Question

Determine the work input and entropy generation during the compression of steam from \(100 \mathrm{kPa}\) to \(1 \mathrm{MPa}\) in \((a)\) an adiabatic pump and \((b)\) an adiabatic compressor if the inlet state is saturated liquid in the pump and saturated vapor in the compressor and the isentropic efficiency is 85 percent for both devices.

Short answer

Based on the mentioned step by step solution, calculate the work input and entropy generation during the compression of steam for both the adiabatic pump and the adiabatic compressor, given the isentropic efficiency is 85%.

Step-by-step solution

Initial and Final Enthalpy and Entropy

Using the saturation tables for water at initial pressure (100 kPa) and final pressure (1 MPa), find values for enthalpy and entropy. For Pump (Saturated liquid): \(h_1 = h_\text{f_{100kPa}}\) (use saturation table) \(s_1 = s_\text{f_{100kPa}}\) (use saturation table) For Compressor (Saturated vapor): \(h_2 = h_\text{g_{100kPa}}\) (use saturation table) \(s_2 = s_\text{g_{100kPa}}\) (use saturation table) For both devices at final pressure (1 MPa): - Saturated liquid values: \(h_\text{f_{1MPa}}\), \(s_\text{f_{1MPa}}\) (use saturation table) - Saturated vapor values: \(h_\text{g_{1MPa}}\), \(s_\text{g_{1MPa}}\) (use saturation table)

Isentropic Enthalpy

Using the isentropic efficiency (85%) for both devices, determine the isentropic enthalpy: For Pump: \(h_{1s} = h_1 - \frac{h_1 - h_{f_{1MPa}}}{\text{Isentropic Efficiency}}\) For Compressor: \(h_{2s} = h_2 - \frac{h_2 - h_{g_{1MPa}}}{\text{Isentropic Efficiency}}\)

Work Input

Calculate the work input for both the adiabatic pump and the adiabatic compressor: For Pump: \(W_\text{input, pump} = h_{1s} - h_1\) For Compressor: \(W_\text{input, compressor} = h_{2s} - h_2\)

Entropy Generation

Determine the entropy generation for both the adiabatic pump and the adiabatic compressor: For Pump: \(\Delta s_{\text{gen, pump}} = \frac{h_1 - h_{1s}}{T_\text{inlet, pump}}\) For Compressor: \(\Delta s_{\text{gen, compressor}} = \frac{h_2 - h_{2s}}{T_\text{inlet, compressor}}\) Calculate the final values and you will have the work input and entropy generation during the compression of steam for both the adiabatic pump and the adiabatic compressor.

Key concepts

Adiabatic Process

An adiabatic process is a thermodynamic process in which no heat transfer takes place between the system and its surroundings. This means that the process is insulated from any heat exchange, so any work done on or by the system results in a change in the internal energy of the system. During compression, if we consider it to be adiabatic, there's no heat input or output; the energy provided to the system via work input is entirely used to increase the internal energy, which can manifest as increased temperature or pressure.

In the context of our exercise, both the pump and the compressor are adiabatic, indicating that all work input goes into compressing the steam from one pressure level to a higher one without any energy loss as heat.

Entropy Generation

Entropy generation refers to the increase in entropy within a system as it undergoes a thermodynamic process. It is a measure of the irreversibility of the process and the energy that is not available to do work. In an ideal, reversible process, entropy generation is zero. However, in real-world applications, due to factors such as friction and turbulence, some entropy is always generated.

In our exercise, even though the compression processes are adiabatic, entropy is generated because real processes are not perfectly isentropic. The entropy generation for both the pump and the compressor indicates the level of irreversibility within each device.

Isentropic Efficiency

Isentropic efficiency is a measure of the deviation of a real process from the ideal isentropic process. It is defined as the ratio of the work input for an ideal isentropic process to the work input for the actual process. The isentropic efficiency allows us to account for inefficiencies in real devices and calculate the actual enthalpy in non-ideal conditions. It typically ranges from 0 to 1 (or 0% to 100%).

In the given exercise, an isentropic efficiency of 85% means that the actual devices are less efficient compared to an ideal situation, with 15% of the work input lost to irreversibilities such as friction. This efficiency is used to calculate the actual enthalpy changes in the pump and compressor.

Saturated Liquid and Vapor

In thermodynamics, a saturated liquid is a liquid that is at its boiling point for a given pressure, and a saturated vapor is its corresponding gaseous phase. Any additional heat transfer to a saturated liquid will not increase its temperature but will cause it to start vaporizing. Conversely, cooling a saturated vapor will not decrease its temperature until it starts to condense.

In the exercise, the initial states of steam for the pump and the compressor are a saturated liquid and a saturated vapor, respectively; these states correspond to the quality of x=0 for the saturated liquid and x=1 for the saturated vapor. These serve as the starting points for our enthalpy calculations using saturation tables at the given pressures.

Enthalpy Calculations

Enthalpy is a property used in thermodynamics to quantify the total heat content of a system. Calculations involving enthalpy help us determine the energy changes during thermodynamic processes. For processes involving phase changes such as boiling or condensing, steam tables provide values of enthalpy for saturated liquid and vapor at specific pressures.

In our exercise, the enthalpy values for the initial and final states are crucial for calculating the work input. The steam tables list values for specific enthalpies of the saturated liquid and vapor, which we use to determine the isentropic and actual enthalpies during the adiabatic compression processes. The enthalpy changes, along with the isentropic efficiency, offer insight into required work input and are indicators of the performance of the adiabatic devices.