Solution

12.89
Borgnakke and Sonntag
At the beginning of compression in a diesel cycle T = 300 K, P = 200 kPa and
after combustion (heat addition) is complete T = 1500 K and P = 7.0 MPa. Find
the compression ratio, the thermal efficiency and the mean effective pressure.
Solution:
Standard Diesel cycle. See P-v and T-s diagrams for state numbers.
Compression process (isentropic) from Eqs.8.33-8.34
P 2 = P 3 = 7000 kPa => v 1 / v 2 = (P 2/P 1) 1/ k = (7000 / 200) 0.7143 = 12.67
T 2 = T 1(P 2 / P 1) (k-1) / k = 300(7000 / 200) 0.2857 = 828.4 K
Expansion process (isentropic) first get the volume ratios
v 3 / v 2 = T 3 / T 2 = 1500 / 828.4 = 1.81
v 4 / v 3 = v 1 / v 3 = (v 1 / v 2)( v 2 / v 3) = 12.67 / 1.81 = 7
The exhaust temperature follows from Eq.8.33
T 4 = T 3(v 3 / v 4) k-1 = 1500 (1 / 7) 0.4 = 688.7 K
q L = C vo(T 4 - T 1) = 0.717(688.7 - 300) = 278.5 kJ/kg
q H = h 3 - h 2 ≈ C po(T 3 - T 2) = 1.004(1500 - 828.4) = 674 kJ/kg
Overall performance
η = 1 - q L / q H = 1- 278.5 / 674 = 0.587
w net = q net = q H - q L = 674 - 278.5 = 395.5 kJ/kg
v max = v 1 = R T 1 / P 1 = 0.287×300 / 200 = 0.4305 m 3 /kg
v min = v max / (v 1 / v 2) = 0.4305 / 12.67 = 0.034 m 3 /kg
P meff =
w net
v max – v min = 395.5 / (0.4305 - 0.034) = 997 kPa
P
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Remark: This is a too low compression ratio for a practical diesel cycle.
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