Question

Steamexpands in a turbine steadily at arate of $$18,000 \mathrm{kg} / \mathrm{h}$$ entering at \(7 \mathrm{MPa}\) and \(600^{\circ} \mathrm{C}\) and leaving at \(50 \mathrm{kPa}\) as saturated vapor. Assuming the surroundings to be at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C},\) determine \((a)\) the power potential of the steam at the inlet conditions and \((b)\) the power output of the turbine if there were no irreversibilities present.

Short answer

Answer: (a) The power potential of the steam at inlet conditions is 5203 kW. (b) The power output of the turbine if there were no irreversibilities present is 5094.5 kW.

Step-by-step solution

Convert mass flow rate to SI units

We are given the mass flow rate in $$\frac{kg}{h}$$. Let us first convert this to $$\frac{kg}{s},$$ which is the SI unit for mass flow rate. $$\dot{m} = 18,000 \frac{\text{kg}}{\text{h}} \cdot \frac{1 \text{hour}}{3600 \text{seconds}} = 5 \frac{\text{kg}}{\text{s}}$$

Determine the enthalpy at inlet and exit conditions

We need to find steam properties at the given pressure and temperature at inlet and outlet conditions. We can use the steam table or an online property calculator like REFPROP to find those values: - Inlet enthalpy ($$h_{in}$$) at $$7 \mathrm{MPa}$$ and $$600^{\circ} \mathrm{C}$$: $$h_{in} = 3625.3 \frac{\mathrm{kJ}}{\mathrm{kg}}$$ - Outlet enthalpy ($$h_{out}$$) at $$50 \mathrm{kPa}$$ as saturated vapor: $$h_{out} = 2584.7 \frac{\mathrm{kJ}}{\mathrm{kg}}$$

Calculate the power potential of steam at inlet conditions

Using the mass flow rate and enthalpy at inlet and outlet conditions, we can calculate the power potential of the steam at inlet conditions as follows: $$P = \dot{m} \cdot (h_{in} - h_{out})$$ $$P = 5 \frac{\text{kg}}{\text{s}} \cdot (3625.3 \frac{\text{kJ}}{\text{kg}} - 2584.7 \frac{\text{kJ}}{\text{kg}}) = 5203 \text{kW}$$ So the power potential of the steam at inlet conditions is $$5203 \text{kW}$$ (a).

Determine the entropy and enthalpy at outlet conditions for reversible expansion

For reversible expansion, we perform an entropy balance, i.e., $$s_{in} = s_{out}$$. From the steam table or property calculator, Entropy at inlet ($$s_{in}$$) at $$7 \mathrm{MPa}$$ and $$600^{\circ} \mathrm{C}$$: $$s_{in} = 6.902 \frac{\mathrm{kJ}}{\mathrm{kg \cdot K}}$$ Now, we use the given exit pressure of $$50 \mathrm{kPa}$$ and the entropy $$s_{out} = s_{in} = 6.902 \frac{\text{kJ}}{\text{kg}\cdot \text{K}}$$ to find the temperature and enthalpy at the outlet for reversible expansion. From the steam table or property calculator, Enthalpy at outlet conditions for reversible expansion ($$h_{out,rev}$$): $$h_{out,rev} = 2606.4 \frac{\mathrm{kJ}}{\mathrm{kg}}$$

Calculate the power output of turbine for reversible expansion

Using the mass flow and outlet enthalpy for reversible expansion, we can calculate the power output of the turbine: $$P_{rev} = \dot{m} \cdot (h_{in} - h_{out,rev})$$ $$P_{rev} = 5 \frac{\text{kg}}{\text{s}} \cdot (3625.3 \frac{\text{kJ}}{\text{kg}} - 2606.4 \frac{\text{kJ}}{\text{kg}}) = 5094.5 \text{kW}$$ So the power output of the turbine if there were no irreversibilities present is $$5094.5 \text{kW}$$ (b).

Key concepts

Mass Flow Rate

The mass flow rate is a measure of how much mass passes through a given point in a system per unit time. Simply put, it's like the speed at which mass moves, similar to how we view the flow of water through a pipe. It is typically calculated using the formula: \( \text{mass flow rate} = \frac{\text{mass}}{\text{time}} \). In our example, the mass flow rate of steam into the turbine is 18,000 kg/h, which translates to 5 kg/s when converted to the standard international (SI) unit (kg/s). This is crucial in calculating the power potential and output because it directly affects how much energy is carried by the steam over time, acting as a multiplier of energy per unit mass in the equation to derive power.

Understanding and working with the SI units ensures that calculations are performed correctly and that the results are comparable with norms established in scientific and engineering practices. The mass flow rate helps us to quantify the turbine's capacity to handle steam and ultimately generate power, making it a foundational element in thermodynamics.

Enthalpy

Enthalpy represents the total heat content of a system and is a combination of internal energy plus the work done by the system as it expands or contracts. It's often denoted \( h \) in thermodynamic equations. For steam and other working fluids in a turbine, enthalpy is indicative of the energy available to be converted into work. The higher the enthalpy, the more energy the steam has.

When we talk about the enthalpy of the steam at the inlet and outlet of a turbine (\( h_{in} \) and \( h_{out} \), respectively), we are referring to the energy content of the steam before and after passing through the turbine. These values, often found using steam tables or software like REFPROP, provide the necessary input to calculate the potential power output of the turbine, which hinges on the change in enthalpy as steam expands through the turbine.

Saturated Vapor

Saturated vapor is a term used to describe a vapor that is at the temperature at which vaporization (boiling) happens at a given pressure. At this point, the vapor is in equilibrium with its liquid; adding heat would convert more liquid to vapor without changing temperature.

In our example, the steam leaving the turbine is mentioned as 'saturated vapor', implying it is at the temperature and pressure where the water and steam are in equilibrium. This is important because it defines the state of the steam exiting the turbine and affects the enthalpy value used in power output calculations. The turbine brings the steam to the brink of condensation without actual condensation occurring, maximizing energy extraction from the steam.

Irreversibilities

Irreversibilities refer to the inherent inefficiencies that occur in real-world processes, which prevent them from being completely reversible. This means in practice some energy is always lost due to factors like friction, heat loss, and turbulence, and can't be recaptured to do work.

In a perfect, reversible process, none of the energy is wasted, and the entropy (a measure of disorder or randomness) remains constant. In the context of turbines, the absence of irreversibilities would mean all of the steam's enthalpy could be converted to work. Hence, part (b) of our exercise explores the ideal scenario, where the power output is maximized as if there were no energy losses or irreversibilities present. In reality, the actual power output would be less due to these inefficiencies. Understanding irreversibilities is critical in designing more efficient systems and predicting actual performance versus theoretical maximums.