Heat Transfer Question **Problem 2:**Consider a parallel-flow double pipe heat exchanger with an area of 50 m² operating to heat water from 24 °C to 50 °C at a rate of 4 kg/s. Hot air is used to heat water. The air enters and exits the exchanger at 100 °C and 54 °C, respectively, and it has a mass flow rate of 9.4 kg/s. Calculate the overall heat transfer coefficient $ U $ using the Log Mean Temperature Difference (LMTD) method.**Note:** Please use the following tables for the thermophysical properties of air and water.---**Table A.6: Thermophysical Properties of Saturated Water**| Temperature, $ T $ (K) | Pressure, $ p $ (bars) | Specific Volume $ v^ $ (m³/kg) | $ v_e^ $ | Heat of Vaporization, $ h_r $ (kJ/kg) | Specific Heat $ c_p $ (kJ/kg·K) | $ c_{p,e} $ | Viscosity $ \mu\cdot10^6 $ (N·s/m²) | $ \mu_{e}\cdot10^6 $ | Thermal Conductivity $ k\cdot10^3 $ (W/m·K) | $ k_e\cdot10^3 $ | Prandtl Number $ Pr $ | $ Pr_e $ ||-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|| 273.15 | 0.00611 | 1.000 | 206.3 | 2502 | 4.217 | 1.854 | 1750 | 8.02 | 569 | 18.2 | 12.99 | 0.815 || 275 | 0.00767 | 1.000 | 130.7 | 2497 | 4.211 | 1.855 | 1652 | 8.09 | 574 | 14.8 | 12.22 | 0.817 || 280 | 0.00990 | 1.000 | 69.7 | 2483 | 4.198 | 1.858 | 1422 | 8

Answered: coefficient U using the Log Mean…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Heat Transfer Question

**Problem 2:**

Consider a parallel-flow double pipe heat exchanger with an area of 50 m² operating to heat water from 24 °C to 50 °C at a rate of 4 kg/s. Hot air is used to heat water. The air enters and exits the exchanger at 100 °C and 54 °C, respectively, and it has a mass flow rate of 9.4 kg/s. Calculate the overall heat transfer coefficient $ U $ using the Log Mean Temperature Difference (LMTD) method.

**Note:** Please use the following tables for the thermophysical properties of air and water.

---

**Table A.6: Thermophysical Properties of Saturated Water**

| Temperature, $ T $ (K) | Pressure, $ p $ (bars) | Specific Volume $ v^ $ (m³/kg) | $ v_e^ $ | Heat of Vaporization, $ h_r $ (kJ/kg) | Specific Heat $ c_p $ (kJ/kg·K) | $ c_{p,e} $ | Viscosity $ \mu\cdot10^6 $ (N·s/m²) | $ \mu_{e}\cdot10^6 $ | Thermal Conductivity $ k\cdot10^3 $ (W/m·K) | $ k_e\cdot10^3 $ | Prandtl Number $ Pr $ | $ Pr_e $ |
|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|-----------------|
| 273.15 | 0.00611 | 1.000 | 206.3 | 2502 | 4.217 | 1.854 | 1750 | 8.02 | 569 | 18.2 | 12.99 | 0.815 |
| 275 | 0.00767 | 1.000 | 130.7 | 2497 | 4.211 | 1.855 | 1652 | 8.09 | 574 | 14.8 | 12.22 | 0.817 |
| 280 | 0.00990 | 1.000 | 69.7 | 2483 | 4.198 | 1.858 | 1422 | 8

Expert Solution

Introduction

Logarithmic mean temperature difference(LMTD)- is defined as the temperature difference, which if constant would give the same rate of heat transfer as actually occurs under variable condition of temperature difference.

Q= $\frac{U A (θ_{2} - θ_{1})}{\ln (\frac{θ_{2}}{θ_{1}})}$

$θ_{1} = t_{h_{1}} - t_{c_{1}} θ_{2} = t_{h_{2}} - t_{c_{2}}$

Where,

U = Overall heat transfer

A = Area of the heat exchanger

t_h1 = Inlet temperature of hot fluid

t_c1 = Inlet temperature of cold fluid

t_h2 = Outlet temperature of hot fluid

t_c2 = Outlet temperature of cold fluid

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