/*! 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 3 A \(0.5-\mathrm{m}^{3}\) tank co... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.5-\mathrm{m}^{3}\) tank contains ammonia, initially at \(40^{\circ} \mathrm{C}, 8\) bar. A leak develops, and refrigerant flows out of the tank at a constant mass flow rate of \(0.04 \mathrm{~kg} / \mathrm{s}\). The process occurs slowly enough that heat transfer from the surroundings maintains a constant temperature in the tank. Determine the time, in s, at which half of the mass has leaked out, and the pressure in the tank at that time, in bar.

Short Answer

Expert verified
Time: \frac{m_0}{0.08} \ \seconds\. Final pressure: Pressure for half mass at given temperature.

Step by step solution

01

- Determine Initial Mass of Ammonia

To find the initial mass of ammonia in the tank, use the specific volume. First, look up the specific volume, \( v \), of ammonia at \( 40^{\circ} \mathrm{C} \) and \( 8 \ \bar \). Use appropriate tables or software to find this value.
02

- Calculate Initial Mass

Using the specific volume, calculate the initial mass \( m_0 \) with the formula: \[ m_0 = \frac{V}{v} \] where \( V = 0.5 \mathrm{m}^3 \) is the volume of the tank. Plug in the specific volume value from Step 1.
03

- Determine Half Mass

Now, find half of the initial mass: \[ m_1 = \frac{m_0}{2} \]
04

- Calculate Time to Lose Half Mass

Given that refrigerant flows out at a mass flow rate \( \dot{m} = 0.04 \mathrm{kg/s} \), we can determine the time \( t \) it takes to lose half the mass with the formula: \[ t = \frac{m_0 - m_1}{\dot{m}} \]
05

- Determine Pressure at Half Mass

Now, to find the pressure when the mass is \( m_1 \), use the known temperature (remaining constant at \( 40^{\circ} \mathrm{C} \)) and look up the corresponding pressure for ammonia at this temperature and updated specific volume.
06

- Look Up or Calculate Pressure

Using the updated specific volume value (which can be found by \( \frac{V}{m_1} \)), look up or calculate the pressure in the ammonia tables.

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Key Concepts

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

Specific Volume
Specific volume is an important concept in thermodynamics. It is defined as the volume occupied by a unit mass of a substance. For ammonia in our exercise, we need to find the specific volume at given conditions of temperature and pressure. This is denoted by the symbol \( v \). To find it, you usually refer to thermodynamic tables or use specialized software. Once you have the specific volume, you can easily calculate the initial mass of the ammonia in the tank using the formula: \[ m_0 = \frac{V}{v} \]This translates to dividing the total volume of the tank by the specific volume of ammonia at the given conditions. For example, if the specific volume is found to be 0.25 m³/kg, with a 0.5 m³ tank, you would have:\[ m_0 = \frac{0.5 \, \text{m}^3}{0.25 \, \text{m}^3/\text{kg}} = 2 \, \text{kg} \]Specific volume helps bridge the gap between volume and mass, making it crucial for calculations involving gas or liquid flow.
Pressure Determination
Determining pressure in a tank across various stages of a thermodynamic process involves several steps. Initially, we know the pressure when the tank is full (8 bar at 40°C). As the ammonia leaks out, we need to calculate the remaining pressure when half the mass is left. Since temperature is constant, we use the new specific volume and reference the thermodynamic tables for ammonia.First, recalculate the specific volume at half mass. With the initial specific volume and mass known, find the new mass \( m_1 \):\[ m_1 = \frac{m_0}{2} \]Now calculate the new specific volume \( v_1 \):\[ v_1 = \frac{V}{m_1} \]Use the ammonia property tables to find the pressure corresponding to 40°C and the new specific volume value. These tables allow you to find the pressure for given temperatures and specific volumes accurately, thus solving for the pressure after some of the ammonia has leaked out.
Thermodynamic Properties
Thermodynamic properties are critical for understanding how substances behave under varying conditions of temperature, pressure, and volume. Key properties include temperature, pressure, specific volume, enthalpy, entropy, and internal energy. In our exercise, knowing these properties helps us to compute mass flow rate, pressure changes, and system behavior over time.
  • Temperature: It remains constant in this problem, simplifying calculations.
  • Pressure: Changes as mass leaks out and must be recalculated using updated specific volumes.
  • Specific Volume: It changes as the mass decreases, affecting pressure.
Combining these properties gives a comprehensive view of the system’s state. Understanding how to use thermodynamic tables to find specific values based on conditions is crucial. They help you match known values like temperature with unknowns such as pressure or specific volume.

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Most popular questions from this chapter

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at \(3.2\) bar, \(10^{\circ} \mathrm{C}\), and is compressed at steady state to \(10 \mathrm{bar}, 70^{\circ} \mathrm{C}\). The volumetric flow rate of refrigerant entering is \(3.0 \mathrm{~m}^{3} / \mathrm{min}\). The power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

Propane vapor enters a valve at \(1.6 \mathrm{MPa}, 70^{\circ} \mathrm{C}\), and leaves at \(0.5 \mathrm{MPa}\). If the propane undergoes a throttling process, what is the temperature of the propane leaving the valve, in \({ }^{\circ} \mathrm{C}\) ?

An insulated rigid tank is initially evacuated. A valve is opened and atmospheric air at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) enters until the pressure in the tank becomes 1 bar, at which time the valve is closed. Is the final temperature of the air in the tank equal to, greater than, or less than \(20^{\circ} \mathrm{C}\) ?

Refrigerant 134 a enters an insulated diffuser as a saturated vapor at 7 bars with a velocity of \(370 \mathrm{~m} / \mathrm{s}\). At the exit, the pressure is 16 bars and the velocity is negligible. The diffuser operates at steady state and potential energy effects can be neglected. Determine the exit temperature, in \({ }^{\circ} \mathrm{C}\).

Ammonia enters a heat exchanger operating at steady state as a superheated vapor at 14 bar, \(60^{\circ} \mathrm{C}\), where it is cooled and condensed to saturated liquid at 14 bar. The mass flow rate of the refrigerant is \(450 \mathrm{~kg} / \mathrm{h}\). A separate stream of air enters the heat exchanger at \(17^{\circ} \mathrm{C}, 1\) bar and exits at \(42^{\circ} \mathrm{C}, 1\) bar. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{min}\).

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