Question

Steam enters a turbine operating at steady state at a pressure of \(3 \mathrm{MPa}\), a temperature of \(400^{\circ} \mathrm{C}\), and a velocity of \(160 \mathrm{~m} / \mathrm{s}\). Saturated vapor exits at \(100^{\circ} \mathrm{C}\), with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Heat transfer from the turbine to its surroundings takes place at the rate of \(30 \mathrm{~kJ}\) per kg of steam at a location where the average surface temperature is \(350 \mathrm{~K}\). (a) For a control volume including only the turbine and its contents, determine the work developed, in \(\mathrm{kJ}\), and the rate at which entropy is produced, in \(\mathrm{kJ} / \mathrm{K}\), each per \(\mathrm{kg}\) of steam flowing. (b) The steam turbine of part (a) is located in a factory where the ambient temperature is \(27^{\circ} \mathrm{C}\). Determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of steam flowing, for an enlarged control volume that includes the turbine and enough of its immediate surroundings so that heat transfer takes place from the control volume at the ambient temperature. Explain why the entropy production value of part (b) differs from that calculated in part (a).

Short answer

The work developed is 533.3 kJ/kg, the rate of entropy production for the turbine alone is 0.729 kJ/K per kg of steam, and 0.7433 kJ/K per kg for the enlarged control volume. Entropy generation slightly increases for the larger control volume due to the assumptions made about heat transfer at ambient temperature.

Step-by-step solution

- Gather initial data

Note the given parameters: Initial pressure, P1 = 3 MPa Initial temperature, T1 = 400°C Initial velocity, v1 = 160 m/sFinal steam condition: Saturated vapor Final temperature, T2 = 100°C Final velocity, v2 = 100 m/s Heat transfer rate, Q_out = 30 kJ/kg Surface temperature, T_surf = 350 KAmbient temperature, T_ambient = 27°C = 300 K

- Determine work developed per kg of steam

Use the Steady Flow Energy Equation: \[\frac{dE_{cv}}{dt} = \frac{d(\text{mass flow rate} \times (h + \frac{v^2}{2} + gz))}{dt} + \frac{dQ_{in}}{dt} - \frac{dW_{out}}{dt}\]Assumption: Change in potential energy (\(dgz\)) is negligible. Therefore,\[\frac{dW_{out}}{dt} = \frac{dQ_{in}}{dt} + (h1 + \frac{v1^2}{2}) - (h2 + \frac{v2^2}{2})\]Given: \(\frac{dQ_{in}}{dt} = -30 kJ/kg\)Table values from the steam tables: Initial enthalpy (\text{h1}) = 3231.6 kJ/kgFinal saturated vapor enthalpy (\text{h2}) at 100°C = 2676.1 kJ/kgSubstitute all known values:\[\frac{dW_{out}}{dt} = -30 + (3231.6 + \frac{160^2}{2 \times 1000}) - (2676.1 + \frac{100^2}{2 \times 1000})\]\[\frac{dW_{out}}{dt} = -30 + (3231.6 + 12.8) - (2676.1 + 5) = 533.3~\text{kJ/kg}\]

- Calculate rate of entropy production for CV containing only the turbine

Apply the entropy balance: Turbine entropy balance: \(\frac{dS_{cv}}{dt} = \frac{dQ_{out}}{dt}/T_{surf} + \text{mass flow rate} \times (s2 - s1) + \frac{dS_{gen}}{dt}\)From steam tables: Initial entropy, S1 = 6.716 kJ/kg.KFinal saturated vapor entropy, S2 = 7.3593 kJ/kg.KHeat transfer rate entropy term: \(\frac{dQ_{out}}{dt}/T_{surf} = - \frac{30}{350} = -0.0857 \text{kJ/kg} \text{K}\)Entropy balance equation: \(\frac{dS_{gen}}{dt} = s2 - s1 - \frac{dQ_{out}}{dt}/T_{surf}\)\(\frac{dS_{gen}}{dt} = 7.3593 - 6.716 + 0.0857 = 0.729 \text{kJ/kg} \text{K}\)

- Determine the rate of entropy production for an enlarged control volume

We have heat transfer taking place at ambient temperature (300 K): Entropy balance for enlarged CV: \(\frac{dS_{gen, cv}}{dt} + \frac{dS_{gen, ambient}}{dt} = s2 - s1 - \frac{dQ_{out}}{dt}/T_{ambient} \)\(\frac{dS_{gen, ambient}}{dt}= s2 - s1 - \frac{30}{300}\)\(\frac{dS_{gen, ambient}}{dt} = 7.3593 - 6.716 + 0.1\)\(\frac{dS_{gen, ambient}}{dt} = 0.7433 \text{kJ/kg} \text{K}\)

- Explain the difference in entropy production values

The difference occurs due to the heat transfer process. When the heat transfer is taken as taking place at the surface temperature, there is less contribution to the entropy generation compared to when it is considered at the lower ambient (300 K) temperature.

Key concepts

steady flow energy equation

Understanding the steady flow energy equation is critical in analyzing steam turbines. This equation expresses the first law of thermodynamics for a control volume where mass flows in and out steadily. The key is to ensure that mass and energy entering the system balanced with what is leaving. For our exercise, this equation helps in determining the work developed by the turbine. The equation is: \frac{dE_{cv}}{dt} = \dot{m}(h + \frac{v^2}{2} + gz) + \dot{Q_{in}} - \dot{W_{out}} Here, \(E_{cv}\) is the energy within the control volume, \(m\) is the mass flow rate, \(h\) represents enthalpy, \(v\) is velocity, \(g\) is gravitational acceleration, and \(z\) is the height from a reference level. The term \dot{Q_{in}} stands for heat added to the system while \dot{W_{out}} represents work output. Potential energy changes (\

entropy balance

Entropy balance in thermodynamic systems helps us understand the irreversibility and inefficiencies. For steam turbines, we need to calculate the entropy generation to measure performance. The entropy balance equation states that the change in entropy of the control volume plus the net entropy transfer due to heat and mass flow equals the entropy generation. In our problem, the entropy balance for the turbine is given by: \[ \frac{dS_{cv}}{dt} = \frac{dQ_{out}}{dt} \/ T_{surf} + \dot{m}(s2 - s1) + \frac{dS_{gen}}{dt} \] The term \(dQ_{out}/T_{surf}\) covers entropy transfer due to heat, where \(T_{surf}\) is surface temperature. \(s2 - s1\) is the change in specific entropy between the entrance and exit states. Finally, \(dS_{gen} \/ dt\) accounts for entropy production, indicating irreversibilities within the system. Lower entropy production shows a more efficient process. For our turbine, the entropy change and production were computed based on given parameters and steam tables.

enthalpy

Enthalpy is a key property in thermodynamics representing the total heat content of a system. Specifically, for steam entering and exiting the turbine, enthalpy changes are crucial for calculating the work developed. Enthalpy combines internal energy (U) and the product of pressure (P) and volume (V), expressed as: \[ H = U + PV \] For our problem, we retrieve enthalpy values \(h1\) and \(h2\) from steam tables. The initial enthalpy \(h1\) refers to the steam at initial temperature and pressure, while \(h2\) represents the enthalpy of saturated vapor at the final temperature. Utilizing these values within the steady flow energy equation helps in determining the work performed by the turbine per kg of steam.

saturated vapor

Saturated vapor refers to steam at a state where any heat loss will result in condensation. It’s at equilibrium between liquid and gas phases. In the turbine, steam transforms from superheated to saturated vapor states. Recognizing properties of saturated vapor is important due to its predictable thermodynamic properties. In our exercise, final steam exits the turbine as saturated vapor at 100°C. We use saturated vapor enthalpy and entropy values from steam tables to complete our calculations in the energy equation and entropy balance. This simplifies understanding the system’s behavior and energy transformations during the process.