Question

A steam power plant operates on an ideal regenerative Rankine cycle. Steam enters the turbine at \(6 \mathrm{MPa}\) and \(450^{\circ} \mathrm{C}\) and is condensed in the condenser at \(20 \mathrm{kPa}\). Steam is extracted from the turbine at \(0.4 \mathrm{MPa}\) to heat the feed water in an open feed water heater. Water leaves the feed water heater as a saturated liquid. Show the cycle on a \(T\) -s diagram, and determine ( \(a\) ) the net work output per kilogram of steam flowing through the boiler and ( \(b\) ) the thermal efficiency of the cycle.

Short answer

#Answer# The net work output per kilogram of steam flowing through the boiler is approximately 173.6 kJ/kg, and the thermal efficiency of the cycle is approximately 49.6%.

Step-by-step solution

Understand Regenerative Rankine Cycle and Draw \(T\) -s Diagram

The regenerative Rankine cycle is an improvement over the simple Rankine cycle because it uses an open feed water heater (OFWH) to increase the efficiency of the cycle. The cycle has the following stages: 1. Steam enters the turbine at high pressure and temperature (Point 1) 2. Steam expands in the turbine and generates work (Point 1 to Point 2) 3. Steam is extracted from the turbine at some intermediate pressure (Point 2) and sent to the open feed water heater. 4. Heat is transferred from extracted steam to the feed water (Point 2 to Point 3). 5. The remaining steam in the turbine continues to expand and generate work (Point 2 to Point 4). 6. Steam exits the turbine and enters the condenser where it is condensed at low pressure (Point 4 to Point 5). 7. Feedwater is pumped from the condenser to the OFWH (Point 5 to Point 6). 8. Feedwater is heated in the OFWH (Point 6 to Point 3). 9. Heated feedwater is further pumped to the boiler, where it gets heated and vaporized (Point 3 to Point 1). With this information, we can draw the \(T\) -s diagram by indicating the initial temperature and pressure of steam and the condenser's pressure, and where steam is extracted for the OFWH.

Determine Enthalpies at Each Point

To determine the net work output and thermal efficiency, we will first find the enthalpies at each point of the cycle. We will use the steam tables to do this. 1. \(h_1\): At \(6 \, \mathrm{MPa}\) and \(450^{\circ} \mathrm{C}\), from the steam tables, \(h_1 = 3314.6 \, \mathrm{kJ/kg}\) and \(s_1 = 6.354 \, \mathrm{kJ/kg \cdot K}\) 2. \(h_2\): At \(0.4 \, \mathrm{MPa}\) and isentropic expansion: \(s_2 = s_1\), we get \(h_2 = 2736.2 \, \mathrm{kJ/kg}\). 3. \(h_3\): Water leaves the OFWH as saturated liquid, so \(h_3 = h_F\) at \(0.4 \, \mathrm{MPa}\), from the steam tables, \(h_3 = 991.7 \, \mathrm{kJ/kg}\). 4. \(h_4\): At \(20 \, \mathrm{kPa}\) and isentropic expansion from point 2 to 4: \(s_4 = s_2\), from steam tables, \(h_4 = 2485.4 \, \mathrm{kJ/kg}\). 5. \(h_5\): At \(20 \, \mathrm{kPa}\) and saturated liquid, from steam tables, \(h_5 = 251.4 \, \mathrm{kJ/kg}\). 6. \(h_6\): Assuming isentropic compression in the pump, from point 5 to 6: \(\frac{v_5}{c_p} = \frac{h_6 - h_5}{T_5}\), using \(v_5 = 0.001017 \, \mathrm{m^3/kg}\), \(c_p = 4.2 \, \mathrm{kJ / kg \cdot K}\) and \(T_5 = 60^{\circ} \mathrm{C}\), we get \(h_6 = 253.2 \, \mathrm{kJ/kg}\). Now we have all the enthalpies needed to calculate the work output and thermal efficiency.

Calculate Net Work Output

First, we need to find the mass fraction of the steam extracted from the turbine and entering the OFWH. Let it be \(y\): \(y = \frac{h_6 - h_5}{h_2 - h_3} = \frac{253.2 - 251.4}{2736.2 - 991.7} \approx 0.110\) Now, we can find the turbine work (\(W_{T}\)), pump work (\(W_{P}\)), and the net work output: \(W_{T} = (1 - y)(h_1 - h_4) + y(h_1 - h_2) = (1 - 0.110)(3314.6 - 2485.4) + 0.110(3314.6 - 2736.2) \approx 912.1 \, \mathrm{kJ/kg}\) \(W_{P} = (h_3 - h_6) = (991.7 - 253.2) \approx 738.5 \, \mathrm{kJ/kg}\) Net work output = \(W_{T} - W_{P} = 912.1 - 738.5 \approx 173.6 \, \mathrm{kJ/kg}\)

Calculate Thermal Efficiency

Finally, we can find the thermal efficiency of the cycle: Thermal efficiency = \(\frac{\text{Net work output}}{\text{Heat input}} = \frac{173.6}{(h_1 - h_3)-(h_2-h_1)y} \approx 0.496 \times 100\% \approx 49.6\%\) In conclusion, the net work output per kilogram of steam flowing through the boiler is approximately \(173.6 \, \mathrm{kJ/kg}\), and the thermal efficiency of the cycle is approximately \(49.6\%\).

Key concepts

Thermodynamics

Thermodynamics is a fundamental branch of physics that deals with heat and temperature and their relation to energy and work. In the context of power plants, thermodynamics principles help us to convert heat into work using heat engines, such as the steam turbines in a Rankine cycle.

Applying thermodynamics, we can analyze the performance of the cycles by calculating the energy transfers involved at each step. Throughout this process, specific thermodynamic properties like enthalpy and entropy play a critical role in understanding how to improve the cycle’s efficiency and output.

Steam Power Plant

A steam power plant utilizes the Rankine cycle to convert heat into mechanical energy, which can then be transformed into electrical energy. The plant consists of four main components: a boiler or steam generator, a steam turbine, a condenser, and a feedwater pump. In a regenerative Rankine cycle, additional components like feedwater heaters are employed.

The use of an open feedwater heater, as mentioned in the exercise, is one such improvement in the cycle which helps raise the temperature of the feedwater before entering the boiler, reducing the heat required for converting water to steam, and increasing overall efficiency.

Thermal Efficiency

Thermal efficiency in a steam power plant is a measure of how effectively the plant converts the heat from fuel into work. Mathematically, it's the ratio of the net work output to the heat input during a cycle, usually expressed as a percentage. The regenerative Rankine cycle mentioned in the exercise is designed to improve this efficiency by taking a portion of the steam from the turbine and using it to pre-warm the feedwater.

Increase in efficiency implies less fuel consumption for the same output, which translates to cost savings and reduced environmental impact.

T-s Diagram

The Temperature-entropy (T-s) diagram is a graphical representation of a thermodynamic cycle. Each point on this diagram reflects the state of the system (i.e., the working fluid, steam in this case) at different stages of the cycle. A T-s diagram makes it easier to visualize the changes in temperature and entropy that occur during the process, and understand how heat is converted to work in a steam power plant.

The area under the curve in a T-s diagram is proportional to the heat transfer taking place in the system. In the given regenerative Rankine cycle, the T-s diagram would show the process of steam extraction and feedwater heating, which is vital for enhancing the cycle’s efficiency.