Question 13.17: The Wärtsilä RT-flex96C, the two-stroke, turbocharged, low-s...

a. The compression ratio \text{CR} = 18.0, and the cutoff ratio \text{CO} = 2.32, so the Diesel cold ASC thermal efficiency is given by Eq. (13.36) as

(η_T)_{\substack{\text{Diesel}\\\text{cold ASC}\\}} = 1 – \frac{\text{CR}^{1-k} (\text{CO}^k – 1)}{k(\text{CO} – 1)}            (13.36)

(η_T)_{\substack{\text{Diesel}\\\text{cold ASC}\\}} = 1 – \frac{(18.0)^{−0.40} [(2.32)^{1.40} – 1]}{1.40 (2.32 − 1)} = 0.617 = 61.7\%

b. The rate of energy provided by burning the fuel is

\dot{Q}_\text{fuel} = (\text{Fuel heating value in kJ/kg-fuel}) × (\text{Fuel flow rate in kg-fuel/s})

= (45.5 × 10^3  \text{kJ/kg-fuel}) × (3.35  \text{kg-fuel/s}) = 152 × 10^3  \text{kJ/s} = 152,000  \text{kW}

Then the actual thermal efficiency of the engine is given by Eqs. (7.5) and (13.33) as

η_T = \frac{\text{Net work output}}{\text{Total heat input}} = \frac{\text{Engine work output − Pump work input}}{\text{Boiler heat input}}                 (7.5)

η_m = \frac{\dot{W}_\text{actual}}{\dot{W}_\text{reversible}} = \frac{(\dot{W}_B)_\text{out}}{\dot{W}_I)_\text{out}} = 1- \frac{\dot{W}_F}{(\dot{W}_I)_\text{out}}                                   (13.33)

(η_T)_{\substack{\text{Diesel}\\\text{actual}\\}}= \frac{(\dot{W}_B)_\text{out}}{\dot{Q}_\text{fuel}} = \frac{80,080}{152,000} = 0.527 = 52.7\%

c. Since (η_T)_\text{actual }=(η_m)(η_T)_\text{cold ASC},

η_m = \frac{(η_T)_\text{actual }}{(η_T)_\text{cold ASC}} = \frac{0.525}{0.617 } = 0.851 = 85.1\%