Heat transfer equation

Heat transfer equation

Heat transfer is a pivotal concept in thermodynamics, affecting everything from weather systems to industrial processes and even biological systems. Heat transfer can occur through conduction, convection, and radiation. The heat transfer equation provides a mathematical framework for understanding these processes. In this article, we’ll delve into the basics of heat transfer equations, considering each method of heat transfer and its corresponding formula.

Basic Equation for Heat Transfer

The general equation for heat transfer, often denoted by $$Q$$, is expressed as follows:

$Q = mc\Delta T$

Where:
– $$Q$$ = Heat transferred
– $$m$$ = Mass of the substance
– $$c$$ = Specific heat capacity of the substance
– $$\Delta T$$ = Change in temperature

Conduction

Conduction is the process of heat transfer through a solid medium without the movement of the medium itself. Fourier’s law describes this process:

$Q = -kA\frac{\Delta T}{\Delta x}$

Where:
– $$k$$ = Thermal conductivity of the material
– $$A$$ = Cross-sectional area
– $$\Delta T$$ = Temperature difference
– $$\Delta x$$ = Thickness of the material

Convection

Convection involves heat transfer between a surface and a liquid or gas moving past it. Newton’s Law of Cooling describes convective heat transfer:

$Q = hA\Delta T$

Where:
– $$h$$ = Convective heat transfer coefficient
– $$A$$ = Surface area of the interface
– $$\Delta T$$ = Temperature difference between the surface and the fluid

Radiation is the transfer of heat in the form of electromagnetic waves, often infrared radiation. The Stefan-Boltzmann Law governs radiative heat transfer:

$Q = \epsilon \sigma A T^4$

Where:
– $$\epsilon$$ = Emissivity of the material
– $$\sigma$$ = Stefan-Boltzmann constant ($$5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4$$)
– $$A$$ = Surface area
– $$T$$ = Absolute temperature in Kelvin

Applications of Heat Transfer Equations

1. Climate Modeling: Understanding heat transfer mechanisms helps in predicting weather patterns.
2. Industrial Processes: Various forms of heat transfer equations are used in chemical engineering, electronics cooling, and food processing.
3. Medical Treatments: Controlled heat transfer is used in treatments like hyperthermia and cryotherapy.

Conclusion

Heat transfer equations provide the backbone for understanding how energy is moved in various systems. With applications ranging from industrial engineering to weather forecasting and medical treatments, the significance of these equations cannot be understated. Knowing which equation to use under different circumstances allows for more precise calculations and predictions, driving advancements in numerous fields.

Understanding the underlying principles of these equations opens the door to innovations in energy efficiency, system optimization, and even life-saving medical procedures.

PROBLEMS AND SOLUTIONS

Conceptual Questions

1. What is the basic equation for heat transfer?
Solution: The basic equation for heat transfer is $$Q = mc\Delta T$$.

2. Can heat be transferred in a vacuum?
Solution: Yes, heat can be transferred in a vacuum through radiation.

3. Name the three main modes of heat transfer.
Solution: The three main modes of heat transfer are conduction, convection, and radiation.

4. Which mode of heat transfer doesn’t require a medium?
Solution: Radiation doesn’t require a medium.

5. What is the thermal conductivity ($$k$$) in Fourier’s law?
Solution: Thermal conductivity ($$k$$) is a material property that indicates the ability of a material to conduct heat.

6. Calculate the heat transfer if 2 kg of water is heated from 20°C to 100°C. (Specific heat of water = $$4.18 \, \text{J/g°C}$$)
Solution:
$Q = (2 \, \text{kg} \times 1000 \, \text{g/kg}) \times 4.18 \, \text{J/g°C} \times (100°C – 20°C) = 2 \times 10^6 \, \text{g} \times 4.18 \, \text{J/g°C} \times 80°C = 668 \times 10^6 \, \text{J} = 668 \, \text{MJ}$

7. A metal rod with a thermal conductivity of $$20 \, \text{W/mK}$$, cross-sectional area of $$0.02 \, \text{m}^2$$, and thickness $$0.1 \, \text{m}$$ has a temperature difference of 30°C across it. Find the heat transfer rate.
Solution:
$Q = -20 \, \text{W/mK} \times 0.02 \, \text{m}^2 \times \frac{30 \, \text{°C}}{0.1 \, \text{m}} = -20 \times 0.02 \times 300 = -120 \, \text{W}$

8. A surface with an emissivity of 0.6 and area of $$2 \, \text{m}^2$$ is at 300 K. Calculate the radiative heat transfer.
Solution:
$Q = 0.6 \times 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \times 2 \, \text{m}^2 \times 300^4 = 0.6 \times 5.67 \times 10^{-8} \times 2 \times 8100 \times 10^{12} \, \text{K}^4 = 5491.44 \, \text{W}$

9. An object of mass 500 g with a specific heat capacity of $$0.9 \, \text{J/g°C}$$ is heated from 25°C to 75°C. Find the heat transferred.
Solution:
$Q = 500 \, \text{g} \times 0.9 \, \text{J/g°C} \times (75 – 25) = 500 \times 0.9 \times 50 = 22500 \, \text{J} = 22.5 \, \text{kJ}$

10. Calculate the heat transfer rate for convection if the convective heat transfer coefficient is $$10 \, \text{W/m}^2\text{K}$$, the area is $$0.5 \, \text{m}^2$$, and the temperature difference is 20°C.
Solution:
$Q = 10 \, \text{W/m}^2\text{K} \times 0.5 \, \text{m}^2 \times 20 \, \text{°C} = 10 \times 0.5 \times 20 = 100 \, \text{W}$

11. A cube of side 1 m and thermal conductivity $$k = 10 \, \text{W/mK}$$ has a temperature difference of 5 K across it. Calculate the heat transfer rate.
Solution:
$Q = -10 \, \text{W/mK} \times 1 \, \text{m}^2 \times \frac{5 \, \text{K}}{1 \, \text{m}} = -10 \times 1 \times 5 = -50 \, \text{W}$

12. How much heat is needed to raise the temperature of 1 kg of copper from 30°C to 70°C? (Specific heat of copper = $$0.39 \, \text{J/g°C}$$)
Solution:
$Q = 1 \, \text{kg} \times 1000 \, \text{g/kg} \times 0.39 \, \text{J/g°C} \times (70 – 30) = 1000 \times 0.39 \times 40 = 15600 \, \text{J} = 15.6 \, \text{kJ}$

13. Calculate the heat loss through a wall that has an area of $$2 \, \text{m}^2$$, thickness of $$0.25 \, \text{m}$$, and thermal conductivity of $$0.8 \, \text{W/mK}$$, given that the interior and exterior temperatures are 20°C and -5°C respectively.
Solution:
$Q = -0.8 \, \text{W/mK} \times 2 \, \text{m}^2 \times \frac{(20 – (-5)) \, \text{°C}}{0.25 \, \text{m}} = -0.8 \times 2 \times \frac{25}{0.25} = -0.8 \times 2 \times 100 = -160 \, \text{W}$

14. What is the convective heat transfer if the heat transfer coefficient is $$25 \, \text{W/m}^2\text{K}$$, the area is $$1 \, \text{m}^2$$, and the temperature difference is 15 K?
Solution:
$Q = 25 \, \text{W/m}^2\text{K} \times 1 \, \text{m}^2 \times 15 \, \text{K} = 25 \times 1 \times 15 = 375 \, \text{W}$

15. A sphere with a surface area of $$0.5 \, \text{m}^2$$ is at a temperature of 350 K. If its emissivity is 0.7, find the rate of heat transfer due to radiation.
Solution:
$Q = 0.7 \times 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \times 0.5 \, \text{m}^2 \times 350^4 = 0.7 \times 5.67 \times 10^{-8} \times 0.5 \times 1500625 \times 10^{12} = 2663.45 \, \text{W}$