Question

An adiabatic air compressor is to be powered by a direct-coupled adiabatic steam turbine that is also driving a generator. Steam enters the turbine at \(12.5 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) at a rate of \(25 \mathrm{kg} / \mathrm{s}\) and exits at \(10 \mathrm{kPa}\) and a quality of \(0.92 .\) Air enters the compressor at \(98 \mathrm{kPa}\) and \(295 \mathrm{K}\) at a rate of \(10 \mathrm{kg} / \mathrm{s}\) and exits at \(1 \mathrm{MPa}\) and \(620 \mathrm{K}\) Determine \((a)\) the net power delivered to the generator by the turbine and \((b)\) the rate of entropy generation within the turbine and the compressor during this process.

Short answer

Answer: To find the net power delivered to the generator (a) and the rate of entropy generation within the turbine and the compressor (b), follow these steps: 1. Calculate the work output of the steam turbine using the specific enthalpies at the inlet and exit states and the mass flow rate of the steam. 2. Calculate the work input of the air compressor using the specific heat capacity at constant pressure, the change in temperature, and the mass flow rate of the air. 3. Subtract the work input of the compressor from the work output of the turbine to find the net power delivered to the generator (a). 4. Determine the specific entropy changes for both the steam and the air and multiply them by their respective mass flow rates to find the rate of entropy generation within the turbine and the compressor (b).

Step-by-step solution

Calculate the work output of the steam turbine

To calculate the work output of the steam turbine, we first need to find the specific enthalpy of the steam at both the inlet and exit states. Using the given steam table or appropriate thermodynamics software, we can find the specific enthalpies: \(h_1\) (inlet) = Specific enthalpy at \(12.5 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) \(h_2\) (exit) = Specific enthalpy at \(10 \mathrm{kPa}\) and a quality of \(0.92\) Now we can calculate the work output of the steam turbine using the mass flow rate of the steam and the difference in specific enthalpies: \(W_{turbine} = \dot{m}_{steam}(h_1 - h_2)\)

Calculate the work input of the air compressor

Using the given data, we can find the specific heat capacity at constant pressure \(c_p\) for the air and apply the adiabatic process equation to determine the work input of the air compressor: \(W_{compressor} = \dot{m}_{air} c_p (T_2 - T_1)\)

Calculate the net power delivered to the generator

Now we can calculate the net power delivered to the generator by the turbine by subtracting the work input of the compressor from the work output of the turbine: \(W_{net} = W_{turbine} - W_{compressor}\) This will give us the net power delivered to the generator (a).

Calculate the rate of entropy generation within the turbine and the compressor

To calculate the rate of entropy generation within the turbine and the compressor, we will first determine the specific entropy changes for both the steam and the air and then multiply them by their respective mass flow rates: \(\Delta s_{steam} = s_2 - s_1\) \(\Delta s_{air} = c_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1}\right)\) Now we can calculate the rate of entropy generation for both the turbine and the compressor: \(\dot{S}_{gen,turbine} = \dot{m}_{steam} \Delta s_{steam}\) \(\dot{S}_{gen,compressor} = \dot{m}_{air} \Delta s_{air}\) The sum of these two values will give us the rate of entropy generation within the turbine and the compressor (b).

Key concepts

Steam Turbine

A steam turbine is a device that converts thermal energy in steam to mechanical energy, which in turn can be used to produce electricity. In our scenario, steam enters the turbine at high pressure and temperature, expands through the turbine stages, and exits at a lower pressure and quality. The quality of steam refers to the proportion of steam that is in vapor phase as opposed to droplets of liquid water. The turbine's ability to do work is expressed by the equation:
\( W_{turbine} = \.m_{steam}(h_1 - h_2) \)
where:\( \.m_{steam} \) is the mass flow rate of the steam, \( h_1 \) is the specific enthalpy of the steam entering the turbine, and \( h_2 \) is the specific enthalpy as it leaves. In essence, the efficiency of a steam turbine is driven by the magnitude of the enthalpy drop and the mass flow rate of the steam.

Air Compressor

An air compressor, conversely, takes in air at a lower pressure and temperature and compresses it to a higher pressure and temperature. It's a crucial component in various industries, playing a vital role in systems such as pneumatic controls, HVAC, and in this case, aiding in the power generation process by providing compressed air to aid combustion or other processes. The work required to power the compressor can be derived using an adiabatic process equation:
\( W_{compressor} = \.m_{air} c_p (T_2 - T_1) \)
This shows that the work input for the compressor is influenced by the mass flow rate of the air \( \.m_{air} \), the specific heat at constant pressure \( c_p \), and the temperature change from \( T_1 \) to \( T_2 \). Optimizing this process is crucial because the air compressor consumes energy, reducing the overall efficiency of the system.

Entropy Generation

Entropy generation is a measure of the irreversible processes occurring within a system. In thermodynamics, it’s often a focal point since it aligns with the Second Law of Thermodynamics which states that the entropy of an isolated system never decreases over time. It's an indicator of lost work potential or wasted energy. For our application involving a steam turbine and an air compressor, we have two main contributors to entropy generation:

• The steam as it expands through the turbine.
• The air as it is compressed in the compressor.

The mathematical formulation to calculate the rate of entropy generation in both cases is based on changes in specific entropy multiplied by the mass flow rates. Reduced entropy generation is generally desired to increase system efficiency.

Specific Enthalpy

Specific enthalpy is a property of a substance, expressed as energy per unit mass, which reflects both the internal energy and the work that would need to be done to bring the substance to its location without acceleration from a specified datum level. It is particularly crucial in energy conversion processes where heat and work are involved. In the steam turbine, for example, the specific enthalpy drop as steam expands is directly related to the work output. Similarly, the increase in specific enthalpy of air as it is compressed in the air compressor relates to the energy input required for compression. Knowing the specific enthalpy values at different states is vital for engineers to calculate and optimize work and efficiency in thermodynamic cycles.

Mass Flow Rate

The mass flow rate is a measure of the amount of mass passing through a cross-section per unit of time and is typically expressed in kilograms per second (kg/s). In our exercise, we analyze mass flow rates for both steam and air because it determines the energy exchange rate. For a steam turbine, the mass flow rate of steam, combined with the specific enthalpy change, reveals how much power the turbine can deliver. In the case of the air compressor, the mass flow rate of air helps determine the power requirement to achieve the desired compression. As such, the control and measurement of mass flow rates are critical for the design and operation of thermodynamic systems such as power plants and HVAC systems.