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Chapter 1: Problem 109

1.109 Air enters the converging nozzle shown in Fig. P1.72 at \(T_{1}=70^{\circ} \mathrm{F}\) ard \(V_{1}=50 \mathrm{ft} / \mathrm{s} .\) At the exit of the nozzle, \(V_{2}\) is given by \\[ V_{2}=\sqrt{V_{1}^{2}+2 c_{p}\left(T_{1}-T_{2}\right)} \\] where \(c_{p}=187 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\) and \(T_{2}\) is the air temperature at the exit of the nozzle, Find the temperature \(T_{2}\) for which \(V_{2}=\) \\[ 1000 \mathrm{ftfs} \\]

Short Answer

Expert verified
Calculate the resulting expression for \(T_{2}\). This returns the air temperature at the exit of the nozzle given that the air exit velocity is \(1000 \mathrm{ft} /\mathrm{s}\).

Step by step solution

01

Rearrange the Equation for \(T_2\)

We first need to rearrange the equation \(V_{2} = \sqrt{V_{1}^{2} + 2c_{p}(T_{1} - T_{2})}\) to isolate \(T_2\) on one side of the equation. Squaring both sides, and then solving for \(T_2\), we get \[T_{2} = T_{1} - \frac{{V_{2}^{2} - V_{1}^{2}}}{{2c_{p}}}\] This allows us to insert the given or known values into the equation.
02

Insert Known Values Into the Equation

We're given \(T_{1} = 70^{\circ} \mathrm{F}\), \(c_{p} = 187 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\), \(V_{1} = 50 \mathrm{ft} / \mathrm{s}\) and \(V_{2} = 1000 \mathrm{ft} / \mathrm{s}\). We substitute these given values into the equation.\[T_{2} = 70^{\circ} \mathrm{F} - \frac{{(1000 \mathrm{ft} / \mathrm{s})^{2} - (50 \mathrm{ft} / \mathrm{s})^{2}}}{{2 \times 187 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}}}\]
03

Evaluate the Expression for \(T_2\)

Next we perform the mathematical operations to solve for \(T_{2}\). This gives us the final temperature at the nozzle exit. Only one answer will be produced by these calculations.

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Most popular questions from this chapter

Assume that the air volume in a small automobile tire is constant and equal to the volume between two concentric cylinders \(13 \mathrm{cm}\) high with diameters of \(33 \mathrm{cm}\) and \(52 \mathrm{cm}\). The air in the tire is initially at \(25^{\circ} \mathrm{C}\) and \(202 \mathrm{kPa}\). Immediately after air is pumped into the tire, the temperature is \(30^{\circ} \mathrm{C}\) and the pressure is 303 kPa. What mass of air was added to the tire? What would be the air pressure after the air has cooled to a temperature of \(0^{\circ} \mathrm{C} ?\)

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be \(-50^{\circ} \mathrm{C}\) and \(900 \mathrm{Pa}\). respectively. (a) Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the Earth's atmosphere during a spring day when the temperature is \(18^{\circ} \mathrm{C}\) and the pressure \(101.6 \mathrm{kPa}(\mathrm{abs})\).

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A compressed air tank in a service station has a volume of \(10 \mathrm{ft}^{3} .\) It contains air at \(70^{\circ} \mathrm{F}\) and 150 psia. How many tubeless tires can it fill to 44.7 psia at \(70^{\circ} \mathrm{F}\) if each tire has a volume of 1.5 \(\mathrm{ft}^{3}\) and the compressed air tank is not refilled? The tank air temperature remains constant at \(70^{\circ} \mathrm{F}\) because of heat transfer through the tank's large surface area.

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