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Chapter 4: Problem 118

A rigid, insulated tank, initially containing \(0.4 \mathrm{~m}^{3}\) of saturated water vapor at \(3.5\) bar, is connected by a valve to a large vessel holding steam at 15 bar, \(320^{\circ} \mathrm{C}\). The valve is opened only as long as required to bring the tank pressure to 15 bar. For the tank contents, determine the final temperature, in \({ }^{\circ} \mathrm{C}\), and final mass, in \(\mathrm{kg}\).

Short Answer

Expert verified
Final temperature: 320掳C, Final mass: 3.013 kg

Step by step solution

01

Identify the Initial Conditions

Initial volume of the tank, \(V_i\) is \(0.4 \, m^3\).Initial pressure of the tank, \(P_i\) is \(3.5 \, \text{bar}\).Since the tank initially contains saturated water vapor, use the steam tables to find the corresponding temperature and specific volume.
02

Find the Initial Specific Volume

From steam tables, for \(P_i = 3.5 \, \text{bar}\), the specific volume of saturated vapor, \(v_g\) is approximately \(0.501 \, \frac{m^3}{kg}\).
03

Calculate Initial Mass of Steam in Tank

Mass of steam, \(m_i = \frac{V_i}{v_g} = \frac{0.4 \, m^3}{0.501 \, \frac{m^3}{kg}} \approx 0.798 \, kg\).
04

Determine Final Pressure and Use the Steam Tables for Final Conditions

Final pressure, \(P_f\) is \(15 \, \text{bar}\). Use steam tables to get the final temperature, \(T_f\), and specific volume, \(v_f\), for steam at 15 bar.
05

Read the Steam Tables for 15 bar

For \(P_f = 15 \, \text{bar}\), use superheated steam tables to find the specific volume, \(v_f\), and the temperature. Given the large vessel at 15 bar and 320掳C, use superheated steam properties for accurate calculations.
06

Calculate Final Mass of Steam in the Tank

The final specific volume of steam at 15 bar and 320掳C: \(v_f = 0.1327 \, \frac{m^3}{kg}\). Then, the final mass of steam in the tank: \(m_f = \frac{V_i}{v_f} = \frac{0.4 \, m^3}{0.1327 \, \frac{m^3}{kg}} \approx 3.013 \, kg\).
07

Determine the Final Temperature

The final temperature, \(T_f\), is 320掳C, as determined from the superheated steam tables at 15 bar.

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Key Concepts

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

rigid insulated tank
In thermodynamics, a rigid insulated tank is a container where the volume remains constant, and no heat exchange occurs with the environment. This means:
  • The volume does not change, so the internal space remains the same.
  • Insulation prevents heat transfer into or out of the tank.
These conditions simplify calculations because the volume (V) is a fixed value, and we don't need to consider heat loss or gain. In this problem, the tank's rigid nature ensures that the volume remains at 0.4 m鲁 throughout the process, and its insulation means no heat is exchanged. This helps in applying the properties found in steam tables directly without additional thermal considerations.
saturated water vapor
Saturated water vapor is a state where water exists in the vapor phase but at a specific pressure and temperature where it is about to condense back into liquid. Key points about saturated water vapor:
  • It is in thermal equilibrium with liquid water at the same temperature.
  • Steam tables help us find the thermodynamic properties of saturated vapor at different pressures.
In this exercise, initially, the water vapor inside the tank is saturated at a pressure of 3.5 bar. Using steam tables, we can determine the corresponding temperature and specific volume, which are critical for subsequent calculations.
steam tables
Steam tables provide critical thermodynamic data of water in both liquid and vapor states at various temperatures and pressures. They include:
  • Specific volume (v): Volume occupied by a unit mass of steam.
  • Temperature (T): The temperature at which phase changes occur at specified pressures.
  • Other properties: Enthalpy, entropy, etc.
Using steam tables, for an initial pressure of 3.5 bar, we identified the specific volume of saturated vapor as 0.501 m鲁/kg. For the final pressure of 15 bar and given temperature of 320掳C, we use the superheated steam tables to find specific volumes and other properties required for final calculations.
specific volume
Specific volume (v) is defined as the volume occupied by a unit mass of a substance. It's given by:
\( v = \frac{V}{m} \)
where V is the volume and m is the mass. In steam tables, specific volume helps in finding the mass of water vapor when volume and specific volume are known, as shown:
\( m = \frac{V}{v} \)
Initially, for saturated vapor at 3.5 bar, the specific volume was 0.501 m鲁/kg and the volume was 0.4 m鲁. Using the relationship, we calculated the initial mass of steam as approximately 0.798 kg. Similarly, for the final condition at a specific volume of 0.1327 m鲁/kg and the same tank volume, we calculated the final mass.
mass calculation
Mass calculation in thermodynamics often involves using specific volume and volume. The formula used is:
\[ m = \frac{V}{v} \]
where:
  • m is the mass (in kg).
  • V is the volume of the tank (in m鲁).
  • v is the specific volume (in m鲁/kg).
For our exercise, we started with an initial specific volume of 0.501 m鲁/kg and volume of 0.4 m鲁, resulting in an initial mass of 0.798 kg. After the final condition change (15 bar, 320掳C, specific volume 0.1327 m鲁/kg), the final mass was calculated to be approximately 3.013 kg. Comparing these values helps verify the amount of steam added during the process.

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Most popular questions from this chapter

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