Question

Water enters the boiler of a steady-flow Carnot engine as a saturated liquid at 400 psia and leaves with a quality of \(0.95 .\) Steam leaves the turbine at a pressure of 20 psia. Show the cycle on a \(T\) -s diagram relative to the saturation lines, and determine ( \(a\) ) the thermal efficiency, ( \(b\) ) the quality at the end of the isothermal heat-rejection process, and \((c)\) the net work output.

Short answer

(a) The thermal efficiency of the steady-flow Carnot engine is 28.7%. (b) The quality at the end of the isothermal heat-rejection process is 72.45%. (c) The net work output of the cycle is 721.17 kJ/kg.

Step-by-step solution

Identify the states and processes

We must break down the cycle into its essential states and processes. The Carnot cycle consists of four reversible processes: two isothermal processes (at constant temperature) and two adiabatic processes (where there is no heat transfer). State 1: Boiler inlet (saturated liquid, 400 psia) State 2: Boiler outlet (quality = 0.95, 400 psia) State 3: Turbine outlet (20 psia) State 4: Completion of the heat-rejection process (20 psia) Process 1-2: Isothermal heat addition in the boiler Process 2-3: Adiabatic reversible expansion in the turbine Process 3-4: Isothermal heat rejection Process 4-1: Adiabatic reversible compression

Draw the T-s diagram

On the temperature-entropy (T-s) diagram, plot the saturation lines and locate each state on the diagram: - State 1: Saturated liquid, on the left side of the saturated line at 400 psia - State 2: Quality 0.95, on the same isotherm as State 1 (400 psia), to the right of State 1 - State 3: Turbine outlet, on the same isotherm as the given pressure (20 psia) - State 4: Completion of the heat-rejection process, on the same isotherm as State 3 (20 psia)

Calculate the thermal efficiency

The thermal efficiency formula for a Carnot engine is given by: Thermal efficiency = 1 - (T_low / T_high) Considering that the temperature remains constant during isothermal processes, we can use the saturation temperature at 400 psia as the high temperature (T_high) and the saturation temperature at 20 psia as the low temperature (T_low). Using steam tables, we find: T_high = 449.9°F T_low = 228.2°F Now we must convert these temperatures into the same unit (K) T_high = 263.277 °C = 536.427 K T_low = 108.999 °C = 382.149 K Plugging the values into the formula: Thermal efficiency = 1 - (382.149 / 536.427) = 0.287 The thermal efficiency of the cycle is 28.7%.

Calculate the quality at the end of the isothermal heat-rejection process

The quality at the end of the isothermal heat-rejection process (State 4) can be found using the isothermal heat rejection process (Process 3-4) since it is a Carnot engine: Entropy change = 0 s4 - s3 = 0 s4 = s3 Quality (x) at State 4 can be found using the definition of entropy for a wet steam: s4 = sf + x4 * (sg - sf) Using steam tables, we find sf, sg values at State 3 (20 psia): sf = 0.4448 sg = 1.7760 Now, we can solve for the quality x4: x4 = (s4 - sf) / (sg - sf) Since s4 = s3, we can use the quality of 0.95 at State 2 to find s3: s3 = sf + x3 * (sg - sf) s3 = sf + 0.95 * (sg - sf) Then, we plug s3 into the formula for x4: x4 = (s3 - sf) / (sg - sf) x4 = 0.7245 The quality at the end of the isothermal heat-rejection process is 0.7245 or 72.45%.

Calculate the net work output

Net work output can be found using the thermal efficiency and heat input: Net work output = Thermal efficiency * Heat input The heat input is the heat added during the isothermal heat addition process (Process 1-2). This can be found using enthalpy values: Heat input = h2 - h1 Enthalpy for wet steam: h = hf + x * (hg - hf) Using steam tables, we find hf and hg values at State 1 and State 2 (400 psia): hf1 = 137.9 kJ/kg hg1 = 2757.1 kJ/kg hf2 = 137.9 kJ/kg hg2 = 2757.1 kJ/kg Enthalpies at State 1 (saturated liquid) and State 2 (quality of 0.95): h1 = hf1 = 137.9 kJ/kg h2 = hf2 + 0.95 * (hg2 - hf2) = 2652.185 kJ/kg Heat input = 2652.185 - 137.9 = 2514.285 kJ/kg Now, we can find the net work output: Net work output = 0.287 * 2514.285 = 721.17 kJ/kg The net work output of the cycle is 721.17 kJ/kg.