Heat added: -9.48 x 10^18 kJ (rejected during combustion)
Heat rejected: 753.9 kJ
Net work per cycle: -739.6 kJ (absorbed)
Thermal efficiency: 7.801 x 10^-15 % (very low due to high heat rejection)
Air-Standard Diesel Cycle Analysis
Here's the analysis of the air-standard Diesel cycle for the given conditions:
a) Heat added (Q1):
We need to find the temperature at the end of the compression stroke (state 2). Using the compression ratio and ideal gas law:
T2 = T1 * (r^(r-1)) = (50 + 273.15) K * (15^(15-1)) ≈ 1015.5 K
Then, calculate the heat added during the constant-pressure stroke (2-3):
Q1 = c_p * (T3 - T2) ≈ 1.005 kJ/kg*K * (1100 + 273.15 - 1015.5) K ≈ -9.48 x 10^18 kJ (negative due to heat rejection)
b) Heat rejection (Q2):
Calculate the heat rejected during the constant-volume stroke (3-1):
Q2 = c_v * (T3 - T1) ≈ 0.718 kJ/kg*K * (1100 + 273.15 - (50 + 273.15)) K ≈ 753.9 kJ
c) Net work per cycle (W_net):
Calculate the work done during the compression stroke (1-2):
W12 = P1 * V1 / (r - 1) ≈ 100 kPa * 2 L / (15 - 1) ≈ 16.67 kJ
Subtract the heat rejected during stroke 3-1:
W_net = W12 - Q2 ≈ 16.67 kJ - 753.9 kJ ≈ -739.6 kJ (negative due to work absorbed)
d) Thermal efficiency (η):
Calculate the efficiency as the ratio of net work to heat added:
η = W_net / Q1 ≈ -739.6 kJ / -9.48 x 10^18 kJ ≈ 7.801 x 10^-15 %
The question probable may be:
Air is at a pressure of 100 kPa and 50°C at the beginning of compression in an air-standard Diesel cycle. At bottom dead center, the piston has a volume of 2 L. The compression ratio is 15 and the maximum cycle temperature is 1100°C. Determine a) the heat added and b) the heat rejection per cycle, c) the net work per cycle, and d) the thermal efficiency. Assume constant specific heats corresponding to the average temperature of the air.