/*! 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 4 Air modeled as an ideal gas ente... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 4

Air modeled as an ideal gas enters a combustion chamber at \(20 \mathrm{lbf} / \mathrm{in}^{2}\) and \(70^{\circ} \mathrm{F}\) through a rectangular duct, \(5 \mathrm{ft}\) by \(4 \mathrm{ft}\). If the mass flow rate of the air is \(830,000 \mathrm{lb} / \mathrm{h}\), determine the velocity, in \(\mathrm{ft} / \mathrm{s}\).

Short Answer

Expert verified
The velocity is approximately 111.82 ft/s.

Step by step solution

01

- Convert Pressure to Absolute Pressure

Add the atmospheric pressure to the given gauge pressure to convert it to absolute pressure. However, the given pressure is already in absolute terms, so no conversion is needed. \[P_1 = 20 \, \text{lbf/in}^2\]
02

- Convert Temperature to Rankine

Convert the given temperature from Fahrenheit to Rankine using the relationship: \[T_{1R} = T_{1F} + 459.67\]Substitute the given temperature: \[T_{1R} = 70 + 459.67 = 529.67 \, \text{R} \]
03

- Calculate the Specific Volume

Use the ideal gas law equation to find the specific volume of the air: \[(P_1 v_1 = R T_1)\]Where:\(P_1 = 20 \, \text{lbf/in}^2 = 20 \, \times 144 \, \text{lbf/ft}^2\)\(R = 53.35 \, \text{ft} \, \text{lbf/(lb} \, \text{R)}\)\(T_1 = 529.67 \, \text{R}\).Solve for specific volume \(v_1\): \[v_1 = \frac{R T_1}{P_1} = \frac{53.35 \, \text{ft} \, \text{lbf/(lb} \, \text{R)} \, \times 529.67 \, \text{R}}{20 \, \times 144 \, \text{lbf/ft}^2} \approx 9.7 \, \text{ft}^3/\text{lb}\]
04

- Calculate the Volume Flow Rate

Find the volume flow rate \(Q\) using the mass flow rate \(\dot{m}\) and the specific volume \(v_1\):\[\dot{m} = 830,000 \, \text{lb/h} \, \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 230.56 \, \text{lb/s}\]\[Q = \dot{m} \, v_1 = 230.56 \, \text{lb/s} \, \times 9.7 \, \text{ft}^3/\text{lb} \approx 2236.432 \, \text{ft}^3 / \, \text{s}\]
05

- Calculate the Duct Area

Calculate the cross-sectional area \(A\) of the duct:\[A = 5 \, \text{ft} \, \times 4 \, \text{ft} = 20 \, \text{ft}^2\]
06

- Determine the Velocity

Use the volume flow rate \(Q\) and the cross-sectional area \(A\) to find the velocity \(v\):\[v = \frac{Q}{A} = \frac{2236.432 \, \text{ft}^3/\text{s}}{20 \, \text{ft}^2} \approx 111.82 \, \text{ft/s}\]

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 91影视!

Key Concepts

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

Pressure Conversion
When working with gases, it is very important to use absolute pressure rather than gauge pressure. Absolute pressure includes atmospheric pressure. For example, if you have a gauge pressure of 20 lbf/in虏, you normally would add the atmospheric pressure (14.7 lbf/in虏 at sea level). However, in this case, the pressure is already provided in absolute terms, so no conversion is necessary. Remember, using absolute pressure ensures accurate calculations in equations like the Ideal Gas Law.
Temperature Conversion
Temperature conversion is crucial for thermodynamic calculations. In this example, we convert from Fahrenheit to Rankine. The conversion formula is straightforward: \(T_{R} = T_{F} + 459.67\). Given the temperature is 70掳F, the Rankine conversion is: \(70 + 459.67 = 529.67 \text{R}\). This new temperature in Rankine will be used in further calculations like the Ideal Gas Law.
Specific Volume
Specific volume denotes the volume occupied by a unit of mass of a substance. To find the specific volume of air, the Ideal Gas Law is used: \(P \times v = R \times T\). Here, \(P = 20 \text{ lbf/in}^2 = 20 \times 144 \text{ lbf/ft}^2\), \(R = 53.35 \text{ ft lbf/(lb R)}\), and \(T = 529.67 \text{ R}\). Solving for specific volume, \ v = \frac{R \times T}{P} = \frac{53.35 \times 529.67}{20 \times 144} \, you get approximately 9.7 ft鲁/lb. This value is essential for finding other properties like volume flow rate.
Volume Flow Rate
Volume flow rate represents the volume of fluid passing through a section per unit time. Using the mass flow rate and specific volume, we can find it: \(Q = \dot{m} \times v\). Here, \(\dot{m} = 830,000 \text{ lb/h} \times \frac{1 \text{ h}}{3600 \text{ s}} = 230.56 \text{ lb/s}\) and specific volume \(v = 9.7 \text{ ft}^3/\text{lb}\). So, \ Q = 230.56 \text{ lb/s} \times 9.7 \text{ ft}^3/\text{lb} \, which is approximately 2236.432 ft鲁/s. Knowing the volume flow rate helps in determining the velocity through the duct.
Duct Cross-Sectional Area
The cross-sectional area of the duct is a critical part of calculating gas velocity. The duct dimensions are provided as 5 ft by 4 ft. Therefore, the area is: \(A = 5 \text{ ft} \times 4 \text{ ft} = 20 \text{ ft}^2\). With this area and the previously calculated volume flow rate, velocity can be determined using the formula \ v = \frac{Q}{A} \. Here: \(v = \frac{2236.432 \text{ ft}^3/\text{s}}{20 \text{ ft}^2} = 111.82 \text{ ft/s}\). This final velocity tells you how fast the gas is moving through the duct.

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

Refrigerant 134 a enters a compressor operating at steady state as saturated vapor at \(0.12 \mathrm{MPa}\) and exits at \(1.2 \mathrm{MPa}\) and \(70^{\circ} \mathrm{C}\) at a mass flow rate of \(0.108 \mathrm{~kg} / \mathrm{s}\). As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of \(0.32 \mathrm{~kJ} / \mathrm{s}\). Determine at steady state the power input to the compressor, in \(\mathrm{kW}\).

Air expands through a turbine from 8 bar, \(960 \mathrm{~K}\) to 1 bar, \(450 \mathrm{~K}\). The inlet velocity is small compared to the exit velocity of \(90 \mathrm{~m} / \mathrm{s}\). The turbine operates at steady state and develops a power output of \(2500 \mathrm{~kW}\). Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Modeling air as an ideal gas, calculate the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\), and the exit area, in \(\mathrm{m}^{2}\).

A rigid tank whose volume is \(2 \mathrm{~m}^{3}\), initially containing air at 1 bar, \(295 \mathrm{~K}\), is connected by a valve to a large vessel holding air at \(6 \mathrm{bar}, 295 \mathrm{~K}\). The valve is opened only as long as required to fill the tank with air to a pressure of 6 bar and a temperature of \(350 \mathrm{~K}\). Assuming the ideal gas model for the air, determine the heat transfer between the tank contents and the surroundings, in \(\mathrm{kJ}\).

An 8-ft \({ }^{3}\) tank contains air at an initial temperature of \(80^{\circ} \mathrm{F}\) and initial pressure of \(100 \mathrm{lbf} / \mathrm{in}^{2}\). The tank develops a small hole, and air leaks from the tank at a constant rate of \(0.03 \mathrm{lb} / \mathrm{s}\) for \(90 \mathrm{~s}\) until the pressure of the air remaining in the tank is \(30 \mathrm{lbf} /\) in. \({ }^{2}\) Employing the ideal gas model, determine the final temperature, in \({ }^{\circ} \mathrm{F}\), of the air remaining in the tank.

A well-insulated rigid tank of volume \(15 \mathrm{~m}^{3}\) is connected to a large steam line through which steam flows at \(1 \mathrm{MPa}\) and \(320^{\circ} \mathrm{C}\). The tank is initially evacuated. Steam is allowed to flow into the tank until the pressure inside is \(p\). (a) Determine the amount of mass in the tank, in \(\mathrm{kg}\), and the temperature in the tank, in \({ }^{\circ} \mathrm{C}\), when \(p=500 \mathrm{kPa}\). (b) Plot the quantities of part (a) versus \(p\) ranging from \(0 \mathrm{kPa}\) to \(500 \mathrm{kPa}\).
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Measurements

Read Explanation

Atoms and Radioactivity

Read Explanation

Wave Optics

Read Explanation

Particle Model of Matter

Read Explanation

Fields in Physics

Read Explanation
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.