Question

A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at \(6000 \mathrm{kPa}\), the turbine inlet at \(600^{\circ} \mathrm{C},\) and the condenser at 50 kPa. Compare the thermal efficiency of this cycle when it is operated so that the liquid enters the pump as a saturated liquid against that when the liquid enters the pump \(11.3^{\circ} \mathrm{C}\) cooler than a saturated liquid at the condenser pressure.

Short answer

Answer: Cooling the liquid by 11.3°C before entering the pump will affect the thermal efficiency of the steam power plant. Comparing the thermal efficiencies for both cases (\(\eta '\) for 11.3°C cooler case and \(\eta\) for the saturated liquid case) will help determine the effect of cooling. If \(\Delta \eta = \eta ' - \eta\) is positive, the 11.3°C cooler case has greater thermal efficiency, whereas a negative value indicates the saturated liquid case has greater thermal efficiency.

Step-by-step solution

Determine the enthalpy values for the saturated liquid case

We need to find enthalpy values at each point of the Rankine cycle when the liquid enters the pump as a saturated liquid. Let's label the points as follows: 1 - entering the pump, 2 - leaving the pump, 3 - leaving the boiler, and 4 - leaving the turbine. At point 1, the liquid is entering the pump as a saturated liquid at 50 kPa. We can find the enthalpy value, \(h_1\), and specific volume, \(v_1\), from the steam table. At point 2, the fluid is leaving the pump and entering the boiler. We can find the enthalpy value, \(h_2\), using the equation \(h_2 = h_1 + v_1\cdot (P_2 - P_1)\), where \(P_2\) is the pressure at point 2 (6000 kPa) and \(P_1\) is the pressure at point 1 (50 kPa). At point 3, the fluid is leaving the boiler at 600°C and 6000 kPa. We can find the enthalpy value, \(h_3\), from the steam table. At point 4, the fluid is leaving the turbine and entering the condenser at 50 kPa. We can find the enthalpy value, \(h_4\), from the steam table, assuming isentropic expansion in the turbine.

Calculate heat input, work output, and thermal efficiency for the saturated liquid case

Heat input in the boiler: \(q_{in} = h_3 - h_2\) Turbine work output: \(w_t = h_3 - h_4\) Pump work input: \(w_p = h_2 - h_1\) Net work output: \(w_{net} = w_t - w_p\) Thermal efficiency: \(\eta = \frac{w_{net}}{q_{in}}\)

Determine the enthalpy values for the 11.3°C cooler case

Now we will determine the enthalpy values when the liquid enters the pump 11.3°C cooler than a saturated liquid at the condenser pressure. At point 1, the fluid is 11.3°C cooler than a saturated liquid at 50 kPa. We can find the enthalpy value, \(h_1'\), and specific volume, \(v_1'\), using interpolation in the steam table. At point 2, we can find the enthalpy value, \(h_2'\), using the equation \(h_2' = h_1' + v_1'\cdot (P_2 - P_1)\). At point 3, the enthalpy value remains the same as in the saturated liquid case, \(h_3\). At point 4, the enthalpy value remains the same as in the saturated liquid case, \(h_4\).

Calculate heat input, work output, and thermal efficiency for the 11.3°C cooler case

Heat input in the boiler: \(q_{in}' = h_3 - h_2'\) Turbine work output: \(w_t' = h_3 - h_4\) Pump work input: \(w_p' = h_2' - h_1'\) Net work output: \(w_{net}' = w_t' - w_p'\) Thermal efficiency: \(\eta ' = \frac{w_{net}'}{q_{in}'}\)

Compare the thermal efficiencies

Now that we have the thermal efficiencies for both cases, we can compare them to determine the effect of cooling the liquid by 11.3°C before entering the pump. \(\Delta \eta = \eta ' - \eta\) A positive value of \(\Delta \eta\) means the thermal efficiency is greater for the 11.3°C cooler case, whereas a negative value means the saturated liquid case has greater thermal efficiency.

Key concepts

Enthalpy in Thermodynamics

Enthalpy, represented by the symbol 'H', is a fundamental thermodynamic quantity that combines internal energy with the product of pressure and volume. It is a state property, meaning it is defined at any particular instant and does not depend on the path taken to reach that state.

In the Rankine cycle, which is the focus of steam power plants, enthalpy plays a critical role. As steam flows through the various components of a power plant—turbines, condensers, pumps, and boilers—the changes in enthalpy associated with phase changes and temperature variations are meticulously tracked. These changes are the backdrop to calculating work done by the system and the heat transferred into or out of the system, leading to the determination of the cycle's thermal efficiency.

Enthalpy is measured in units of energy, typically joules or kilojoules. It is obtained from thermodynamic tables, commonly known as steam tables, at specific temperatures and pressures relevant to the operation of steam power plants. Understanding and calculating enthalpy changes are integral steps in evaluating the performance of the Rankine cycle.

Saturated Liquid

A saturated liquid is a state of a substance where it is about to start the phase transformation from liquid to vapor without any increase in temperature. In other words, it is at its boiling point for a given pressure. This concept is essential in a steam power plant because the Rankine cycle often involves alternating liquid-vapor phase changes.

In a steam power plant, the liquid initially enters the pump as a saturated liquid before it's compressed to a higher pressure and heated to become superheated steam, which is then expanded in a turbine. When you look at a phase diagram, the line separating the liquid and vapor regions represents saturation conditions; any point on this line represents a state where liquid and vapor phases coexist in equilibrium.

The choice of whether the water enters the pump as a saturated liquid—as in the provided exercise—or slightly subcooled (cooler than the saturated liquid) has practical implications for the operation and efficiency of the cycle. Thermodynamic tables provide the necessary properties, such as enthalpy and specific volume, for saturated liquids at particular pressures, which are indispensable for calculating the cycle's thermal performance.

Steam Power Plant

A steam power plant, also known as a thermal power station, is a facility designed to convert the energy stored in fuel into electricity. The heart of this plant is the Rankine cycle, where water is boiled to create steam that drives a turbine connected to an electric generator.

The basic components of a steam power plant include the boiler (or steam generator), turbine, condenser, feedwater pump, and the interconnecting piping. Coal, natural gas, nuclear material, or renewables like solar thermal energy can be the source of heat required to produce steam. In the ideal Rankine cycle, high-pressure and high-temperature steam expands through the turbine, converting thermal energy to mechanical energy. The expanded steam is then condensed back into water, which is subsequently pumped into the boiler to complete the cycle.

Efficiency is a primary concern in the design and operation of steam power plants. Every step in the cycle has been optimized to extract the maximum possible amount of energy from the working fluid, typically water or steam. Consequently, understanding concepts like enthalpy, saturated liquid, and phase changes is crucial to enhance the operating conditions and improve the thermodynamic efficiency of the plant.