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A rigid, well-insulated tank consists of two compartments, each having the same volume, separated by a valve. Initially, one of the compartments is evacuated and the other contains \(0.25 \mathrm{lbmol}\) of a gas at \(50 \mathrm{lbf} / \mathrm{in} .^{2}\) and \(100^{\circ} \mathrm{F}\). The valve is opened and the gas expands to fill the total volume, eventually achieving an equilibrium state. Using the ideal gas model (a) determine the final temperature, in \({ }^{\circ} \mathrm{F}\), and final pressure, in lbf/in. \({ }^{2}\) (b) evaluate the exergy destruction, in Btu. (c) What is the cause of exergy destruction in this case? Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

Short Answer

Expert verified
Final temperature is 100°F, final pressure is 25 lbf/in². Exergy destruction occurs due to irreversibility in gas expansion.

Step by step solution

01

Identify Initial Conditions

One compartment initially contains 0.25 lbmol of gas at a pressure of 50 lbf/in² and a temperature of 100°F, while the other compartment is evacuated. The volumes of both compartments are the same.
02

Use Ideal Gas Law to Determine Initial Volume

Using the ideal gas law, calculate the initial volume: \[PV = nRT\] where: \(P = 50\ \text{lbf/in}^2\)\(n = 0.25\ \text{lbmol}\)\(R = 10.7316\ \text{psia}\cdot\text{ft}^3/\text{lbmol}\text{R}\)\(T = (100 + 459.67)\ \text{R}\)Compute volume using: \[V = \frac{nRT}{P}\].
03

Compute the Total Volume After Expansion

Since the gas expands into an evacuated compartment of the same volume, the new total volume \(V_{\text{total}}\) is twice the initial volume calculated in Step 2.
04

Determine Final Temperature

Since the process is adiabatic and insulated, no heat transfer occurs. For an ideal gas in an adiabatic process where volume changes but temperature is constant initially: \[T_{\text{final}} = T_{\text{initial}} = 100^\text{°F}\]
05

Use Ideal Gas Law to Find Final Pressure

Using the final volume \(V_{\text{total}}\) and final temperature \(T_{\text{final}}\), calculate the final pressure using the ideal gas law: \[P_{\text{final}} = \frac{nRT_{\text{final}}}{V_{\text{total}}}\]
06

Convert Final Pressure to lbf/in²

Ensure the final pressure result is expressed in lbf/in².
07

Evaluate Exergy Destruction (Part (b))

Exergy destruction is given by: \[X_{\text{destroyed}} = T_0 \Delta S_{\text{universe}}\] where \(T_0\) is the reference temperature \(70^\text{°F}\). Calculate the entropy change using the gas properties and conditions before and after the expansion.
08

Determine Causes of Exergy Destruction (Part (c))

The main cause of exergy destruction in this process is the irreversibility associated with the rapid expansion of the gas into the evacuated compartment.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics, expressed as \(PV = nRT\). It relates the pressure \(P\), volume \(V\), and temperature \(T\) of an ideal gas, with \(n\) representing the amount of substance (in moles), and \(R\) being the gas constant. This law is essential in determining the behavior of gases under various conditions. In the given exercise, we use the Ideal Gas Law to calculate the initial volume of the gas. By rearranging the formula as \(V = \frac{nRT}{P}\), we can input the initial conditions to find the volume. This calculation is crucial as it sets the stage for understanding how the gas will behave when it expands into the evacuated compartment.
Adiabatic Process
An adiabatic process is one in which no heat transfer occurs between the system and its surroundings. This means all changes in a system's state are purely due to internal energy variations, particularly work done by or on the system. In the given problem, the tank is well-insulated, indicating an adiabatic process as no heat can enter or leave. Consequently, the gas's temperature remains constant during expansion \(T_{\text{initial}} = T_{\text{final}}\). This principle helps simplify the calculation since we only consider work done by the expanding gas without accounting for heat transfer.
Exergy Destruction
Exergy represents the maximum useful work obtainable from a system. Exergy destruction occurs due to irreversibilities in real processes, signifying wasted potential. It quantifies energy that could not be converted into useful work. To evaluate exergy destruction for the solution, we use: \[X_{\text{destroyed}} = T_0 \times \text{Δ}S_{\text{universe}} \]. Here, \(T_0\) is the reference temperature provided, and \(ΔS_{\text{universe}}\) is the total entropy change. The reference temperature \(T_0 = 70^\text{°F}\) is used for this calculation, which gives a clear picture of how much potential work was lost due to the rapid, irreversible expansion of the gas.
Entropy Change
Entropy change \( \text{Δ}S \) is a measure of disorder or randomness in a system. In thermodynamics, it helps indicate the irreversibility of processes and the amount of useful energy converted into waste. In the exercise, the gas expands into an evacuated volume leading to an increase in entropy. Since the process is adiabatic, changes in entropy are related to changes in volume and pressure. Calculating \( \text{Δ}S_{\text{universe}} \) helps in determining the exergy destruction and understanding the extent of irreversibility. Using the entropy balance and the initial and final states of the gas, we determine the increase in entropy due to expansion. This step is critical for evaluating the thermal efficiency and environmental impact of the process.

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Most popular questions from this chapter

Air enters an insulated turbine operating at steady state with a pressure of 5 bar, a temperature of \(500 \mathrm{~K}\), and a volumetric flow rate of \(3 \mathrm{~m}^{3} / \mathrm{s}\). At the exit, the pressure is 1 bar. The isentropic turbine efficiency is \(76.7 \%\). Assuming the ideal gas model and ignoring the effects of motion and gravity, determine (a) the power developed and the exergy destruction rate, each in \(\mathrm{kW}\). (b) the exergetic turbine efficiency. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar.

\(7.92\) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas enters a turbine operating at steady state at 50 bar, \(500 \mathrm{~K}\) with a velocity of \(50 \mathrm{~m} / \mathrm{s}\). The inlet area is \(0.02 \mathrm{~m}^{2}\). At the exit, the pressure is 20 bar, the temperature is \(440 \mathrm{~K}\), and the velocity is \(10 \mathrm{~m} / \mathrm{s}\) The power developed by the turbine is \(3 \mathrm{MW}\), and heat transfer occurs across a portion of the surface where the average temperature is \(462 \mathrm{~K}\). Assume ideal gas behavior for the carbon dioxide and neglect the effect of gravity. Let \(T_{0}=298 \mathrm{~K}, p_{0}=1\) bar. (a) Determine the rate of heat transfer, in \(\mathrm{kW}\). (b) Perform a full exergy accounting, in \(\mathrm{kW}\), based on the net rate exergy is carried into the turbine by the carbon dioxide.

Steam enters a turbine operating at steady state at \(4 \mathrm{MPa}\), \(500^{\circ} \mathrm{C}\) with a mass flow rate of \(50 \mathrm{~kg} / \mathrm{s}\). Saturated vapor exits at \(10 \mathrm{kPa}\) and the corresponding power developed is \(42 \mathrm{MW}\). The effects of motion and gravity are negligible. (a) For a control volume enclosing the turbine, determine the rate of heat transfer, in MW, from the turbine to its surroundings Assuming an average turbine outer surface temperature of \(50^{\circ} \mathrm{C}\), determine the rate of exergy destruction, in MW. (b) If the turbine is located in a facility where the ambient temperature is \(27^{\circ} \mathrm{C}\), determine the rate of exergy destruction for an enlarged control volume including the turbine and its immediate surroundings so heat transfer takes place at the ambient temperature. Explain why the exergy destruction values of parts (a) and (b) differ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

Oxygen \(\left(\mathrm{O}_{2}\right)\) enters a well-insulated nozzle operating at steady state at \(80 \mathrm{lbf}\) in. \({ }^{2}, 1100^{\circ} \mathrm{R}, 90 \mathrm{ft} / \mathrm{s}\) At the nozle exit, the pressure is 1 lbf/in. \({ }^{2}\) The isentropic nozle efficiency is \(85 \%\). For the nozle, determine the exit velocity, in \(\mathrm{m} / \mathrm{s}\), and the exergy destruction rate, in Btu per lb of oxygen flowing. Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}\)

One lbmol of carbon monoxide gas is contained in a \(90-\mathrm{ft}^{3}\) rigid, insulated vessel initially at 5 atm. An electric resistor of negligible mass transfers energy to the gas at a constant rate of \(10 \mathrm{Btu} / \mathrm{s}\) for \(2 \mathrm{~min}\). Employing the ideal gas model and ignoring the effects of motion and gravity, determine (a) the change in exergy of the gas, (b) the electrical work, and (c) the exergy destruction, each in Btu. Let \(T_{0}=70^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm}\).

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