From TecQuipment TE93 User Guide (2009) | Source: TecQuipment Ltd User Guide, Experiment 2
Tags: cooling curve, heat transfer coefficient, transient cooling, Newton's law of cooling, lumped capacitance, log-linear plot, gradient method, forced convection cooling, TE93
Experiment 2 measures how quickly the pre-heated copper rod cools in the airstream. Plotting log₁₀(T₂ − T₁) against time gives a straight line whose gradient yields the heat transfer coefficient (α). Repeating at different valve settings and rod positions shows how air velocity and position within the rod bank affect cooling rate.
Cooling curve
A plot of log₁₀(T₂ − T₁) against time in seconds. For a lumped-capacitance body cooling by forced convection, this is a straight line with a negative slope.
Heat transfer coefficient (α)
The proportionality constant relating heat flux to temperature difference between the rod surface and the surrounding air. Units: W/m²K (or equivalently J/m²s°K). Higher α means faster cooling.
Lumped capacitance assumption
The assumption that temperature gradients within the solid body are negligible, so the entire body is at a uniform temperature at any instant. Valid when the Biot number is small (Bi << 1). Copper's high thermal conductivity makes this a reasonable assumption for the Heated Rod.
Gradient M
The slope of the log₁₀(T₂ − T₁) vs time line. Always negative (temperature difference decreases over time). Used to calculate α via the conversion formula.
Newton's law of cooling
The rate of heat loss from a body is proportional to the temperature difference between the body and its surroundings: q̇ = α × A₁ × (T₂ − T₁).
Determine the cooling rate of the Heated Rod, both within a full bank of rods and by itself
Plot cooling curves and extract the heat transfer coefficient (α) for the Heated Rod at various positions and air velocities
Newton's law of cooling gives the rate of heat transfer:
q̇ = α × A₁ × (T₂ − T₁)
The heat lost by the rod over a small time interval dt causes a temperature drop dT:
q̇ × dt = −m × c × dT
Combining these:
−dT / (T₂ − T₁) = (α × A₁) / (m × c) × dt
Integrating from T₀ (initial temperature at t = 0) to T₂ (temperature at time t):
logₑ(T₂ − T₁) − logₑ(T₀ − T₁) = −(α × A₁ × t) / (m × c)
This is the equation of a straight line in logₑ(T₂ − T₁) vs t, with slope:
M_e = −α × A₁ / (m × c)
Since it is more practical to plot log₁₀ values, and logₑN = 2.3026 × log₁₀N, the conversion gives:
α = −2.3026 × (m × c) / A₁ × M
Where M is the gradient of the log₁₀(T₂ − T₁) vs t plot.
c (specific heat of copper at room temperature) = 380 J/kg°C
m = mass stamped on the Heated Rod end plate (copper part only)
A₁ = π × d × L₁ (effective surface area, using L₁ = 0.1034 m)
d = measured rod diameter (nominal 12.4 mm)
Fit all rods except the upstream centre position in column 1, which is left free for the Heated Rod
Switch on the heater and insert the Heated Rod into the heater compartment
Wait at least two minutes for the rod to reach a stable temperature (~75 °C)
Set the air valve to 100 % and switch on the fan
Record ambient temperature T₁
Remove the Heated Rod and insert it into the working section
Record T₂ every 10 seconds until the rod approaches ambient temperature
Repeat at valve settings 90 %, 80 %, 70 %, 60 %, 50 %, 40 %, 30 %, 20 %, 10 %
Repeat with the Heated Rod in each of the four column positions
The single-rod method is the same but with all other rods replaced by blanking plugs.
From the sample results in the guide:
One rod alone, column 1, 100 % valve: gradient M ≈ −0.0082, intercept ≈ 1.75
All rods, Heated Rod in column 1, 100 % valve: gradient M ≈ −0.0091, intercept ≈ 1.77
All rods, Heated Rod in column 4, 100 % valve: gradient M ≈ −0.0126, intercept ≈ 1.76
The rod cools at roughly the same rate when alone as when in the first column of a full bank. This makes sense: column 1 faces undisturbed oncoming flow in both cases.
Column 4 shows a steeper gradient (faster cooling). The upstream rods increase turbulence, which enhances heat transfer for downstream rods.
As the Heated Rod moves from column 1 (upstream) to column 4 (downstream):
Turbulence intensity increases because of wake effects from the preceding rods
Higher turbulence enhances convective heat transfer
The heat transfer coefficient α increases with column number
Cooling is fastest in column 4, slowest in column 1
Higher air valve openings produce higher velocities. Higher velocity increases the heat transfer coefficient, steepening the cooling curve gradient. The relationship between α and velocity feeds into the dimensionless analysis (covered in Part 4).
Newton's law of cooling:
q̇ = α × A₁ × (T₂ − T₁)
Heat balance over time interval dt:
−m × c × dT = α × A₁ × (T₂ − T₁) × dt
Integrated cooling equation (log form):
logₑ(T₂ − T₁) = logₑ(T₀ − T₁) − (α × A₁ × t) / (m × c)
Heat transfer coefficient from log₁₀ gradient:
α = −2.3026 × (m × c / A₁) × M
Where M = gradient of the log₁₀(T₂ − T₁) vs time plot.
⚠️ The formula for α uses log₁₀ gradient M, not logₑ gradient. The factor 2.3026 handles the base conversion. If you use natural log gradient directly, drop the 2.3026.
⚠️ M is always negative (temperature difference is falling). The negative sign in the formula for α cancels M's sign, giving a positive heat transfer coefficient.
⚠️ Use the effective surface area A₁ (based on L₁ = 0.1034 m), not the exposed copper area. Using the wrong length is a common calculation error.
⚠️ The rod must be at a stable, uniform temperature before removal from the heater. If it has not equilibrated, early data points will not sit on the expected straight line.
⚠️ The cooling curve should be a straight line on a log₁₀(ΔT) vs t plot. Curvature suggests the lumped capacitance assumption is breaking down, or that ambient temperature drifted during the test.
Q: Why does plotting log₁₀(T₂ − T₁) against time produce a straight line?
A: The integrated form of Newton's cooling law for a lumped-capacitance body is an exponential decay in temperature difference. Taking the logarithm of an exponential gives a linear function of time, with slope proportional to −α × A₁ / (m × c).
Q: A cooling curve has gradient M = −0.0091. The Heated Rod has mass 0.080 kg, effective surface area 4.02 × 10⁻³ m², and copper specific heat 380 J/kg°C. What is α?
A: α = −2.3026 × (0.080 × 380 / 4.02 × 10⁻³) × (−0.0091) = 2.3026 × 7562.2 × 0.0091 ≈ 158.5 W/m²K.
Q: Why does the Heated Rod cool faster in column 4 than in column 1?
A: The upstream rods generate turbulent wakes. By column 4, the airflow is highly turbulent, which disrupts the thermal boundary layer around the Heated Rod, enhancing convective heat transfer and increasing the heat transfer coefficient.
Q: What assumption allows us to treat the Heated Rod as having a single uniform temperature?
A: The lumped capacitance assumption. Copper has high thermal conductivity (~385 W/mK), so internal temperature gradients are negligible compared to the surface-to-air temperature difference. Formally, this holds when the Biot number (α × characteristic length / k_copper) is much less than 1.
cooling curve, transient cooling, Newton's law of cooling, heat transfer coefficient, lumped capacitance method, Biot number, forced convection, thermal boundary layer, turbulence enhancement, log-linear plot, exponential decay, specific heat of copper, TE93 Experiment 2, rod bank heat transfer, wake turbulence