Question

A steel casting cools to 90 percent of the original temperature difference in \(30 \mathrm{~min}\) in still air. The time it takes to cool this same casting to 90 percent of the original temperature difference in a moving air stream whose convective heat transfer coefficient is 5 times that of still air is (a) \(3 \mathrm{~min}\) (b) \(6 \mathrm{~min}\) (c) \(9 \mathrm{~min}\) (d) \(12 \mathrm{~min}\) (e) \(15 \mathrm{~min}\)

Short answer

Answer: 6 minutes

Step-by-step solution

Understand the given information

In still air, the steel casting cools to 90% of the original temperature difference in 30 minutes. The convective heat transfer coefficient in a moving air stream is 5 times that of the still air.

Use Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings. Mathematically, we can write it as: \(\frac{dT}{dt} = -h(T - T_s)\), where \(T\) is the temperature of the object, \(T_s\) is the temperature of the surroundings, \(h\) is the convective heat transfer coefficient, and \(t\) is the time.

Integrate the differential equation

To find the relationship between the temperature of the casting and time, we need to integrate the above equation. Let's denote the original temperature difference as \(ΔT_0\). So, we want to find the time it takes for the casting to cool to \(0.9ΔT_0\). Integrating the equation, we have: \(\int_{ΔT_0}^{0.9ΔT_0} \frac{dT}{T - T_s} = -\int_0^t h dt\)

Evaluate the integrals for still and moving air

Let's denote the time it takes for the casting to cool to \(0.9ΔT_0\) in still air as \(t_s\) and in moving air as \(t_m\). Similarly, let's denote the heat transfer coefficient for still air as \(h_s\) and for moving air as \(h_m\). Using the given information that \(t_s = 30\,\text{min}\) and \(h_m = 5h_s\), we can write the equation for still and moving air as: \(\int_{ΔT_0}^{0.9ΔT_0}\frac{dT}{T - T_s} = -30h_s\) \(\int_{ΔT_0}^{0.9ΔT_0}\frac{dT}{T - T_s} = -t_m(5h_s)\)

Divide and solve for the time in moving air

Divide the second equation by the first equation to eliminate the integral term: \(\frac{-t_m(5h_s)}{-30h_s} = 1\) \(t_m = 6\,\text{min}\) So, the time it takes for the steel casting to cool to 90% of the original temperature difference in a moving air stream is \(6\,\text{min}\), which corresponds to the answer option (b).

Key concepts

Convective Heat Transfer Coefficient

The convective heat transfer coefficient (

What is "Convective Heat Transfer Coefficient"?

The convective heat transfer coefficient, often denoted by the symbol \(h\), is a measure of the rate at which heat is transferred from a solid surface to a fluid or vice versa. Higher values indicate that heat is transferred more efficiently.

• In still air, heat transfer is slower due to lower convective heat transfer coefficients.
• In moving air, as provided in the problem, the coefficient is five times that of still air, resulting in faster cooling.

Role in Newton's Law of Cooling

Newton's Law of Cooling uses this coefficient in its equation:\[\frac{dT}{dt} = -h(T - T_s)\]Here:

• \(T\) is the object's temperature at time \(t\).
• \(T_s\) is the surrounding temperature.
• \(h\) is the convective heat transfer coefficient.

The presence of \(h\) in the equation makes it critical in determining the rate at which the temperature changes over time.

Differential Equation Integration

The integration of a differential equation helps us understand the change in temperature over time given the coefficients present.

Integrating Newton's Law of Cooling

The equation \(\frac{dT}{dt} = -h(T - T_s)\) describes the rate of temperature change and needs to be integrated to relate temperature change over time.To find how long it takes for the temperature to decrease to a certain percentage of its original difference, we integrate as follows:\[\int_{\Delta T_0}^{0.9\Delta T_0} \frac{dT}{T - T_s} = -\int_0^t h \, dt\]

• The left side of the equation integrates the temperature difference.
• The right side focuses on time and the convective coefficient.

This integration process enables us to find specific cooling times, like when the temperature of the casting falls to 90% of its original difference.

Temperature Difference Cooling

Temperature difference cooling refers to the process of an object decreasing its temperature relative to its surroundings, and it is significantly influenced by both environmental conditions and physical properties.

Understanding Cooling to a Percentage

In the problem, cooling occurs until the object reaches 90% of the original temperature difference with its environment. Let's define this difference as \(\Delta T_0\):- Initial temperature difference: \(\Delta T_0\)- Target temperature difference: \(0.9\Delta T_0\)This implies that the temperature drop is sufficient to account for 10% of the original gap between the object's initial temperature and its surroundings.

• Still air requires 30 minutes for this cooling process.
• In moving air, this time is reduced proportionally based on the increased convective coefficient.

By comparing how different conditions affect this cooling process, as in moving air, we deduce that certain environments can dramatically improve heat dissipation rates.