Understanding Heat Transfer Mechanisms
Before diving into the sample problems, it is crucial to grasp the three main mechanisms of heat transfer:
- Conduction: The transfer of heat through a solid material, where energy is passed from one molecule to another through direct contact.
- Convection: The transfer of heat through fluids (liquids or gases) caused by the motion of the fluid itself, which can be natural or forced.
- Radiation: The transfer of heat through electromagnetic waves, which can occur in a vacuum and does not require a medium.
Sample Problems in Heat Transfer
In this section, we will present sample problems related to conduction, convection, and radiation, along with step-by-step solutions.
1. Conduction Problems
Problem 1: A metal rod of length 2 m and cross-sectional area 0.01 m² is heated at one end to a temperature of 100°C while the other end is maintained at 20°C. If the thermal conductivity of the metal is 200 W/m·K, calculate the rate of heat transfer through the rod.
Solution:
To find the rate of heat transfer via conduction, we use Fourier's Law of Heat Conduction, which is given by:
\[
Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L}
\]
Where:
- \(Q\) = rate of heat transfer (W)
- \(k\) = thermal conductivity (W/m·K)
- \(A\) = cross-sectional area (m²)
- \(T_1\) = temperature at one end (°C)
- \(T_2\) = temperature at the other end (°C)
- \(L\) = length of the rod (m)
Plugging in the values:
- \(k = 200 \, \text{W/m·K}\)
- \(A = 0.01 \, \text{m}^2\)
- \(T_1 = 100 \, \text{°C}\)
- \(T_2 = 20 \, \text{°C}\)
- \(L = 2 \, \text{m}\)
\[
Q = \frac{200 \cdot 0.01 \cdot (100 - 20)}{2} = \frac{200 \cdot 0.01 \cdot 80}{2} = \frac{160}{2} = 80 \, \text{W}
\]
Thus, the rate of heat transfer through the rod is 80 W.
Problem 2: A wall with a thickness of 0.15 m separates two rooms. The temperature in one room is 30°C, while in the other it is 10°C. If the wall has a thermal conductivity of 0.5 W/m·K and an area of 10 m², calculate the heat loss through the wall.
Solution:
Using Fourier's Law again:
\[
Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L}
\]
Where:
- \(k = 0.5 \, \text{W/m·K}\)
- \(A = 10 \, \text{m}^2\)
- \(T_1 = 30 \, \text{°C}\)
- \(T_2 = 10 \, \text{°C}\)
- \(L = 0.15 \, \text{m}\)
\[
Q = \frac{0.5 \cdot 10 \cdot (30 - 10)}{0.15} = \frac{0.5 \cdot 10 \cdot 20}{0.15} = \frac{100}{0.15} \approx 666.67 \, \text{W}
\]
Therefore, the heat loss through the wall is approximately 666.67 W.
2. Convection Problems
Problem 3: A hot water radiator has a surface temperature of 80°C, and the surrounding air is at 20°C. If the heat transfer coefficient between the radiator's surface and the air is 10 W/m²·K and the area of the radiator is 2 m², calculate the rate of heat transfer to the air.
Solution:
To calculate the rate of heat transfer due to convection, we use Newton's Law of Cooling:
\[
Q = h \cdot A \cdot (T_s - T_a)
\]
Where:
- \(Q\) = rate of heat transfer (W)
- \(h\) = heat transfer coefficient (W/m²·K)
- \(A\) = surface area (m²)
- \(T_s\) = surface temperature (°C)
- \(T_a\) = ambient temperature (°C)
Substituting the values:
- \(h = 10 \, \text{W/m²·K}\)
- \(A = 2 \, \text{m}^2\)
- \(T_s = 80 \, \text{°C}\)
- \(T_a = 20 \, \text{°C}\)
\[
Q = 10 \cdot 2 \cdot (80 - 20) = 10 \cdot 2 \cdot 60 = 1200 \, \text{W}
\]
Thus, the rate of heat transfer to the air is 1200 W.
Problem 4: A 1.5 m tall person stands in a room at 25°C, with their skin temperature at 33°C. If the heat transfer coefficient for the person is 15 W/m²·K and the surface area of the person is 1.8 m², calculate the heat loss from the person to the surrounding air by convection.
Solution:
Using Newton's Law of Cooling again:
\[
Q = h \cdot A \cdot (T_s - T_a)
\]
Where:
- \(h = 15 \, \text{W/m²·K}\)
- \(A = 1.8 \, \text{m}^2\)
- \(T_s = 33 \, \text{°C}\)
- \(T_a = 25 \, \text{°C}\)
\[
Q = 15 \cdot 1.8 \cdot (33 - 25) = 15 \cdot 1.8 \cdot 8 = 216 \, \text{W}
\]
Therefore, the heat loss from the person to the surrounding air is 216 W.
3. Radiation Problems
Problem 5: A black body at a temperature of 600 K emits radiation. Calculate the total power radiated per unit area using the Stefan-Boltzmann law.
Solution:
The Stefan-Boltzmann law states:
\[
P = \sigma T^4
\]
Where:
- \(P\) = power per unit area (W/m²)
- \(\sigma\) = Stefan-Boltzmann constant \(= 5.67 \times 10^{-8} \, \text{W/m²·K}^4\)
- \(T\) = absolute temperature (K)
Substituting the values:
\[
P = 5.67 \times 10^{-8} \cdot (600)^4
\]
Calculating \(600^4\):
\[
600^4 = 1.296 \times 10^{11}
\]
Now substituting back:
\[
P = 5.67 \times 10^{-8} \cdot 1.296 \times 10^{11} \approx 7350.72 \, \text{W/m}^2
\]
Thus, the total power radiated per unit area is approximately 7350.72 W/m².
Problem 6: A surface at 350 K is exposed to an environment at 300 K. If the emissivity of the surface is 0.9, calculate the net radiative heat transfer.
Solution:
The net radiative heat transfer can be calculated using:
\[
Q = \epsilon \sigma (T_s^4 - T_a^4)
\]
Where:
- \(Q\) = net radiative heat transfer (W/m²)
- \(\epsilon\) = emissivity of the surface
- \(T_s\) = temperature of the surface (K)
- \(T_a\) = temperature of the ambient environment (K)
- \(\sigma = 5.67 \
Frequently Asked Questions
What is the formula for calculating heat transfer by conduction?
The formula for heat transfer by conduction is Q = k A (T1 - T2) / d, where Q is the heat transfer, k is the thermal conductivity, A is the area, T1 and T2 are the temperatures, and d is the thickness of the material.
How do you calculate the heat transfer in a heat exchanger?
The heat transfer in a heat exchanger can be calculated using the equation Q = U A ΔT_m, where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the heat transfer area, and ΔT_m is the log mean temperature difference.
What is the principle behind convection heat transfer?
Convection heat transfer occurs due to the movement of fluid and can be calculated using Q = h A (Ts - Tf), where Q is the heat transfer, h is the convection heat transfer coefficient, A is the surface area, Ts is the surface temperature, and Tf is the fluid temperature.
How do you solve a problem involving radiant heat transfer?
Radiant heat transfer can be solved using the Stefan-Boltzmann law: Q = ε σ A (T1^4 - T2^4), where Q is the heat transfer, ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the area, and T1 and T2 are the absolute temperatures of the surfaces.
What is the effect of insulation on heat transfer?
Insulation reduces heat transfer by conduction, convection, and radiation. The effectiveness of insulation can be calculated using the R-value, where higher R-values indicate better insulating properties, thereby reducing heat loss.
How can you calculate the rate of heat loss from a building?
The rate of heat loss from a building can be calculated using the equation Q = U A ΔT, where Q is the heat loss, U is the overall heat transfer coefficient of the building envelope, A is the surface area, and ΔT is the temperature difference between indoors and outdoors.
What are the units of measurement for heat transfer?
The units of measurement for heat transfer are typically in watts (W) for power or joules (J) for energy. In the context of heat transfer problems, specific heat capacity can be measured in J/(kg·K) or cal/g·°C.
What is a common example of a phase change affecting heat transfer?
A common example is the melting of ice. The heat required to change ice at 0°C to water at 0°C without changing temperature is called latent heat, calculated using Q = m L_f, where m is mass and L_f is the latent heat of fusion.
How do you determine the thermal conductivity of a material?
Thermal conductivity can be determined experimentally using a guarded hot plate or a transient line source method. It is often expressed in watts per meter-kelvin (W/m·K).
In a heat transfer problem, how do you approach finding unknown temperatures?
To find unknown temperatures in a heat transfer problem, you can apply the energy balance method, where the heat lost by one body equals the heat gained by another, often leading to a system of equations that can be solved simultaneously.
