A Rigid Uninsulated Tank Of Volume 20 M³ Is Connected To A S

1 A Rigid Uninsulated Tank Of Volume 20 M3 Is Connected To A Supply

Analyze the thermodynamic process involving a rigid, uninsulated tank of volume 2.0 m³ initially evacuated, connected to a supply pipe containing air at 150 kPa and 290 K. When the valve is opened, air flows into the tank until the pressure equilibrates at 150 kPa, with a final tank temperature of 380 K. Determine the total heat removed from the tank during this process.

Paper For Above instruction

The problem involves understanding the thermodynamic behavior of the air as it fills the rigid, uninsulated tank and the associated heat transfer during the process. Since the tank is rigid (fixed volume) and uninsulated (heat exchange with surroundings is possible), the process occurs at constant volume, with heat transfer and changes in internal energy and enthalpy to be considered. The initial and final states of the air, as well as properties such as temperature, pressure, and specific heats, are essential to evaluate the heat transfer involved.

Initially, the tank is evacuated, meaning it contains no air, thus zero mass, and no internal energy. When the valve opens, air from the supply pipe flows into the tank until the pressure reaches 150 kPa. It is given that the final temperature within the tank is 380 K. The main goal is to determine the total heat transferred out of the tank during this filling process. This problem involves applying the principles of the first law of thermodynamics at constant volume with heat transfer, considering the mass flow during the process, and using air properties for calculations.

Analysis of the Thermodynamic Process

Since the tank is initially empty and the process occurs until the final pressure and temperature are known, we need to find the mass of air in the tank at the final state. The process involves ideal gas behavior, so the properties of air can be approximated as those of an ideal gas with constant specific heats. The specific gas constant for air is R = 0.287 kJ/kg·K.

At the final state, the pressure and temperature are given as P₂ = 150 kPa and T₂ = 380 K. Using the ideal gas law, the mass of air in the tank at this state is:

m = (P₂ V) / (R T₂) = (150 kPa 2.0 m³) / (0.287 kJ/kg·K 380 K)

Converting units properly:

First, note that 150 kPa = 150 kJ/m³ (since 1 kPa·m³ = 1 kJ), so:

m = (150 kJ/m³ 2.0 m³) / (0.287 kJ/kg·K 380 K) = (300 kJ) / (0.287 * 380) kJ/kg

Calculating denominator: 0.287 * 380 ≈ 108.86

Thus, m ≈ 300 / 108.86 ≈ 2.75 kg

Energy Analysis and Heat Transfer

Applying the first law of thermodynamics for a steady flow process at constant volume, considering that the initial internal energy in the empty tank is zero (since no air), the energy balance for the process gives:

Q = ΔU + W

Since the tank is rigid, no work is done on or by the tank (W = 0), and the change in internal energy ∆U is given by:

∆U = m c_v (T₂ - T_initial)

The initial temperature of the air entering the tank is not specified directly, but considering the process, the supply air has T₁ = 290 K, and as the air fills and equilibrates, the final temperature is given as 380 K inside the tank.

Given the supply conditions and the initial evacuation, the initial state can be considered as zero mass and zero internal energy. The air entering raises the internal energy from zero to that associated with the mass m at T₂:

U_final = m c_v T₂

The heat removed from the tank (Q) during the process is equal to the change in the internal energy of the air in the tank if we consider the system as isolated besides heat transfer (no work done). Since the process involves heat exchange with the surroundings (as the tank is uninsulated), the actual amount of heat exchanged can be calculated by considering the energy balance:

Final Calculation

Using the specific heats of air: c_v ≈ 0.718 kJ/kg·K and c_p ≈ 1.005 kJ/kg·K. The change in internal energy of the air is:

∆U = m c_v (T₂ - T_initial)

But since the initial internal energy is zero (empty tank), and assuming all heat exchange occurs during the filling process, the total heat removed from the tank is:

Q = m c_v (T₂ - T_initial)

here, T_initial is essentially the temperature of the incoming air, assumed to be 290 K. Therefore:

Q = 2.75 kg 0.718 kJ/kg·K (380 K - 290 K) = 2.75 0.718 90 ≈ 177.3 kJ

Conclusion

Therefore, the total heat removed from the tank during the process is approximately 177.3 kJ. This indicates that heat was expelled during the filling process, likely due to temperature gradients and uninsulated conditions, allowing heat to flow out of the tank as the air temperature inside rises from the incoming air temperature to the final state at 380 K.

References

• Bejan, A. (2016). Advanced Engineering Thermodynamics. Wiley.