91Ó°ÊÓ

\(7.92\) 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.

Step by step solution

01

- Apply the First Law of Thermodynamics

Start by applying the First Law of Thermodynamics to the control volume (the turbine) for a steady-state process: \[ W_{\text{turbine}} = \frac{dQ}{dt} + \frac{dE_{\text{ke}}}{dt} + \frac{dE_{\text{pe}}}{dt} + m \times (h_1 - h_2) \]Since the effect of gravity is neglected, the term \(\frac{dE_{\text{pe}}}{dt}\) is zero. Therefore, \[ W_{\text{turbine}} = \frac{dQ}{dt} + m \times (\frac{v_1^2 - v_2^2}{2}) + m \times (h_1 - h_2) \]Rearrange to solve for the rate of heat transfer: \[ \frac{dQ}{dt} = W_{\text{turbine}} - m \times (h_1 - h_2) - m \times (\frac{v_1^2 - v_2^2}{2}) \]

02

- Determine mass flow rate

The mass flow rate \( m \) can be determined using the ideal gas equation: \[ m = \frac{P_1 A_1}{R T_1} \]Given in the problem: - Inlet pressure \( P_1 = 50 \text{ bar} \) - Inlet temperature \( T_1 = 500 \text{ K} \) - Inlet area \( A_1 = 0.02 \text{ m}^2 \)Using \( R = 0.1889 \text{ kJ/(kg·K)} \) for CO_2, \[ m = \frac{50 × 10^5 \times 0.02}{0.1889 \times 500} = 10.57 \text{ kg/s} \]

03

- Compute specific enthalpies

Assuming ideal gas behavior, we can use specific heat capacities to determine specific enthalpy: \[ h_1 - h_2 = c_p (T_1 - T_2) \]Given, - \( T_1 = 500 \text{ K} \) - \( T_2 = 440 \text{ K} \)Using the specific heat capacity \( c_p = 0.844 \text{ kJ/(kg·K)} \) for CO_2, \[ h_1 - h_2 = 0.844 \times (500 - 440) = 50.64 \text{ kJ/kg} \]

04

- Evaluate kinetic energy terms

Compute the kinetic energy term in the energy equation: \[ \frac{v_1^2 - v_2^2}{2} \]Given in the problem: - Inlet velocity \( v_1 = 50 \text{ m/s} \) - Exit velocity \( v_2 = 10 \text{ m/s} \) \[ \frac{50^2 - 10^2}{2} = 1200 \text{ J/kg} = 1.2 \text{ kJ/kg} \]

05

- Calculate heat transfer rate

Now substitute into the First Law equation to find the heat transfer rate: \[ 3 \times 10^3 = \frac{dQ}{dt} + 10.57 \times (50.64) + 10.57 \times (1.2) \]This simplifies to: \[ 3000 = \frac{dQ}{dt} + 540.04 + 12.684 \] \[ \frac{dQ}{dt} = 3000 - 552.724 = 2447.276 \text{ kW} \]

06

- Exergy rate

Using the definitions for exergy, compute the exergy rate: \[ \dot{E}_{in} = m \Big( h_1 - h_2 - T_0 (s_1 - s_2) + \frac{{v_1^2}}{2} - \frac{{v_2^2}}{2} \Big) \]To simplify: T_1/T_2 parts: \[ \Big( c_p(T_1 - T_2) - T_0 c_p ln(T_1/T_2) \Big) \]

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Key Concepts

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

steady-state process

In thermodynamics, a steady-state process refers to a situation where the properties at any given point in the system do not change over time. This means there's no accumulation of mass or energy within the system boundaries. The conditions are stable. For example, in the turbine problem, since it's a steady-state operation, the mass flow rate, energy flow, and other thermodynamic properties such as temperature and pressure at both the inlet and the outlet, remain constant over time.
This concept simplifies our calculations, as we don’t have to consider changes within the system, but only the exchanges at the boundaries.

enthalpy

Enthalpy, often represented as 'h', is a measure of the total energy of a thermodynamic system, including internal energy and the product of its pressure and volume. It's a useful concept since it simplifies the energy analysis of open systems like turbines and compressors.
For instance, in the turbine problem, the change in specific enthalpy \(h_1 - h_2\) is a key term in the energy balance equation. Assuming ideal gas behavior, we used the specific heat capacity at constant pressure \(c_p\) to compute enthalpy changes with the formula:
\[h = c_p \times T \]
This helps us understand the energy transformation as the gas expands through the turbine.

mass flow rate

The mass flow rate is the amount of mass passing through a cross-section of a system per unit of time. It’s often denoted as 'm˙'.
In the problem, the mass flow rate of the COâ‚‚ entering the turbine is calculated using the ideal gas law in conjunction with the given inlet conditions. The formula used is:
\[ m = \frac{P_1 A_1}{R T_1} \]
With all the given values, we computed the mass flow rate as 10.57 kg/s. This is important as it helps to determine how much matter and energy enter and leave the system.

kinetic energy

Kinetic energy (KE) of a fluid flow is the energy due to its velocity. In thermodynamic systems like turbines, kinetic energy changes are considered when calculating work and heat transfer.
The kinetic energy per unit mass is given by:
\[ \frac{v^2}{2} \]
In our turbine problem, the difference in kinetic energy between the inlet and outlet is \[ \frac{50^2 - 10^2}{2} = 1200 \text{ J/kg} \].
While this change might seem negligible in larger systems, it can significantly impact the overall energy balance in precision calculations.

ideal gas behavior

Under ideal gas behavior, gases follow the ideal gas law:
\[ PV = nRT \]
Where P is pressure, V is volume, T is temperature, n is the number of moles, and R is the universal gas constant.
Assuming ideal gas behavior simplifies our calculations because it allows us to use straightforward relationships between pressure, volume, and temperature. For COâ‚‚ in our turbine, ideal gas assumptions help in determining properties like enthalpy and mass flow rate effectively.

turbine efficiency

Turbine efficiency describes how effectively a turbine converts the energy in the working fluid to mechanical work. It's given by the ratio of the actual work produced by the turbine to the maximum possible work:
\[ \text{Efficiency} = \frac{W_{\text{actual}}}{W_{\text{ideal}}} \]
In our problem, though efficiency was not directly calculated, the power output of 3 MW indicates the real work output, which is crucial for understanding the practical performance of the turbine.

exergy analysis

Exergy analysis identifies the maximum usable work obtainable from a system. Unlike energy, exergy also accounts for environmental factors and irreversibilities. In our turbine, the exergy change considers enthalpy, entropy, and kinetic energy changes to provide a full picture of the available work potential:
\[ \text{Net Exergy Rate} = m \times \bigg( h_1 - h_2 - T_0 (s_1 - s_2) + \frac{{v_1^2}}{2} - \frac{{v_2^2}}{2} \bigg) \]This comprehensive approach ensures we evaluate the system's performance based on maximum efficiency relative to the surroundings.