Question

Several geothermal power plants are in operation in the United States and more are being built since the heat source of a geothermal plant is hot geothermal water, which is "free energy." An 8 -MW geothermal power plant is being considered at a location where geothermal water at \(160^{\circ} \mathrm{C}\) is available. Geothermal water is to serve as the heat source for a closed Rankine power cycle with refrigerant-134a as the working fluid. Specify suitable temperatures and pressures for the cycle, and determine the thermal efficiency of the cycle. Justify your selections.

Short answer

Based on the given problem, design a geothermal power plant using a closed Rankine power cycle with refrigerant-134a as the working fluid. Specify suitable temperatures and pressures for the cycle and determine the thermal efficiency of the power plant.

Step-by-step solution

Determine the maximum temperature at the evaporator

As the geothermal water is given with a temperature of \(160^{\circ} \mathrm{C}\) , we need to select a slightly lower temperature for the evaporator to ensure efficient heat transfer. We can select the evaporator temperature to be \(154^{\circ} \mathrm{C}\).

Find the saturation pressure at the evaporator temperature

Using a refrigerant-134a property table, look up the saturation pressure at \(154^{\circ} \mathrm{C}\) evaporator temperature. We find that the saturation pressure is approximately \(1.6 \,\text{MPa}\).

Determine the minimum temperature at the condenser

To select a suitable minimum temperature at the condenser, it is necessary to select a value such that the cooling water can condense the refrigerant efficiently. A typical value for the condenser temperature is around \(30^{\circ} \mathrm{C}\).

Find the saturation pressure at the condenser temperature

Using the refrigerant-134a property table, look up the saturation pressure at \(30^{\circ} \mathrm{C}\) condenser temperature. We find that the saturation pressure is approximately \(0.12 \,\text{MPa}\).

Calculate enthalpy and entropy values at the main points of the cycle

For a Rankine cycle, four main points (states) should be identified: 1. At the inlet of the turbine, after the evaporator 2. At the outlet of the turbine, before the condenser 3. At the inlet of the pump, after the condenser 4. At the outlet of the pump, before the evaporator We need to find the enthalpy (\(h\)) and entropy (\(s\)) values for these points using the refrigerant-134a property tables. - For point 1: \(h_1\) is the saturated vapor enthalpy at \(1.6 \,\text{MPa}\). We find \(h_1= 405.17 \,\text{kJ/kg}\), and \(s_1 = 1.823 \,\text{kJ/(kg.K)}\). - For point 3: \(h_3\) is the saturated liquid enthalpy at \(0.12 \,\text{MPa}\). We find \(h_3 = 96.01 \,\text{kJ/kg}\).

Determine the isentropic enthalpy at the outlet of the turbine

Assuming an isentropic process in the turbine, the entropy at the outlet of the turbine will be the same as the entropy at the inlet: \(s_2 = s_1\). Using the refrigerant-134a property table, we find that for \(s_2 = 1.823 \,\text{kJ/(kg.K)}\) at \(0.12 \,\text{MPa}\), the enthalpy is \(h_2=325.49 \,\text{kJ/kg}\) (interpolating between the liquid and vapor values if necessary).

Calculate the isentropic enthalpy at the outlet of the pump

Assuming an isentropic process in the pump, the entropy at the outlet of the pump will be the same as the entropy at the inlet: \(s_4 = s_3\). We calculate the enthalpy increase by multiplying the specific volume at point 3 (\(v_3\)) with the increase in pressure between points 3 and 4: \(\Delta h = v_3(P_4 - P_3)\). From the property tables for refrigerant-134a, we have \(v_3 = 0.001 \,\text{m}^3/\text{kg}\), \(P_4=1.6 \,\text{MPa}\), and \(P_3=0.12 \,\text{MPa}\). Thus, \(\Delta h = 0.001(1600-120) = 1.48 \,\text{kJ/kg}\). Finally, we find the enthalpy at point 4 by adding the enthalpy increase to the enthalpy at point 3: \(h_4 = h_3 + \Delta h = 96.01 + 1.48 = 97.49 \,\text{kJ/kg}\).

Determine the thermal efficiency of the cycle

The thermal efficiency of the Rankine cycle can be calculated using the formula: \(\eta_{th} = \frac{W_{net}}{Q_{in}}\) where: \(W_{net} = W_{turbine} - W_{pump}\) (Net work output) \(Q_{in} = h_1 - h_4\) (Heat input) From our enthalpy calculations, we can determine the work output by the turbine as \(W_{turbine} = h_1 - h_2 = 405.17 - 325.49 = 79.68 \,\text{kJ/kg}\). The work input by the pump is simply \(W_{pump} = h_4 - h_3 = 97.49 - 96.01 = 1.48 \,\text{kJ/kg}\). Thus, the net work output is \(W_{net} = 79.68 - 1.48 = 78.2 \,\text{kJ/kg}\). Our heat input for the cycle is \(Q_{in} = h_1 - h_4 = 405.17 - 97.49 = 307.68 \,\text{kJ/kg}\). Using these values, we can calculate the thermal efficiency of the cycle as: \(\eta_{th} = \frac{78.2}{307.68} \approx 0.254\) or \(25.4 \%\) The thermal efficiency of the considered geothermal power plant with a closed Rankine cycle using refrigerant-134a as the working fluid is approximately \(25.4\%\).