91Ó°ÊÓ

For any thermodynamic system, figuring out the initial mass is crucial. In this exercise, we have a tank which we know to have a specific volume and a specific pressure. First, note the given data:

• Volume of the tank, \( V = 0.38 \, \text{m}^3 \).
• Temperature, \( T = 400^{\text{\degree}} \text{C} \).
• Pressure, \( P = 3 \, \text{bar} \).
• Mass flow rate, \( \text{\dot{m}} = 0.005 \, \text{kg/s} \).

Using the steam tables for superheated steam at the given conditions, we find the specific volume, \( v_1 = 0.2164 \, \text{m}^3/\text{kg} \). The initial mass of steam in the tank can be calculated with the formula: \( m = \frac{V}{v} \). Plugging in the values, we get: \[ m_1 = \frac{0.38}{0.2164} \approx 1.756 \, \text{kg} \]Hence, the initial mass of steam in the tank is approximately 1.756 kg.

Specific volume is a property of steam that needs to be calculated at different states. Initially, the specific volume is given by steam tables for superheated steam at the conditions of \( 400^{\text{\degree}} \text{C} \) and \( 3 \, \text{bar} \), which is \( 0.2164 \, \text{m}^3/\text{kg} \). After a percentage of the mass is removed, we must recalculate this specific volume.
At the end of the discharge process, 75% of the initial mass remains. Therefore, we have: \[ m_2 = 0.75 \times m_1 = 0.75 \times 1.756 \approx 1.317 \, \text{kg} \]Now, the specific volume at the new condition, \( v_2 \), is: \[ v_2 = \frac{V}{m_2} = \frac{0.38}{1.317} \approx 0.288 \, \text{m}^3/\text{kg}: \]. This helps in determining the new pressure.

In thermodynamics, a constant temperature process is called an isothermal process. In this exercise, as steam is discharged, a heater maintains the temperature at exactly 400°C. This means the temperature does not change throughout; only the specific volume and pressure do. The heating ensures the system's internal energy remains constant despite the mass outflow.
Isothermal processes are significant because they simplify many calculations and make it easier to refer to standard tables (or charts) like steam tables.

The rate at which mass leaves the system is critical in solving many thermodynamics problems. In this problem, steam leaves the tank at a constant mass flow rate of \( 0.005 \, \text{kg/s} \).
Knowing this, we calculate the discharge time with the formula:\( t = \frac{m_{discharged}}{\text{\dot{m}}} \). Here, \( m_{discharged} = m_1 - m_2 \approx 0.439 \, \text{kg} \). Therefore, the calculation for the time is:\[ t = \frac{0.439}{0.005} = 87.8 \, \text{s} \]Thus, it takes approximately 87.8 seconds to discharge enough steam so that 75% of the initial mass remains in the tank.

After specific volume and temperature, the final step is to find the new pressure of the tank. Specific volume is retrieved earlier as \( 0.288 \, \text{m}^3/\text{kg} \). We can look up the steam tables again for superheated steam at \( 400^{\text{\degree}} \text{C} \) and this specific volume.
This allows us to find the new pressure, which is approximately: \( P_2 \approx 1.74 \, \text{bar} \).
This drop in pressure is connected to the volume remaining constant while the mass inside the tank is reduced.