Question

Consider a steady-flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 60°C. The quality of water is 0.891 at the beginning of the heat-rejection process and 0.1 at the end. Determine (a) the thermal efficiency,(b) the pressure at the turbine inlet, and (c) the net work output.

ηₜₕC= ___%

Answer 1

The thermal efficiency of a Carnot cycle with given temperatures can be found using the equation ηₜₕC = 1 - (Tc/Th). The provided temperatures give a thermal efficiency of 46.57%. The pressure at the turbine inlet and net work output cannot be determined without additional information.

Step-by-step explanation:

The thermal efficiency of a Carnot cycle is given by the equation:

ηₜₕC = 1 - (Tc/Th)

Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. In this case, Tc = 60°C + 273 = 333 K, and Th = 350°C + 273 = 623 K. Plugging these values into the equation, we get:

ηₜₕC = 1 - (333/623) = 0.4657

So the thermal efficiency of the Carnot cycle is 46.57%.

To find the pressure at the turbine inlet, we need to know the specific volume at that point. Without that information, we cannot determine the pressure. Similarly, without the specific volume at the turbine inlet, we cannot determine the net work output.