Question

Consider a steady flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 50 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 (how much percent).

Answer 1

η = 48.1 %

Step-by-step explanation:

Given that

The maximum temperature ,T(max) = 350 C

T(max) = 350+ 273 = 623 K

The minimum temperature ,T(min) = 50 C

T(min) = 50 + 273 = 323 K

We know that efficiency of Carnot cycle is given as

`\eta=1-(T_(min))/(T_(max))`

Now by putting the values in the above equation we get

`\eta=1-(50+273)/(350+273)\\\eta=0.481`

The efficiency of Carnot cycle will be 48.1 %.

Therefore the answer will be 48.1 %.

η = 48.1 %