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Air enters a turbine operating at steady state at \(500 \mathrm{kPa}\), \(860 \mathrm{~K}\) and exits at \(100 \mathrm{kPa}\). A temperature sensor indicates that the exit air temperature is \(460 \mathrm{~K}\). Stray heat transfer and kinetic and potential energy effects are negligible, and the air can be modeled as an ideal gas. Determine if the exit temperature reading can be correct. It yes, determine the power developed by the turbine for an expansion between these states, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of air flowing. If no, provide an explanation with supporting calculations.

Short Answer

Expert verified
The exit temperature reading of 460 K is not correct. The power developed by the turbine is 402 kJ/kg.

Step by step solution

01

- Given Data

List the provided data:- Inlet pressure: \(P_1 = 500 \text{ kPa}\)- Inlet temperature: \(T_1 = 860 \text{ K}\)- Exit pressure: \(P_2 = 100 \text{ kPa}\)- Exit temperature: \(T_2 = 460 \text{ K}\)- The process is adiabatic (no heat transfer)- Kinetic and potential energy effects are negligible- Air behaves as an ideal gas
02

- Use Ideal Gas Properties

The specific heat capacities for air can be approximated as constant over the given temperature range. Use the following specific heat values for air:\(c_p = 1.005 \text{ kJ/(kg·K)}\)\(c_v = 0.718 \text{ kJ/(kg·K)}\)Also, remember \(\text{R} = 0.287 \text{ kJ/(kg·K)}\)
03

- Apply the Isentropic Relation for Ideal Gas

Check if the process is isentropic using the relation for temperatures and pressures:\[\left( \frac{T_2}{T_1} \right) = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}}\]where \( k = \frac{c_p}{c_v} \approx 1.4\). Plug in the given values:\[\left( \frac{460}{860} \right) \stackrel{?}{=} \left( \frac{100}{500} \right)^{\frac{0.4}{1.4}}\]Calculate both sides of the equation to see if they're equal.
04

- Verify the Given Exit Temperature

Calculate the right-hand side of the isentropic relation:\[\left( \frac{100}{500} \right)^{\frac{0.4}{1.4}} \approx 0.528\]Now calculate the left-hand side:\[\left( \frac{460}{860} \right) \approx 0.535\]Since 0.528 is not equal to 0.535, the given exit temperature reading of 460 K cannot be correct for an ideal gas undergoing an isentropic process.
05

- Power Developed by the Turbine

Given the discrepancy in the temperature reading, we now can use the energy equation to calculate the actual power developed by the turbine. For an adiabatic process, the change in enthalpy equals the work done:\[W_t = \frac{h_1 - h_2}{1 \text{ kg}}\]Using \( h = c_p T \):\[W_t = c_p (T_1 - T_2) = 1.005 \text{ kJ/(kg·K)} (860 \text{ K} - 460 \text{ K}) = 1.005 (400) = 402 \text{ kJ/kg}\]
06

Conclusion

The exit temperature reading of 460 K is not consistent with the pressures provided under the assumption of an isentropic process. The actual power developed by the turbine, using the exit temperature in the calculation, is 402 kJ/kg.

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Key Concepts

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

Isentropic Process
An isentropic process is a type of thermodynamic process that is both adiabatic (no heat transfer) and reversible. In practical terms, this means the system's entropy remains constant throughout the process. For ideal gases, isentropic processes follow specific relationships between pressure and temperature. These relationships can help us determine how systems behave under ideal conditions. The equation relating pressure and temperature for an isentropic process in ideal gases is \(\frac{T2}{T1} = \frac{P2}{P1}^\frac{k-1}{k}\) where \(k\) is the ratio of specific heats \((k = c_p/c_v)\). This formula helps us predict output states of the system if one set of conditions is known.
Specific Heat Capacities
Specific heat capacities tell us how much energy is needed to change the temperature of a substance. For an ideal gas, we often deal with two specific heat capacities: \(c_p\) and \(c_v\). The specific heat capacity at constant pressure, \(c_p\), is the amount of heat added per unit mass per unit temperature increase when the pressure is constant. Meanwhile, \(c_v\) is the specific heat capacity at constant volume. These values are essential for calculations in thermodynamics because they allow us to relate temperature changes to energy changes. For air, typical values are \(c_p = 1.005 \text{kJ/(kg·K)}\) and \(c_v = 0.718 \text{kJ/(kg·K)}\). Knowing these helps us use equations in exercises like finding work done or energy transferred.
Adiabatic Process
An adiabatic process is one where no heat is transferred into or out of the system. In an adiabatic expansion or compression, the change in a system's energy comes purely from work done by or on the system. This makes adiabatic processes important in understanding idealized engines like gas turbines. For example, in our problem, we assume the turbine process is adiabatic, which simplifies equations by eliminating heat transfer terms. Adiabatic processes can be either reversible or irreversible. When reversible and adiabatic, the process is also isentropic. This interrelation allows us to apply formulas like \(W = c_p (T1 - T2)\), estimating performance under ideal conditions.
Enthalpy Change
In thermodynamics, enthalpy is a measure of the total energy of a system, including its internal energy and the work done by the system when changing volume at constant pressure. The change in enthalpy, \(\triangle h\), is crucial for understanding energy transfer in processes. During an adiabatic process in a turbine, the work done can be directly related to the change in enthalpy of the gas. The equation \(W_t = \frac{h1 - h2}{1 \text{kg}}\) is used to calculate the work developed by the turbine. By substituting \(h\) with \(c_pT\), we calculate how much power is generated by temperature changes alone, resulting in formulas like \(W_{t} = c_p (T1 - T2)\). In our example, the power developed by the turbine is 402 kJ/kg, demonstrating energy conversion in thermodynamic processes.

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Most popular questions from this chapter

One-tenth kmol of carbon monoxide \((\mathrm{CO})\) in a pistoncylinder assembly undergoes a process from \(p_{1}=150 \mathrm{kPa}\), \(T_{1}=300 \mathrm{~K}\) to \(p_{2}=500 \mathrm{kPa}, T_{2}=370 \mathrm{~K}\). For the process, \(\mathrm{W}=-300 \mathrm{~kJ}\). Employing the ideal gas model, determine (a) the heat transfer, in kJ. (b) the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Show the process on a sketch of the \(T-s\) diagram.

An electrically driven pump operating at steady state draws water from a pond at a pressure of 1 bar and a rate of \(50 \mathrm{~kg} / \mathrm{s}\) and delivers the water at a pressure of 4 bar. There is no significant heat transfer with the surroundings, and changes in kinetic and potential energy can be neglected. The isentropic pump efficiency is \(75 \%\). Evaluating electricity at \(8.5\) cents per \(\mathrm{kW} \cdot \mathrm{h}\), estimate the hourly cost of running the pump.

Water at 20 bar, \(400^{\circ} \mathrm{C}\) enters a turbine operating at steady state and exits at \(1.5\) bar. Stray heat transfer and kinetic and potential energy effects are negligible. A hard-to-read data sheet indicates that the quality at the turbine exit is \(98 \%\). Can this quality value be correct? If no, explain. If yes, determine the power developed by the turbine, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of water flowing.

A cylindrical copper rod of base area A and length \(L\) is insulated on its lateral surface. One end of the rod is in contact with a wall at temperature \(T_{\mathrm{H}}\). The other end is in contact with a wall at a lower temperature \(T_{\mathrm{C}}\). At steady state, the rate at which energy is conducted into the rod from the hot wall is $$ \dot{Q}_{\mathrm{H}}=\frac{\kappa \mathrm{A}\left(T_{\mathrm{H}}-T_{\mathrm{C}}\right)}{L} $$ where \(\kappa\) is the thermal conductivity of the copper rod. (a) For the rod as the system, obtain an expression for the time rate of entropy production in terms of \(\mathrm{A}, L, T_{\mathrm{H}}, T_{\mathrm{C}}\), and \(\kappa\). (b) If \(T_{\mathrm{H}}=327^{\circ} \mathrm{C}, T_{\mathrm{C}}=77^{\circ} \mathrm{C}, \kappa=0.4 \mathrm{~kW} / \mathrm{m} \cdot \mathrm{K}, \mathrm{A}=0.1 \mathrm{~m}^{2}\), plot the heat transfer rate \(\dot{Q}_{\mathrm{H}}\), in \(\mathrm{kW}\), and the time rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), each versus \(L\) ranging from \(0.01\) to \(1.0 \mathrm{~m}\). Discuss.

Refrigerant 22 enters the heat exchanger of an airconditioning system at \(80 \mathrm{lbf} / \mathrm{in}^{2}\) with a quality of \(0.2\). The refrigerant stream exits at \(80 \mathrm{lbf} / \mathrm{in}^{2}, 60^{\circ} \mathrm{F}\). Air flows in counterflow through the heat exchanger, entering at \(14.9 \mathrm{lbf}\) in. \(^{2}, 80^{\circ} \mathrm{F}\), with a volumetric flow rate of \(100,000 \mathrm{ft}^{3} / \mathrm{min}\) and exiting at \(14.5 \mathrm{lbf} / \mathrm{in}^{2}, 65^{\circ} \mathrm{F}\). Operation is at steady state, stray heat transfer from the outside of the heat exchanger to the surroundings can be neglected, and kinetic and potential energy effects are negligible. Assuming ideal gas behavior for the air, determine the rate of entropy production in the heat exchanger, in Btu/min \({ }^{\circ}{ }^{\circ} \mathrm{R}\).

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