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Convection is a fundamental mode of heat transfer that occurs when heat is carried away from a surface by the movement of a fluid, such as air or water. Unlike conduction, where heat transfers through a stationary medium, convection involves fluid motion that enhances heat transfer. This motion can be caused by external forces like fans or pumps, or it can arise naturally due to temperature differences within the fluid itself.
In mechanical engineering, understanding convection is crucial for designing efficient cooling systems, heating devices, and thermal management solutions. For example, cooling electronic components in computers or managing heat loss in buildings relies heavily on convection principles. In this section, we will explore the two main types of convection: forced convection, where fluid motion is driven externally, and natural convection, where buoyancy forces caused by temperature differences induce fluid flow.
Convection is the heat transfer process involving the combined effects of conduction within the fluid and the bulk motion of the fluid itself. The fluid motion transports thermal energy away from the hot surface, enhancing heat transfer beyond what conduction alone can achieve.
There are two types of convection:
In the diagram above, the orange arrows represent natural convection currents rising due to buoyancy, while the blue arrow shows forced flow driven by a fan blowing air over the heated plate.
To quantify convection heat transfer, we use Newton's Law of Cooling, which states that the rate of heat transfer \( Q \) from a solid surface to a fluid is proportional to the temperature difference between the surface and the fluid far away from the surface:
The convective heat transfer coefficient \( h \) is a measure of how effectively heat is transferred between the surface and the fluid. It depends on fluid properties, flow velocity, and the nature of convection (forced or natural).
Physically, convection involves the formation of a thermal boundary layer near the surface, where temperature gradients exist, and fluid velocity changes from zero at the surface (due to the no-slip condition) to the free stream velocity away from the surface. The thickness and behavior of this boundary layer strongly influence the heat transfer rate.
Dimensionless numbers are essential tools in convection heat transfer. They allow us to generalize experimental results and apply them to different fluids and conditions. The key dimensionless numbers are:
| Dimensionless Number | Formula | Physical Meaning | Typical Range |
|---|---|---|---|
| Reynolds Number (Re) | \( Re = \frac{\rho u L}{\mu} = \frac{u L}{ u} \) | Ratio of inertial to viscous forces; indicates flow regime (laminar or turbulent) | Laminar: <2300; Turbulent: >4000 |
| Prandtl Number (Pr) | \( Pr = \frac{ u}{\alpha} = \frac{c_p \mu}{k} \) | Ratio of momentum diffusivity to thermal diffusivity; relates velocity and thermal boundary layers | Typically 0.7 (air) to 7 (water) |
| Grashof Number (Gr) | \( Gr = \frac{g \beta (T_s - T_\infty) L^3}{ u^2} \) | Ratio of buoyancy to viscous forces; important in natural convection | Varies widely; >10^9 indicates turbulent natural convection |
| Nusselt Number (Nu) | \( Nu = \frac{h L}{k} \) | Ratio of convective to conductive heat transfer; dimensionless heat transfer coefficient | 1 (pure conduction) to several hundred (strong convection) |
In forced convection, empirical correlations relate the Nusselt number to Reynolds and Prandtl numbers to estimate the convective heat transfer coefficient. One widely used correlation for turbulent flow inside smooth pipes is the Dittus-Boelter equation:
This correlation applies for Reynolds numbers between 10,000 and 120,000 and Prandtl numbers between 0.7 and 160. It is commonly used in designing heat exchangers and pipe flow cooling systems.
For natural convection, the Nusselt number depends on the Grashof and Prandtl numbers combined into the Rayleigh number \( Ra = Gr \times Pr \). The Churchill-Chu correlation is a reliable formula for vertical plates:
This correlation covers a wide range of Rayleigh numbers and is useful for estimating heat loss from walls, solar collectors, and other vertical surfaces.
Convection heat transfer plays a vital role in many practical scenarios:
Step 1: Calculate Reynolds number \( Re \):
\( Re = \frac{\rho u D}{\mu} = \frac{990 \times 2 \times 0.05}{0.0006} = 165,000 \)
Note: Since \( Re > 4000 \), flow is turbulent, so Dittus-Boelter equation applies.
Step 2: Use Dittus-Boelter equation for heating (fluid temperature rising): \( n = 0.4 \)
\( Nu = 0.023 Re^{0.8} Pr^{0.4} \)
Calculate each term:
\( Re^{0.8} = (165,000)^{0.8} \approx 20,700 \)
\( Pr^{0.4} = (3.5)^{0.4} \approx 1.7 \)
Therefore,
\( Nu = 0.023 \times 20,700 \times 1.7 = 810 \)
Step 3: Calculate heat transfer coefficient \( h \):
\( h = \frac{Nu \times k}{D} = \frac{810 \times 0.64}{0.05} = 10,368 \, \mathrm{W/m^2 \cdot K} \)
Answer: The convective heat transfer coefficient is approximately \( 10,400 \, \mathrm{W/m^2 \cdot K} \).
Step 1: Calculate Grashof number \( Gr \):
\( Gr = \frac{g \beta (T_s - T_\infty) L^3}{ u^2} = \frac{9.81 \times \frac{1}{323} \times (75 - 25) \times 1^3}{(1.6 \times 10^{-5})^2} \)
\( Gr = \frac{9.81 \times 0.0031 \times 50}{2.56 \times 10^{-10}} = \frac{1.52}{2.56 \times 10^{-10}} = 5.94 \times 10^9 \)
Step 2: Calculate Rayleigh number \( Ra = Gr \times Pr = 5.94 \times 10^9 \times 0.7 = 4.16 \times 10^9 \)
Step 3: Use Churchill-Chu correlation:
\( Nu = \left(0.825 + \frac{0.387 Ra^{1/6}}{[1 + (0.492/Pr)^{9/16}]^{8/27}}\right)^2 \)
Calculate \( Ra^{1/6} = (4.16 \times 10^9)^{1/6} \approx 63.5 \)
Calculate denominator term:
\( (0.492/0.7)^{9/16} = (0.703)^{0.5625} \approx 0.82 \)
\( [1 + 0.82]^{8/27} = (1.82)^{0.296} \approx 1.19 \)
Now,
\( Nu = \left(0.825 + \frac{0.387 \times 63.5}{1.19}\right)^2 = (0.825 + 20.65)^2 = (21.475)^2 = 461.5 \)
Step 4: Calculate heat transfer coefficient \( h \):
\( h = \frac{Nu \times k}{L} = \frac{461.5 \times 0.026}{1} = 12.0 \, \mathrm{W/m^2 \cdot K} \)
Step 5: Calculate heat loss per unit width:
Area per unit width \( A = L \times 1 = 1 \times 1 = 1 \, \mathrm{m^2} \)
\( Q = h A (T_s - T_\infty) = 12.0 \times 1 \times (75 - 25) = 600 \, \mathrm{W} \)
Answer: Heat loss from the plate is approximately 600 W per meter width.
Step 1: Identify individual heat transfer coefficients:
Forced convection coefficient \( h_f = 25 \, \mathrm{W/m^2 \cdot K} \)
Natural convection coefficient \( h_n = 10 \, \mathrm{W/m^2 \cdot K} \)
Step 2: Use root-sum-square method to combine:
\( h_{total} = \sqrt{h_f^2 + h_n^2} = \sqrt{25^2 + 10^2} = \sqrt{625 + 100} = \sqrt{725} = 26.9 \, \mathrm{W/m^2 \cdot K} \)
Answer: The combined heat transfer coefficient is approximately 27 W/m²·K.
Step 1: Calculate the thermal time constant \( \tau \):
\( \tau = \frac{\rho V c_p}{h A} \)
Calculate volume \( V = \frac{m}{\rho} \). Since density \( \rho \) is not given, use mass and specific heat directly:
Thermal capacity \( C = m c_p = 0.2 \times 900 = 180 \, \mathrm{J/K} \)
\( \tau = \frac{C}{h A} = \frac{180}{15 \times 0.01} = \frac{180}{0.15} = 1200 \, \mathrm{s} \)
Step 2: Use Newton's law cooling formula for transient temperature:
\( \frac{T - T_\infty}{T_i - T_\infty} = e^{-t/\tau} \)
Rearranged to solve for time \( t \):
\( t = -\tau \ln \left(\frac{T - T_\infty}{T_i - T_\infty}\right) \)
Substitute values:
\( t = -1200 \ln \left(\frac{50 - 30}{90 - 30}\right) = -1200 \ln \left(\frac{20}{60}\right) = -1200 \ln (0.333) \)
\( \ln(0.333) = -1.099 \)
\( t = -1200 \times (-1.099) = 1319 \, \mathrm{s} \approx 22 \, \mathrm{minutes} \)
Answer: The component will cool to 50°C in approximately 22 minutes.
Step 1: Calculate Reynolds number for both velocities:
\( Re_1 = \frac{u L}{ u} = \frac{1 \times 0.5}{1.5 \times 10^{-5}} = 33,333 \)
\( Re_2 = \frac{4 \times 0.5}{1.5 \times 10^{-5}} = 133,333 \)
Since \( Re < 5 \times 10^5 \), flow is laminar.
Step 2: Use laminar flow Nusselt number correlation for flat plate:
\( Nu_x = 0.332 Re_x^{1/2} Pr^{1/3} \)
Calculate \( Nu_1 \):
\( Nu_1 = 0.332 \times (33,333)^{0.5} \times (0.7)^{1/3} = 0.332 \times 182.57 \times 0.887 = 53.7 \)
Calculate \( Nu_2 \):
\( Nu_2 = 0.332 \times (133,333)^{0.5} \times 0.887 = 0.332 \times 365.15 \times 0.887 = 107.6 \)
Step 3: Calculate heat transfer coefficients:
\( h = \frac{Nu \times k}{L} \)
\( h_1 = \frac{53.7 \times 0.026}{0.5} = 2.79 \, \mathrm{W/m^2 \cdot K} \)
\( h_2 = \frac{107.6 \times 0.026}{0.5} = 5.59 \, \mathrm{W/m^2 \cdot K} \)
Answer: Increasing velocity from 1 m/s to 4 m/s doubles the heat transfer coefficient from 2.79 to 5.59 W/m²·K.
When to use: Estimating heat transfer in pipes, ducts, or over surfaces with fans or pumps.
When to use: During problem solving to choose correct heat transfer correlations efficiently.
When to use: Solving buoyancy-driven convection problems on vertical or horizontal surfaces.
When to use: Mixed convection scenarios such as outdoor surfaces with wind and buoyancy effects.
When to use: Calculating Reynolds, Prandtl, and Grashof numbers in convection problems.
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