problem no. 3 in an air - standard brayton cycle the compressor inlet conditions are 100kpa and 280°k. the…

Explanation:

Step1: Identify the specific - heat formula

For an air - standard Brayton cycle, the work output of the turbine $W_T$ is given by $W_T=\dot{m}c_p(T_3 - T_4)$. For air, the specific heat at constant pressure $c_p = 1.005\ kJ/(kg\cdot K)$.

Step2: Rearrange the formula for mass - flow rate

We need to find the mass - flow rate $\dot{m}$. Rearranging the formula $W_T=\dot{m}c_p(T_3 - T_4)$ gives $\dot{m}=\frac{W_T}{c_p(T_3 - T_4)}$. Here, $W_T = 11190\ kW=11190\ kJ/s$, $T_3 = 1167\ K$, and we assume an isentropic process for simplicity (if not given otherwise, and we are not concerned with the actual $T_4$ value as we are using the basic power - specific heat relation).

Step3: Calculate the mass - flow rate

Substitute the values into the formula: $\dot{m}=\frac{11190\ kJ/s}{1.005\ kJ/(kg\cdot K)\times(1167\ K - T_4)}$. In an ideal Brayton cycle, if we assume no heat loss in the turbine and use the basic power - specific heat relation for work output of turbine, and since we are not given details about non - idealities, we can assume the work is solely due to the temperature drop across the turbine. $\dot{m}=\frac{11190}{1.005\times(1167 - 280)}=\frac{11190}{1.005\times887}=\frac{11190}{891.435}\approx12.55\ kg/s$.

Answer:

$12.55\ kg/s$