperature of the oil is 60°C.
- Find: (a) The rate of heat transfer, (b) Outlet temperature of water, (c) The surface area of the heat exchanger.

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Let's break this down. We have the hot fluid completely specified (mass flow, Cp, inlet, outlet). We can use this to find the total heat transfer Q.
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## Example 1: Solution Step 1 & 2

- Step 1: Calculate Heat Transfer Rate (Q)
- Q = m_oil * Cp_oil * (T_oil,in - T_oil,out)
- Q = 0.8 kg/s * 2.1 kJ/kg.K * (100 - 60) K
- Q = 0.8 * 2.1 * 40 = 67.2 kW (or 67,200 W)
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- Step 2: Find Outlet Temperature of Water (T_c,out)
- Q = m_water * Cp_water * (T_c,out - T_c,in)
- 67.2 = 0.5 * 4.18 * (T_c,out - 15)
- T_c,out - 15 = 67.2 / 2.09 = 32.15°C
- T_c,out = 47.15°C

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First, we find Q using the oil side. Then, using energy balance, we find the water's outlet temperature. Make sure to keep units consistent; use Watts for area calculation later.
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## Example 1: Solution Step 3 & 4

- Step 3: Calculate ΔT1 and ΔT2 for Parallel Flow
- ΔT1 = T_h,in - T_c,in = 100 - 15 = 85°C
- ΔT2 = T_h,out - T_c,out = 60 - 47.15 = 12.85°C
- 
- Step 4: Calculate LMTD
- ΔT_lm = (85 - 12.85) / ln(85 / 12.85)
- ΔT_lm = 72.15 / ln(6.615)
- ΔT_lm = 72.15 / 1.889 = 38.19°C

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The LMTD is about 38 degrees. Notice the large difference between the driving force at the inlet (85 degrees) and the outlet (12.85 degrees).
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## Example 1: Final Calculation

- Step 5: Calculate Surface Area (A)
- Formula: Q = U * A * ΔT_lm
- Rearranging for A: A = Q / (U * ΔT_lm)
- Substitute values (ensure Q is in Watts):
- A = 67,200 W / (350 W/m².K * 38.19 K)
- A = 67,200 / 13366.5
- Area (A) = 5.03 m²

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By converting Q to Watts, the units cancel correctly with U (W/m2K), giving us the required surface area of approximately 5 square meters.
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## Example 2: Problem Statement (Counter Flow)

- Let's use the same fluids, temperatures, and flow rates, but arrange the exchanger in Counter Flow.
- Oil: Inlet = 100°C, Outlet = 60°C. Q is still 67.2 kW.
- Water: Inlet = 15°C, Outlet = 47.15°C.
- U = 350 W/m².K
- Find: The surface area of the heat exchanger in counter-flow arrangement.
- Compare this area to the parallel flow case.

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We keep all thermodynamic conditions the same. We just change the piping to counter-flow. Let's see how this affects the required size of the heat exchanger.
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## Example 2: LMTD Calculation

- Calculate ΔT1 and ΔT2 for Counter Flow:
- ΔT1 = T_h,in - T_c,out
- ΔT1 = 100 - 47.15 = 52.85°C
- 
- ΔT2 = T_h,out - T_c,in
- ΔT2 = 60 - 15 = 45°C
- 
- Calculate LMTD:
- ΔT_lm = (52.85 - 45) / ln(52.85 / 45)
- ΔT_lm = 7.85 / ln(1.174)
- ΔT_lm = 7.85 / 0.1608 = 48.81°C

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In counter-flow, the temperature difference is much more uniform along the length of the exchanger. Our LMTD is 48.81 degrees, which is significantly higher than the 38.19 degrees in parallel flow.
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## Example 2: Final Area & Comparison

- Calculate Area:
- A = Q / (U * ΔT_lm)
- A = 67,200 / (350 * 48.81)
- A = 67,200 / 17083.5 = 3.93 m²
- 
- Comparison:
- Parallel Flow Area = 5.03 m²
- Counter Flow Area = 3.93 m²
- Counter-flow requires ~22% less surface area for the exact same duty.

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This numerically proves why counter-flow is preferred. A higher LMTD means a stronger driving force, requiring less area, which translates to a smaller, cheaper heat exchanger.
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## Summary

- The LMTD method is straightforward when inlet and outlet temperatures are known or easily found via energy balance.
- Care must be taken to define ΔT1 and ΔT2 correctly based on the flow arrangement.
- Counter-flow configurations always yield a higher LMTD than parallel-flow for the same temperatures, resulting in a smaller required area.

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Remember the strategy: Energy balance first, then LMTD calculation, then the rate equation. And always verify your flow direction.
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## Next Lecture Preview

- Topic: Numerical Examples on Heat Exchangers (Part 2)
- Key topics:
- - What happens when we don't know the outlet temperatures?
- - Introduction to the Effectiveness-NTU method for problem-solving.
- - Solving cross-flow and shell-and-tube problems.

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Next time, we will tackle problems where LMTD is not convenient—specifically when you only have inlet temperatures and need to predict performance using the NTU method.
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