7.18. The compressed-air-driven water rocket shown in Fig. 7.26 is ejecting water vertically downward through a frictionless nozzle. The exit area of the nozzle is 1 in². When the pressure and elevation are as shown, how much thrust does the rocket produce? Air 5 psig h 8 ft FIGURE 7.26
7.18. The compressed-air-driven water rocket shown in Fig. 7.26 is ejecting water vertically downward through a frictionless nozzle. The exit area of the nozzle is 1 in². When the pressure and elevation are as shown, how much thrust does the rocket produce? Air 5 psig h 8 ft FIGURE 7.26
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:### Example 7.18: Calculating Thrust in a Compressed-Air Water Rocket
**Problem Statement:**
A compressed-air-driven water rocket, depicted in Figure 7.26, is ejecting water vertically downward through a frictionless nozzle. The nozzle has an exit area of 1 in². Given the pressure and elevation conditions as illustrated, calculate the thrust produced by the rocket.
**Details from Figure 7.26:**
- The rocket contains compressed air at a pressure of 5 psig (pounds per square inch gauge).
- Water is present below the compressed air.
- The height of the water column, \( h \), is 8 feet.
- The exit area of the nozzle is \( 1 \, \text{in}^2 \).
**Diagram Explanation:**
The diagram shows a vertical section of the water rocket with two distinct areas:
- **Upper Section:** Contains compressed air at 5 psig.
- **Lower Section:** Filled with water, which is being forced out through the nozzle.
The nozzle at the bottom is designed to eject water vertically downward, and the challenge is to determine the thrust this setup produces.
Understanding these factors is crucial to solving for the thrust using principles of fluid dynamics and physics.
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hi i am not seeing your unit conversion for step 2 can u plz clarify
![The expression shown is a calculation of the pressure exerted by water at a certain height. The equation is detailed as follows:
\[
\rho_{\text{water}} = \rho gh = \frac{62.4 \, \text{lbm}}{\text{ft}^3} \times \frac{32.2 \, \text{ft} \cdot \text{lbm}}{\text{s}^2 \cdot \text{lbf}} \times 8 \, \text{ft} = \frac{\text{lbm}}{\text{ft} \cdot \text{s}^2 \cdot \text{lbf}}
\]
This expression simplifies further to:
\[
16074.2 \, \frac{\text{lb}}{\text{in}^2} = 16.1 \, \text{Kpsi}
\]
**Explanation of Terms:**
- \(\rho\) represents the density of water.
- \(g\) is the acceleration due to gravity.
- \(h\) is the height of the water column.
- The units are converted from a mix of pounds mass (lbm), feet (ft), seconds (s), and pounds force (lbf), eventually resulting in pressure in pounds per square inch (psi).
- The final result is in kilopounds per square inch (Kpsi).](https://content.bartleby.com/qna-images/question/f4dfc165-1571-419f-99fc-a4587d76c340/f86985f6-9543-4e41-82bd-21b3f4a234fd/apkbcm_processed.png)
Transcribed Image Text:The expression shown is a calculation of the pressure exerted by water at a certain height. The equation is detailed as follows:
\[
\rho_{\text{water}} = \rho gh = \frac{62.4 \, \text{lbm}}{\text{ft}^3} \times \frac{32.2 \, \text{ft} \cdot \text{lbm}}{\text{s}^2 \cdot \text{lbf}} \times 8 \, \text{ft} = \frac{\text{lbm}}{\text{ft} \cdot \text{s}^2 \cdot \text{lbf}}
\]
This expression simplifies further to:
\[
16074.2 \, \frac{\text{lb}}{\text{in}^2} = 16.1 \, \text{Kpsi}
\]
**Explanation of Terms:**
- \(\rho\) represents the density of water.
- \(g\) is the acceleration due to gravity.
- \(h\) is the height of the water column.
- The units are converted from a mix of pounds mass (lbm), feet (ft), seconds (s), and pounds force (lbf), eventually resulting in pressure in pounds per square inch (psi).
- The final result is in kilopounds per square inch (Kpsi).
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