/*! 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 11 An 8-ft \({ }^{3}\) tank contain... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 11

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.

Short Answer

Expert verified
The final temperature of the air remaining in the tank is -22.51掳F.

Step by step solution

01

- List the Given Information

Identify and list all the given information: - Initial temperature \( T_i = 80^{\circ} F \) - Initial pressure \( P_i = 100 \ \text{lbf/in}^2 \) - Final pressure \( P_f = 30 \ \text{lbf/in}^2 \) - Volume of the tank \( V = 8 \ \text{ft}^3 \) - Constant rate of air leakage \( \dot{m} = 0.03 \ \text{lb/s} \) - Time period of leakage \( t = 90 \ \text{s} \)
02

- Convert Units Where Necessary

Convert initial and final temperatures to Rankine since we are using an ideal gas model: \[ T_i (\text{R}) = 80 + 459.67 = 539.67 \ \text{R} \]
03

- Calculate the Mass of Air Leaked

Calculate the total mass of air leaked using the equation \( m_{\text{leaked}} = \dot{m} \cdot t \): \[ m_{\text{leaked}} = 0.03 \ \frac{\text{lb}}{\text{s}} \times 90 \ \text{ s} = 2.7 \ \text{ lb} \]
04

- Apply the Ideal Gas Law

Use the ideal gas law in the form \( PV = mRT \). For the initial state: - \( P_i = 100 \ \text{lbf/in}^2 \) - \( V = 8 \ \text{ft}^3 \) - \( T_i = 539.67 \ \text{R} \) - \( R = 53.34 \ \text{ft}^3/\text{R}/\text{lbmol} \)
05

- Solve for Initial Mass

Rearrange the ideal gas equation to solve for the initial mass (m_i): \[ m_i = \frac{P_i V}{RT_i} \] \[ m_i = \frac{100 \ \text{lbf/in}^2 \times 8 \ \text{ft}^3}{53.34 \ \text{ft}^3/\text{R}/\text{lbmol} \times 539.67 \ \text{R}} \]
06

- Calculate Final Mass

Calculate the mass of air remaining after leakage using the equation \( m_f = m_i - m_{\text{leaked}} \): \[ m_f = m_i - 2.7 \ \text{lb} \]
07

- Apply the Ideal Gas Law for Final State

Use the ideal gas law for the final state: - \( P_f = 30 \ \text{lbf/in}^2 \) - \( V = 8 \ \text{ft}^3 \) - \( m_f \)
08

- Solve for Final Temperature

Rearrange the ideal gas equation to solve for the final temperature (T_f): \[ T_f = \frac{P_f V}{m_f R} \] Convert the final temperature from Rankine to Fahrenheit: \[ T_f (^{\circ} F) = T_f (\text{R}) - 459.67 \]

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Key Concepts

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

Initial and Final States
In the problem, we begin by understanding the initial and final states of the air inside the tank. Initially, the air has a temperature of 80掳F and a pressure of 100 lbf/in虏. The volume of the tank remains constant at 8 ft鲁 throughout the process. After the leak, the pressure drops to 30 lbf/in虏, and we need to determine the final temperature. Knowing these states helps us apply the ideal gas law effectively.
In problems involving gas leakage or any thermodynamic process, recording initial and final states accurately is crucial. Without these baseline measurements, it would be impossible to track how the system behaves over time.
Leakage Rate
The problem describes a leakage rate of air at 0.03 lb/s for a period of 90 seconds. This highlights a constant rate of mass loss over time. The total mass lost (\Delta m) is a product of the rate and the time duration of the leakage:
    \[ m_{\text{leaked}} = \dot{m} \cdot t \]

By multiplying the leakage rate (0.03 lb/s) by the time (90 s), we get a total mass lost of 2.7 lb. A consistent and accurate understanding of leakage rates allows us to make correct mass balance calculations for the initial and final states of the gas.
Unit Conversion
Before performing any calculations using the ideal gas law, it is essential to ensure all units are consistent for the calculations. This involves converting the initial and final temperatures into Rankine. The conversion from Fahrenheit to Rankine is given by:
    \[ T(\text{R}) = T(\text{掳F}) + 459.67 \]

Therefore, the initial temperature of 80掳F converts to 539.67 R. This unit conversion ensures compatibility with the gas constant used in the ideal gas law equation.
Mass Calculation
Using the ideal gas law, we convert given parameters to find the mass of air at various states. Initially, we apply the formula:
    \[PV = mRT\]
Rearranging to solve for mass:
    \[m_i = \frac{P_i V}{R T_i}\]
Using the given values, calculate the initial mass. Then, subtract the mass lost due to leakage:
    \[m_f = m_i - m_{\text{leaked}}\]
Finally, substitute into the ideal gas law to find the final temperature:
    \[T_f = \frac{P_f V}{m_f R}\]
Ensuring all units and values are appropriately applied in the calculation is key.

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Most popular questions from this chapter

Steam enters a turbine operating at steady state at \(2 \mathrm{MPa}\), \(360^{\circ} \mathrm{C}\) with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Saturated vapor exits at \(0.1 \mathrm{MPa}\) and a velocity of \(50 \mathrm{~m} / \mathrm{s}\). The elevation of the inlet is \(3 \mathrm{~m}\) higher than at the exit. The mass flow rate of the steam is \(15 \mathrm{~kg} / \mathrm{s}\), and the power developed is \(7 \mathrm{MW}\). Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine (a) the area at the inlet, in \(\mathrm{m}^{2}\), and (b) the rate of heat transfer between the turbine and its surroundings, in \(\mathrm{kW}\).

Air enters a compressor operating at steady state at \(1.05\) bar, \(300 \mathrm{~K}\), with a volumetric flow rate of \(12 \mathrm{~m}^{3} / \mathrm{min}\) and exits at 12 bar, \(400 \mathrm{~K}\). Heat transfer occurs at a rate of \(2 \mathrm{~kW}\) from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in \(\mathrm{kW}\).

Refrigerant \(134 a\) enters the evaporator of a refrigeration system operating at steady state at \(-4^{\circ} \mathrm{C}\) and quality of \(20 \%\) at a velocity of \(7 \mathrm{~m} / \mathrm{s}\). At the exit, the refrigerant is a saturated vapor at a temperature of \(-4^{\circ} \mathrm{C}\). The evaporator flow channel has constant diameter. If the mass flow rate of the entering refrigerant is \(0.1 \mathrm{~kg} / \mathrm{s}\), determine (a) the diameter of the evaporator flow channel, in \(\mathrm{cm}\). (b) the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\).

Figure P4.8 shows a mixing tank initially containing \(2000 \mathrm{lb}\) of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of \(0.8 \mathrm{lb} / \mathrm{s}\) and the other delivering cold water at a mass flow rate of \(1.2 \mathrm{lb} / \mathrm{s}\). Water exits through a single exit pipe at a mass flow rate of \(2.5 \mathrm{lb} / \mathrm{s}\). Determine the amount of water, in lb, in the tank after one hour.

Refrigerant \(134 \mathrm{a}\) enters a well-insulated nozzle at \(200 \mathrm{lbf} / \mathrm{in} .{ }^{2}, 220^{\circ} \mathrm{F}\), with a velocity of \(120 \mathrm{ft} / \mathrm{s}\) and exits at 20 lbf/in. \({ }^{2}\) with a velocity of \(1500 \mathrm{ft} / \mathrm{s}\). For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in \({ }^{\circ} \mathrm{F}\).
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