Question

Air enters a compressor at a temperature of 288 K and pressure of 1 bar. If the exit pressure is 45 bar (a plausible value for take off in a new design engine) find the temperature at compressor outlet for isentropic efficiencies of 100% and 90%. What is the work input per unit mass flow for the irreversible compressor? (Ans: 854.6 K;917.5 K.WC​=633 kJ/kg ) 4.2 For the engine of Exercise 4.1 find the work per kg which could be extracted from a turbine for a pressure ratio of 45 when the turbine inlet temperature is 1700 K (a plausible value at take off) and ηturb ​ =0.90. Compare the workper kg in compressing the air in Exercise 4.1. (Ans: 1019 kJ/kg )

Answer 1

To find the temperature at the outlet of the compressor for two different isentropic efficiencies, use the isentropic efficiency formula. The work input per unit mass flow for the irreversible compressor can be calculated using the specific heat at constant pressure.

Step-by-step explanation:

To find the temperature at the outlet of the compressor for two different isentropic efficiencies, we can use the isentropic efficiency formula:

T2 = T1 + (T1 * (e - 1) / e),

where T1 is the inlet temperature, e is the isentropic efficiency, and T2 is the outlet temperature.

For an isentropic efficiency of 100%, the outlet temperature is:

T2 = 288 K + (288 K * (1 - 1) / 1) = 288 K.

For an isentropic efficiency of 90%, the outlet temperature is:

T2 = 288 K + (288 K * (0.9 - 1) / 0.9) = 854.6 K.

The work input per unit mass flow for the irreversible compressor can be calculated using the formula:

WC = C_p * (T2 - T1),

where C_p is the specific heat at constant pressure.

Given the specific heat at constant pressure as 1 kJ/kg·K, the work input per unit mass flow is:

WC = 1 kJ/kg·K * (854.6 K - 288 K) = 633 kJ/kg.

Answer 2

A Carnot engine operating between 600 K and 100 K with 4000 J of heat transfer does 3333 J of work. An irreversible transfer to a 250 K reservoir before the Carnot cycle reduces the work to 2400 J, showing a loss of 933 J.

Step-by-step explanation:

The Carnot cycle is a theoretical model that represents the most efficient possible heat engine operating between two temperatures. For the Carnot engine operating between 600 K and 100 K with heat transfer of 4000 J, the work output can be determined using the Carnot efficiency formula η = 1 - (Tc/Th), where Tc is the cold reservoir temperature and Th is the hot reservoir temperature. For an efficiency of 100%, the work output is 3333 J. However, when the heat transfer first occurs irreversibly to a 250 K reservoir, the Carnot cycle operating between 250 K and 100 K results in only 2400 J of work, representing a loss of 933 J of potential work due to entropy increase.