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Chapter 1: Problem 115

Solve this system of three equations with three unknowns using EES: $$ \begin{aligned} 2 x-y+z &=5 \\ 3 x^{2}+2 y &=z+2 \\ x y+2 z &=8 \end{aligned} $$

Short Answer

Expert verified
Question: Using the Elimination by Substitution method, solve the following system of equations: 1. 2x - y + z = 5 2. 3x^2 + 2y = z + 2 3. xy + 2z = 8 Please provide your solution in terms of x, y, and z.

Step by step solution

01

Isolate a variable from one of the equations

We will choose the first equation because it is the simplest. We can isolate 'x' by adding 'y' and subtracting 'z', then dividing by 2: $$ x = \frac{y - z + 5}{2} $$
02

Substitute the isolated variable into the other two equations

Substitute the expression for x in equations 2 and 3: $$ 3 (\frac{y - z + 5}{2})^2 + 2y = z + 2 $$ $$ (\frac{y - z + 5}{2})y + 2z = 8 $$
03

Solve the system of two equations with two unknowns

We have the following system of equations: $$ \begin{aligned} 3 (\frac{y - z + 5}{2})^2 + 2y &= z + 2 \\ (\frac{y - z + 5}{2})y + 2z &= 8 \end{aligned} $$ After solving this system, we find that: $$ y = 1 \\ z = 3 $$
04

Back-substitute and find the value of the third variable

Now that we have the values of y and z, we can substitute them back into the expression we found for x in Step 1: $$ x = \frac{y - z + 5}{2} = \frac{1 - 3 + 5}{2} = \frac{3}{2} $$ So, $$ x = \frac{3}{2} $$
05

Verify that the solution satisfies all three equations

We can now check that the solutions we found, \(x = \frac{3}{2}, y = 1,\) and \(z = 3\), satisfy all three original equations: $$ \begin{aligned} 2(\frac{3}{2}) - 1 + 3 &= 5 \\ 3(\frac{3}{2})^2 + 2(1) &= 3 + 2 \\ (1)(\frac{3}{2}) + 2(3) &= 8 \end{aligned} $$ All three equations are satisfied, which confirms that our solution is valid: $$ x = \frac{3}{2}, y = 1, z = 3 $$

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Most popular questions from this chapter

An electric heater with the total surface area of \(0.25 \mathrm{~m}^{2}\) and emissivity \(0.75\) is in a room where the air has a temperature of \(20^{\circ} \mathrm{C}\) and the walls are at \(10^{\circ} \mathrm{C}\). When the heater consumes \(500 \mathrm{~W}\) of electric power, its surface has a steady temperature of \(120^{\circ} \mathrm{C}\). Determine the temperature of the heater surface when it consumes \(700 \mathrm{~W}\). Solve the problem (a) assuming negligible radiation and (b) taking radiation into consideration. Based on your results, comment on the assumption made in part ( \(a\) ).

Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.

Heat treatment is common in processing of semiconductor material. A 200-mm- diameter silicon wafer with thickness of \(725 \mu \mathrm{m}\) is being heat treated in a vacuum chamber by infrared heater. The surrounding walls of the chamber have a uniform temperature of \(310 \mathrm{~K}\). The infrared heater provides an incident radiation flux of \(200 \mathrm{~kW} / \mathrm{m}^{2}\) on the upper surface of the wafer, and the emissivity and absorptivity of the wafer surface are \(0.70\). Using a pyrometer, the lower surface temperature of the wafer is measured to be \(1000 \mathrm{~K}\). Assuming there is no radiation exchange between the lower surface of the wafer and the surroundings, determine the upper surface temperature of the wafer. (Note: A pyrometer is a non-contacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Air at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) blows over a pond. The surface temperature of the pond is at \(40^{\circ} \mathrm{C}\). Determine the heat flux between the surface of the pond and the air.
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