Question

A stationary gas-turbine power plant operates on an ideal regenerative Brayton cycle \((\epsilon=100 \text { percent })\) with air as the working fluid. Air enters the compressor at \(95 \mathrm{kPa}\) and \(290 \mathrm{K}\) and the turbine at \(880 \mathrm{kPa}\) and \(1100 \mathrm{K}\). Heat is transferred to air from an external source at a rate of \(30,000 \mathrm{kJ} / \mathrm{s}\) Determine the power delivered by this plant (a) assuming constant specific heats for air at room temperature and ( \(b\) ) accounting for the variation of specific heats with temperature.

Short answer

#Short Answer# In case (a), where we assume constant specific heats for air at room temperature, the power delivered by the Brayton cycle power plant can be calculated using the formula: Power delivered = \(\frac{\dot{Q}_{in}}{(T_3-T_2)} \cdot W_{net}\) For case (b), we would need to be given specific heat values at different temperatures or a table to reference to account for the variation of specific heats with temperature. With that information, we could use a similar approach to determine the power delivered.

Step-by-step solution

Analyzing the Brayton Cycle

First, let's analyze the Brayton cycle. A Brayton cycle consists of the following processes: 1-2: Isentropic compression in a compressor (from state 1 to state 2) 2-3: Constant-pressure heat addition in a combustion chamber (from state 2 to state 3) 3-4: Isentropic expansion in a turbine (from state 3 to state 4) 4-1: Constant-pressure heat rejection to complete the cycle (from state 4 to state 1)

Determine the pressures & temperatures at each state

Let's use the given values to find the pressures and temperatures at each state for both cases. Case (a): Assuming constant specific heats for air at room temperature. We can use the isentropic relationships: \(p_2=\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma -1}}p_1\) and \(p_4=\left(\frac{T_4}{T_3}\right)^{\frac{\gamma}{\gamma -1}}p_3\), where \(\gamma = \frac{C_p}{C_v} = \frac{1.005 \text{kJ/kg.K}}{0.718 \text{kJ/kg.K}} \approx 1.4\). We have state 1 and state 3 information: \(p_1=95 \text{kPa} , \ T_1=290\text{K}\) \(p_3=880 \text{kPa} , \ T_3=1100\text{K}\) Calculate \(T_2\) and \(T_4\): \(T_2 = T_1 \cdot \left(\frac{p_2}{p_1}\right)^{\frac{\gamma - 1}{\gamma}}\) \(T_4 = T_3 \cdot \left(\frac{p_4}{p_3}\right)^{\frac{\gamma - 1}{\gamma}}\) As \(p_2 = p_3\) and \(p_4 = p_1\), the equations for \(T_2\) and \(T_4\) become: \(T_2 = T_1 \cdot \left(\frac{p_3}{p_1}\right)^{\frac{\gamma -1}{\gamma}}\) \(T_4 = T_3 \cdot \left(\frac{p_1}{p_3}\right)^{\frac{\gamma -1}{\gamma}}\)

Calculating the work done in compressor and turbine

Next, we need to determine the work done in the compressor and in the turbine. Work done in the compressor: \(W_c = C_p \cdot (T_2 - T_1)\) Work done in the turbine: \(W_t = C_p \cdot (T_3 - T_4)\)

Calculating the net work done in the cycle

We can find the net work done in the cycle as: \(W_{net} = W_t - W_c\)

Calculating the power delivered in each case

Finally, let's calculate the power delivered for each case using the mass flow rate of air. Power delivered = \(mass \ flow \ rate \cdot W_{net}\) The mass flow rate of air can be determined from the given heat addition rate: \(\dot{Q}_{in} = \dot{m}C_p(T_3-T_2)\) We can solve for mass flow rate: \(\frac{\dot{Q}_{in}}{C_p(T_3-T_2)} = \dot{m}\) Now we can calculate the power delivered by the plant in each case. Case (a): Power delivered = \(\frac{\dot{Q}_{in}}{(T_3-T_2)} \cdot W_{net}\) Case (b): Since we do not have specific information for specific heats at higher temperatures, we can't solve case (b). However, if given specific heat values at different temperatures or a table to reference, we could use a similar approach to solve this case.

Key concepts

Isentropic Processes

Isentropic processes are adiabatic (no heat transfer) processes in which entropy remains constant. This is an idealization often used in cycle analysis since real processes involve some irreversible phenomena, leading to entropy changes. In the context of the Brayton cycle, isentropic processes are encountered during the compression (from state 1 to state 2) and expansion (from state 3 to state 4) stages. Since air behaves as an ideal gas under these conditions, we use isentropic relationships to link pressures and temperatures at different states of the cycle, as demonstrated in the exercise solution.

During isentropic compression, air is compressed to a higher pressure with an increase in temperature but no change in entropy. Likewise, during isentropic expansion, air expands, doing work while dropping in temperature without a change in entropy. Understanding isentropic processes is critical for calculating the work done in the compressor and turbine, and consequently the net power output of the Brayton cycle power plant.

Specific Heat Capacity

Specific heat capacity, symbolized by \(C_p\) for constant pressure and \(C_v\) for constant volume, is a measure of the heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). For air and many other gases, these values can vary with temperature but are often considered constant for simplified cycle analysis, as seen in the stated problem where \(C_p\) is assumed constant for case (a).

The ratio of specific heat capacities, \(\gamma = C_p/C_v\), is essential in the equations used to analyze isentropic processes. Specific heat capacity is particularly important not just for calculating the work done by the compressor and the turbine but also for determining the heat transfer rates within the cycle's constant-pressure processes.

Regenerative Brayton Cycle

The regenerative Brayton cycle introduces a heat exchanger between the turbine exit and the compressor exit to preheat the compressed air before it enters the combustion chamber. This process improves the cycle's efficiency by recovering some of the energy from the exhaust gases that would otherwise be lost. The ideal regenerative Brayton cycle assumes 100% effectiveness of this heat exchange process, as noted by \(\epsilon=100 \text{ percent}\) in the given exercise.

Implementing regeneration in a Brayton cycle power plant means the air entering the combustion chamber is already preheated, reducing the external heat necessary to reach the high temperatures required for efficient engine operation. Regeneration is a key factor in increasing the thermal efficiency of power plants, as it effectively uses the thermal energy within the system itself.

Constant-Pressure Heat Addition

Understanding Heat Addition at Constant Pressure

Within the Brayton cycle, constant-pressure heat addition occurs in the combustion chamber after the compression stage. During this process, energy is supplied to the heated compressed air (state 2) by burning fuel, which increases the temperature of the air to a significantly higher level (state 3) while maintaining constant pressure. As given in the problem statement, heat is transferred at a rate of \(30,000 \mathrm{kJ} / \mathrm{s}\).

This constant-pressure process can be represented on a T-s (temperature-entropy) diagram as a horizontal line to the right, indicating an increase in temperature and entropy. The heat addition process is central to the power generation capability of the Brayton cycle, as the high-energy air-fuel mixture is the source of the turbine's driving force.

Isentropic Expansion and Compression

Isentropic expansion refers to the expansion of a gas through a turbine where it does work on the surroundings and experiences a drop in both pressure and temperature without heat exchange, symbolizing no change in entropy. Conversely, isentropic compression involves compressing the gas in a compressor, resulting in increased pressure and temperature, again without any change in entropy.

In practice, these processes are not perfectly isentropic due to inherent irreversibilities. However, for simplicity in calculations and conceptual understanding, it is assumed there are no losses from these processes in the exercise provided. This assumption aids in the calculation of work and efficiency analyses for the power plant cycle. The work done by both the compressor (\(W_c\)) and turbine (\(W_t\)) can be described by the simple relation involving the constant-pressure specific heat capacity \(C_p\) and the temperature changes during these stages, as derived in the exercise.