/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 The temperature, \(T\), in an 11... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The temperature, \(T\), in an 11 -m-wide \(\times 11\) -m-long \(\times 4\) -m-high room varies according to the relation $$ T=20+3 x+4 y-2 z^{\circ} \mathrm{C} $$ where \(x[\mathrm{~m}]\) and \(y[\mathrm{~m}]\) are the coordinates in the horizontal plane and \(z[\mathrm{~m}]\) is the vertical coordinate measured from the floor upward. If a \(1.65-\mathrm{m}\) -tall person walks through the room in the \(x\) - direction at \(2 \mathrm{~m} / \mathrm{s}\), what is the rate of change of temperature at the top of the person's head?

Short Answer

Expert verified
The rate of change of temperature at the top of the person's head is 6 °C/s.

Step by step solution

01

Express the Position of the Person's Head

For a person walking in the room, the top of their head will be at a height of 1.65 m, therefore the vertical position, \(z\), is constant as \(z = 1.65\). Hence, the temperature as a function of \(x\) and \(y\) only can be written as \( T(x, y) = 20 + 3x + 4y - 2(1.65) \).
02

Simplify the Temperature Equation

Simplify the expression \( T(x, y) = 20 + 3x + 4y - 3.3 \) by performing the necessary arithmetic operations: \( T(x, y) = 16.7 + 3x + 4y \).
03

Determine the Direction of Movement

The person is moving in the \(x\)-direction. This implies that \(y\) remains constant as the person walks, specifically \(y = 0\). This further simplifies the equation to \(T(x) = 16.7 + 3x\).
04

Find the Rate of Change of Temperature with Respect to Time

To find the rate of change of temperature, differentiate \(T(x) = 16.7 + 3x\) with respect to \(x\) to get \(\frac{dT}{dx} = 3\). Since the person moves in the \(x\) direction at \(2\text{ m/s}\), the rate of change of temperature over time,\(\frac{dT}{dt}\), can be calculated using the chain rule: \(\frac{dT}{dt} = \frac{dT}{dx} \times \frac{dx}{dt} = 3 \times 2 = 6\text{ °C/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Variation
Understanding temperature variation in a room is crucial, especially in fluid mechanics. In this problem, the temperature in a room is defined by an equation that changes based on the position within the room, represented by coordinates. The formula given is:\[ T = 20 + 3x + 4y - 2z \]Where:
  • \(T\) is the temperature in degrees Celsius.
  • \(x\), \(y\), and \(z\) are the coordinates in meters.
This formula suggests that temperature increases with both the \(x\) and \(y\) coordinates and decreases with the \(z\) coordinate.
For a person moving horizontally through the room, the temperature experienced at the top of their head depends not on all the variables, but on \(x\) and \(y\) as their height, \(z\), remains constant. This understanding helps simplify the problem to focus only on the changing coordinates as the person moves, which is a common method in dealing with spatial variation problems in fluid mechanics.
Differential Equations
Differential equations are mathematical tools used to describe the relationship between functions and their derivatives. They are essential in fluid mechanics for modeling how various factors change over time or space.
In this exercise, we use a differential equation to determine the rate of temperature change as a person walks in the room. Given that our simplified temperature equation is:\[ T(x) = 16.7 + 3x \]We are concerned with how \(T\) changes as \(x\) changes. By taking the derivative of \(T\) with respect to \(x\), we find:\[ \frac{dT}{dx} = 3 \]This tells us that every meter the person moves in the x-direction causes the temperature to increase by 3 degrees Celsius.
Differential equations like these are powerful because they offer a clear way to quantify changes in a system. By understanding them, we can predict behavior in dynamic environments.
Rate of Change
Rate of change is a key concept that describes how one quantity changes in relation to another. In fluid mechanics, it is crucial for understanding how properties like temperature and pressure evolve over time.
In this exercise, the rate of change of temperature at the top of a person's head as they move is the focus. We've determined that \( \frac{dT}{dx} = 3 \), meaning a linear increase in temperature as a function of distance walked in the \(x\) direction.
Since the person walks at a speed of 2 meters per second, we apply the chain rule:\[ \frac{dT}{dt} = \frac{dT}{dx} \times \frac{dx}{dt} = 3 \times 2 = 6 \]So, the rate of change of temperature with respect to time is 6 degrees Celsius per second.
  • This result shows how fast the temperature felt increases as the person moves.
  • Understanding such rates of change is fundamental in fluid mechanics, where conditions often change rapidly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A two-dimensional velocity field is given by $$ \mathbf{v}=5 x \mathbf{i}+2 y \mathbf{j} \mathbf{~ m} / \mathrm{s} $$ where \(x\) and \(y\) are in meters. Determine the equation of the pathline that originates at the point (1,2) . For what range of times since release does the released particle remain within the domain of \(0 \leq x \leq 100 \mathrm{~m}, 0 \leq y \leq 100 \mathrm{~m}\). Plot the pathline within this domain.

Carbon dioxide flows at a rate of \(0.5 \mathrm{~kg} / \mathrm{s}\) through a 200 -mm-diameter conduit such that at a particular section, the absolute pressure is \(300 \mathrm{kPa}\) and the temperature is \(15^{\circ} \mathrm{C}\). Just downstream of this section, the diameter contracts to \(150 \mathrm{~mm}\). Estimate the difference in pressure between the uncontracted and contracted section. Assume that the flow is incompressible.

A fan draws air into a duct at a rate of \(2.5 \mathrm{~m}^{3} / \mathrm{s}\) from a room in which the temperature is \(26^{\circ} \mathrm{C}\) and the pressure is \(100 \mathrm{kPa}\). The diameter of the intake duct is \(310 \mathrm{~mm}\). Estimate the average velocity at which the air enters the duct and the mass flow rate into the duct.

Experiments indicate that the shear stress, \(\tau_{0}\), on the wall of a 200 -mm pipe can be related to the flow in the pipe using the relation $$ \tau_{0}=0.04 \rho V^{2} $$ where \(\rho\) and \(V\) are the density and velocity, respectively, of the fluid in the pipe. If water at \(20^{\circ} \mathrm{C}\) flows in the pipe at a flow rate of \(60 \mathrm{~L} / \mathrm{s}\) and the pipe is horizontal, estimate the pressure drop per unit length along the pipe.

An airplane operates at a low elevation where the pressure is \(101 \mathrm{kPa}\) and the temperature is \(20^{\circ} \mathrm{C}\). (a) Neglecting compressibility effects, estimate the flight speed at which the stagnation pressure is \(15 \%\) higher than the static pressure. (b) Estimate by how much the result in part (a) would differ if compressibility effects were taken into account.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.