Question

At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops 1000 kW of power. Determine the rate of entropy production and then comment on whether this cycle is possible or impossible, and why.

Answer 1

Wmax = 750 kw < power developed ( 1000kw ) for a reversible the cycle is Impossible

Step-by-step explanation:

Hot reservoir Temperature = 1000 K

Cold reservoir Temperature = 500 K

Heat transfer ( energy received by Hot reservoir ) ( Q ) = 1500 kW

Heat transfer ( energy received by Cold reservoir via Hot reservoir ) = 1000 Kw

Calculate the rate of entropy production

The higher the entropy production the less efficient the system

Δs = Cp In ( T2 / T1 )

power developed = 1000 kW

considering that the cycle is reversible and the constant volume or constant pressure of the substance in the thermodynamic cycle is not given we will use the efficiency to determine if the cycle is possible or not

Л = efficiency

∴Л = 1 - T2 / T1 = 1 - ( 500 / 1000 ) = 0.5

note as well that; Л = work output / work input = Wmax / Q

= 0.5 = Wmax / Q

∴ rate of entropy production = Q ( 0.5 ) = 1500 * 0.5 = 750 kw

Given that Wmax = 750 kw < power developed ( 1000kw ) for a reversible the cycle is Impossible