Question

Water vapor enters a turbine operating at steady state at \(30 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) and expands adiabatically to \(6 \mathrm{MPa}\) with no significant change in kinetic or potential energy. If the isentropic turbine efficiency is \(80 \%\), determine the work developed, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of steam flowing, using the generalized property charts. Compare with the result obtained using steam table data. Discuss.

Short answer

Calculate steam properties from charts and tables; determine specific enthalpies and use efficiency to find work developed. Compare both methods and discuss deviations.

Step-by-step solution

Define the problem

We are given an adiabatic turbine process where steam enters at 30 MPa and 600°C and expands to 6 MPa. The isentropic efficiency is 80%. We need to determine the work developed per kg of steam and compare it using property charts and steam tables.

Isentropic process calculation (using property charts)

For an isentropic process, we need to determine the properties of the steam at the initial and final states. Using generalized property charts, locate the state at 30 MPa and 600°C to determine the entropy (s1) and specific enthalpy (h1).

Determine the isentropic final state

Using the entropy found in Step 2 (s1), determine the specific enthalpy (h2s) at 6 MPa under the constraint of constant entropy (isentropic process). This is the theoretical enthalpy if the process were isentropic.

Calculate the actual work using isentropic efficiency

The actual work developed per kg of steam can be calculated using the isentropic efficiency ( \( \text{Efficiency} = \frac{h_1 - h_2}{h_1 - h_{2s}} \)) Rearrange to solve for actual enthalpy (h2): \( h_2 = h_1 - \text{Efficiency} \times (h_1 - h_{2s}) \)

Calculate the work developed per kg

Finally, calculate the work developed per kg by using the enthalpy difference: \( W_{actual} = h_1 - h_2 \)

Repeat calculations using steam tables (verification)

Repeat Steps 2 to 5 using steam tables to find specific enthalpy and entropy values at the given states. Compare the results.

Discuss

Discuss the differences observed between the generalized property charts and steam tables results. Highlight any deviations and their possible reasons such as assumptions, approximations, or inherent inaccuracies in the data.

Key concepts

adiabatic process

In thermodynamics, an adiabatic process is one where no heat is transferred to or from the system due to thermal insulation or rapid execution. For a turbine operating adiabatically, this means that all energy transformations involve only work and no heat. This assumption simplifies our calculations. For example, when steam enters a turbine at high pressure and temperature and expands to a lower pressure, the energy conversion is purely work done by the steam on the turbine blades.
Key points of adiabatic process:

• No heat exchange with surroundings
• Energy transformations limited to work
• Applied in systems like turbines, compressors, and nozzles

specific enthalpy

Specific enthalpy is a measure of the total energy content per unit mass of a substance. It combines internal energy and the energy required to make room for the substance by displacing its surroundings. It’s given by the formula:
\( h = u + Pv \)
where

• \(h\) is specific enthalpy,
• \(u\) is specific internal energy,
• \(P\) is pressure,
• \(v\) is specific volume.

In steam turbine calculations, knowing the specific enthalpy at various points (inlet and outlet) helps in determining how much work is done by the steam as it expands through the turbine. For an isentropic process, this value is used to find the theoretical work that could be extracted if there were no losses.

entropy calculation

Entropy is a measure of disorder or randomness in a system. For an isentropic process, entropy remains constant. In our turbine example, we must find the entropy at the initial state using the given conditions (30 MPa and 600°C). This can be done using steam tables or property charts. Afterward, to find the final state entropy also remains constant, which helps us find the new state variables (pressure and specific enthalpy).
Steps:

• Determine initial entropy (\(s_1\))
• Assume entropy remains constant (isentropic)
• Use initial entropy to find final states’ properties under new pressure

steam tables

Steam tables include data about the properties of water and steam, such as temperature, pressure, specific volume, enthalpy, and entropy. In practice, they are used extensively to determine the state of the steam before and after passing through the turbine. For our example, we look up the values at 30 MPa and 600°C to find initial specific enthalpy and entropy, then do the same for the final pressure using the initial entropy value (assuming an isentropic process). This gives accurate values directly from real-world measurements, useful for checking results obtained by other methods.

property charts

Property charts or Mollier diagrams graphically represent the relationships between various thermodynamic properties like enthalpy, entropy, temperature, and pressure. These charts provide a quick visual way to determine the state properties of steam during an adiabatic expansion. Using property charts:

• Locate the initial state on the chart using the given conditions (pressure and temperature)
• Follow lines of constant entropy to find other state properties at different pressures

These serve as an alternative to steam tables, often allowing for quicker but slightly less precise estimations of thermodynamic properties necessary for work output calculations in turbines.