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Chapter 7: Problem 53

A large condenser in a steam power plant dumps \(15 \mathrm{MW}\) by condensing saturated water vapor at \(45^{\circ} \mathrm{C}\) to saturated liquid. What is the water flow rate and the entropy generation rate with an ambient at \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Water flow rate is approximately 6759 kg/s; entropy generation rate is 0.508 kW/K.

Step by step solution

01

Understand the Given Data

Identify the power dumped, the initial and final states of the water vapor, and the ambient temperature. Specifically, we have- Heat rejected, \(\dot{Q}_{out} = 15\, \text{MW} = 15 \times 10^6\, \text{W}\)- Initial State: saturated vapor at \(45^{\circ} \text{C}\)- Final State: saturated liquid at \(45^{\circ} \text{C}\)- Ambient Temperature: \(T_0 = 25^{\circ} \text{C} = 298 \text{K}\).
02

Calculate the Water Flow Rate

To find the water flow rate, use the specific enthalpy change from saturated vapor to saturated liquid at the given temperature. Look up the steam tables for vapor and liquid enthalpies:- \(h_{\text{vapor}} = 2407 \text{ kJ/kg}\)- \(h_{\text{liquid}} = 188 \text{ kJ/kg}\).The specific enthalpy change is \(\Delta h = h_{\text{vapor}} - h_{\text{liquid}}\).The mass flow rate \(\dot{m}\) is given by: \[ \dot{m} = \frac{\dot{Q}_{out}}{\Delta h} = \frac{15\, \text{MW}}{2407 - 188 \text{ kJ/kg}}\]Converting MW to kJ/s:\[ \dot{m} = \frac{15 \times 10^6 \text{ kJ/s}}{2219 \text{ kJ/kg}} \approx 6759 \text{ kg/s} \].
03

Calculate Entropy Generation Rate

Entropy generation rate can be calculated using the formula:\[ \dot{S}_{gen} = \frac{\dot{Q}_{out}}{T_0} - \dot{m} \Delta s \]where \(\Delta s = s_{\text{vapor}} - s_{\text{liquid}}\). Look up specific entropy values from steam tables:- \(s_{\text{vapor}} = 7.88 \text{ kJ/kg}\cdot\text{K}\)- \(s_{\text{liquid}} = 0.58 \text{ kJ/kg}\cdot\text{K}\).\[ \Delta s = 7.88 - 0.58 = 7.3 \text{ kJ/kg}\cdot\text{K} \]Calculate:\[ \dot{S}_{gen} = \frac{15 \times 10^6}{298} - 6759 \times 7.3 \approx 0.508 \text{ kW/K} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy Generation
Entropy generation is a key concept in the second law of thermodynamics. It measures the irreversibility of a process. In simpler terms, it is a way to quantify the energy dispersal or the waste produced during energy transformation. For our case with the condenser:
  • We calculate the rate at which entropy is generated when thermal energy is rejected to the surroundings.
  • The formula we used is: \[ \dot{S}_{gen} = \frac{\dot{Q}_{out}}{T_0} - \dot{m} \Delta s \]
  • The first term, \(\frac{\dot{Q}_{out}}{T_0}\), represents the theoretical upper limit of how efficient heat dumping can be.
  • The second term accounts for changes in the system鈥檚 entropy due to mass flow and phase changes.
Understanding entropy generation helps to improve system efficiency by identifying where energy losses occur.
Steam Power Plant
A steam power plant is a facility that converts water into steam to drive turbines for electricity generation. The basic components of a steam power plant include:
  • Boiler - heats water to create steam.
  • Turbine - the steam spins the blades to generate mechanical energy.
  • Condenser - cools the steam back into liquid for reuse.
  • Pump - moves the liquid back into the boiler.
In our exercise, the condenser plays a critical role in cooling down the steam which has completed its work cycle. Its efficiency is crucial in maintaining the overall efficiency of the power plant.
Condensation Process
The condensation process involves converting steam back into liquid water. This process takes place in the condenser unit of a power plant. Here鈥檚 what happens:
  • Heat is extracted from the steam causing it to lose energy and transform into a liquid.
  • This conversion process at constant temperature is efficient but not perfect.
  • The saturation at \(45^{\circ} \, \text{C}\) ensures that the steam is fully condensed at the same temperature, transitioning from vapor to liquid.
  • A properly functioning condenser removes latent heat, allowing the cycle to repeat.
This precise control over the phase transition directly impacts the power plant's performance.
Specific Enthalpy
Specific enthalpy is a measure of the total energy per unit mass in a thermodynamic system. It guides the transformation of steam in power plants:
  • Enthalpy changes as water vapor condenses differently, based on the starting and ending phases.
  • Using steam tables, we determine enthalpy values for both vapor and liquid phases, \(h_{\text{vapor}}\) and \(h_{\text{liquid}}\).
  • The change in specific enthalpy, \(\Delta h = h_{\text{vapor}} - h_{\text{liquid}}\), helps compute how much energy is released or absorbed.
Specific enthalpy is thus crucial in calculating the mass flow rate and validating thermodynamic efficiency.
Steam Tables
Steam tables are essential tools in thermodynamics. They provide physical properties of water and steam under various conditions:
  • Tables include information on enthalpy, entropy, temperature, and pressure.
  • In our exercise, we derived specific enthalpies and specific entropies from these tables for both saturated vapor and saturated liquid states.
  • Steam tables are pivotal for accurately calculating changes and processes in systems operating with steam.
Using steam tables allows engineers to perform precise calculations needed for designing and optimizing power systems.

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Most popular questions from this chapter

A rigid \(1.0-\mathrm{m}^{3}\) tank contains water initially at \(120^{\circ} \mathrm{C}\), with \(50 \%\) liquid and \(50 \%\) vapor by volume. A pressure- relief valve on top of the tank is set to \(1.0 \mathrm{MPa}\) (the tank pressure cannot exceed 1.0 MPa; water will be discharged instead). Heat is now transferred to the \(\operatorname{tank}\) from a \(200^{\circ} \mathrm{C}\) heat source until the tank contains saturated vapor at \(1.0 \mathrm{MPa}\). Calculate the heat transfer to the tank and show that this process does not violate the second law.

A mixing chamber receives \(5 \mathrm{~kg} / \mathrm{min}\) ammonia as saturated liquid at \(-20^{\circ} \mathrm{C}\) from one line and ammonia at \(40^{\circ} \mathrm{C}, 250 \mathrm{kPa}\) from another line through a valve. The chamber also receives \(325 \mathrm{~kJ} / \mathrm{min}\) energy as heat transferred from a \(40^{\circ} \mathrm{C}\) reservoir, as shown in Fig. P7.58. This should produce saturated ammonia vapor at \(-20^{\circ} \mathrm{C}\) in the exit line. What is the mass flow rate in the second line, and what is the total entropy generation in the process?

An empty can of \(0.002 \mathrm{~m}^{3}\) is filled with R-134a from a line flowing saturated vapor R-134a at \(10^{\circ} \mathrm{C}\). The filling is done quickly, so it is adiabatic, and the process stops when the pressure is \(200 \mathrm{kPa}\). Find the final mass in the can and the total entropy generation.

An initially empty spring-loaded piston/cylinder requires \(100 \mathrm{kPa}\) to float the piston. A compressor with a line and valve now charges the cylinder with water to a final pressure of \(1.4 \mathrm{MPa}\), at which point the volume is \(0.6 \mathrm{~m}^{3}\), state 2 . The inlet condition to the reversible adiabatic compressor is saturated vapor at \(100 \mathrm{kPa}\). Find the final mass of water, the piston work from 1 to 2 and the required compressor work.

A small water pump at ground level has an inlet pipe down into a well at a depth \(H\) with the water at \(100 \mathrm{kPa}, 15^{\circ} \mathrm{C}\). The pump delivers water at \(400 \mathrm{kPa}\) to a building. The absolute pressure of the water must be at least twice the saturation pressure to avoid cavitation. What is the maximum depth this setup will allow?
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