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Chapter 4: Problem 162

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

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Answer: 6 minutes

Step by step solution

01

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.
02

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.
03

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: \(\int_{ΔT_0}^{0.9ΔT_0} \frac{dT}{T - T_s} = -\int_0^t h dt\)
04

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: \(\int_{ΔT_0}^{0.9ΔT_0}\frac{dT}{T - T_s} = -30h_s\) \(\int_{ΔT_0}^{0.9ΔT_0}\frac{dT}{T - T_s} = -t_m(5h_s)\)
05

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).

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Most popular questions from this chapter

Lumped system analysis of transient heat conduction situations is valid when the Biot number is (a) very small (b) approximately one (c) very large (d) any real number (e) cannot say unless the Fourier number is also known.

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