Heat flow on a cylindrical pipe: T=(T0-TR) sin((y)/(√(x^2+y^2)))+TR ; y ≥0 When a circular pipe

step 1 In this question, we're going to work with heat flow. We're going to determine the temperature o

Heat flow on a cylindrical pipe: T=(T0-TR)...

Heat flow on a cylindrical pipe: $T=\left(T_{0}-T_{R}\right) \sin \left(\frac{y}{\sqrt{x^{2}+y^{2}}}\right)+T_{R} ; y \geq 0$ When a circular pipe is exposed to a fan-driven source of heat, the temperature of the air reaching the pipe is greatest at the point nearest to the source (see diagram). As you move around the circumference of the pipe away from the source, the temperature of the air reaching the pipe gradually decreases. One possible model of this phenomenon is given by the formula shown, where $T$ is the temperature of the air at a point $(x, y)$ on the circumference of a pipe with outer radius $r=\sqrt{x^{2}+y^{2}}, T_{0}$ is the temperature of the air at the source, and $T_{R}$ is the surrounding room temperature. Assuming $T_{0}=220^{\circ} \mathrm{F}, T_{R}=72^{\circ}$ and $r=5 \mathrm{cm}:($ a) Find the temperature of the air at the points $(0,5),(3,4),(4,3),(4.58,2),$ and (4.9,1) (b) Why is the temperature decreasing for this sequence of points? (c) Simplify the formula using $r=5$ and use it to find two points on the pipe's circumference where the temperature of the air is $113^{\circ}$

Heat flow on a cylindrical pipe: $T=\left(T_{0}-T_{R}\right) \sin \left(\frac{y}{\sqrt{x^{2}+y^{2}}}\right)+T_{R} ; y \geq 0$
When a circular pipe is exposed to a fan-driven source of heat, the temperature of the air reaching the pipe is greatest at the point nearest to the source (see diagram). As you move around the circumference of the pipe away from the source, the temperature of the air reaching the pipe gradually decreases. One possible model of this phenomenon is given by the formula shown, where $T$ is the temperature of the air at a point $(x, y)$ on the circumference of a pipe with outer radius $r=\sqrt{x^{2}+y^{2}}, T_{0}$ is the temperature of the air at the source, and $T_{R}$ is the surrounding room temperature. Assuming $T_{0}=220^{\circ} \mathrm{F}, T_{R}=72^{\circ}$ and $r=5 \mathrm{cm}:($ a) Find the temperature of the air at the points $(0,5),(3,4),(4,3),(4.58,2),$ and (4.9,1)
(b) Why is the temperature decreasing for this sequence of points? (c) Simplify the formula using $r=5$ and use it to find two points on the pipe's circumference where the temperature of the air is $113^{\circ}$

Key Concepts

Temperature Distribution Modeling

This concept involves creating a mathematical model that describes how temperature varies over a surface or object. In the context of heat flow on a cylindrical pipe, the model uses a function to relate temperature to position on the pipe’s circumference, thereby quantifying how external heat affects the object at different locations.

Coordinate Transformation in Circular Geometry

Since the pipe is circular, points on its surface are naturally expressed in coordinates that satisfy the equation x² + y² = r². This concept involves understanding how cartesian coordinates relate to circular or cylindrical coordinates, allowing one to simplify the temperature function when the radius is fixed.

Trigonometric Function Behavior

Trigonometric functions, such as the sine function, are used to model periodic variations and angular dependencies in physical phenomena. In this context, the sine function captures how the impact of the heat source decreases as one moves farther around the pipe’s surface, reflecting the angular distribution of temperature.

Solving Trigonometric Equations

Many problems, including finding specific temperature values on the pipe, require solving equations that involve trigonometric functions. This concept includes techniques for isolating the trigonometric term, simplifying the equation, and then determining the angles (or corresponding coordinates) that satisfy the given condition.

Physical Interpretation of Mathematical Models

This idea stresses the importance of connecting mathematical formulations to their physical meanings. Here, the decrease in temperature around the pipe corresponds to the decrease in the sine function value as a function of the angle away from the heat source, thereby providing a concrete explanation of observable heat flow behavior.

Transcript

00:02 In this question, we're going to work with heat flow.

00:06 We're going to determine the temperature of the air reaching the pipe at certain points on the circumference of the pipe.

00:21 Using the equation t is equal to t0 minus t r sign of y over the square root of x squared plus y square plus t r.

00:39 Where our t -0 is equal to 220 or t r is equal to 72 or first point is 0 .05.

01:11 So t -not minus t r, that's 220 minus 72 is equal to 148.

01:26 So we have t being equal to 148 sine of y, or y in this case is 5.

01:41 And because our radius, which is equal to 5 in this case, is equal to the square root of x squared plus y squared, we have 5 over 5 plus t r, which is 72.

02:04 That's equal to 74 .58 degrees fahrenheit.

02:23 For next point is 3 .4.

02:31 Repeating the procedural, t is equal to 148, which represents t0 minus tr, sign of y, or y in this case is 4 over or radius, which is the constant 5, plus 70.

02:54 That's equal to 74 .07 degrees fahrenheit.

03:20 Or other point is for three.

03:30 Repeating the process, t is equal to 148 sign of y, or y in this case is 3 over or radius, which is equal to 5 plus 72, and that gives 73 .55 degrees fahrenheit.

04:05 The final point that we are considering is 4 .582.

04:22 T is equal to 148 sine over y, which is two in this case, over a radius 5, plus 72, and that's equal to 73 .03.

04:50 Degrees fahrenheit.

04:55 Looking at your anso, we notice that as or y decreases.

05:04 See y is moving from 5 to 4 to 3 to 2, as or why decreases, or temperature also decreases.

05:14 And that's because the sign or the argument of where sign function is also decreasing as y decreases.

05:24 Hence why the temperature.

05:26 Is decreasing.

05:32 Simplifying or formula that was given we are going to find the two points on the pipe circumference where which the temperature of the air is 113 degrees...

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