Question 13.15: A six-cylinder, four-stroke Otto cycle internal combustion e...

a. From Eq. (13.30), using k = 1.40 for the cold ASC,

(η_T)_{\substack{\text{Otto}\\\text{cold ASC}\\}} = 1 − \text{CR}^{1−k} = 1 − 9.00^{− 0.40}  = 0.585 = 58.5\%

b. From Figure 13.48a

\dot{Q}_H = \dot{Q}_\text{fuel} = (\dot{m} c _v)_a (T_1 − T_{4s}) = \dot{m}_\text{fuel} (A/F) (c_v)_a (T_1 − T_{4s})

and

T_1 = T_\text{max} = T_{4s} + \frac{(\dot{Q} /\dot{m})_\text{fuel}}{(A/F)_\text{mass} (c_v)_a}

Since process 3 to 4s is isentropic, Eq. (7.38) gives

\frac{T_{os}}{T} = (\frac{p_{os}}{p})^{(k−1)/k} = (\frac{v_{os}}{v})^{1− k} = (\frac{p_{os}}{p})^{k−1}                    (7.38)

T_{4s} = T_3 \text{CR}^{k-1} = (60.0 + 459.67) (9.00)^{0.40} = 1250  \text{R}

Then,

T_\text{max } = \frac{20.0 × 10³  \text{ Btu/lbm fuel}}{(16.0  \text{lbm air/lbm fuel})[0.172  \text{ Btu/(lbm air.R)}]} + 1250  \text{R} = 8520  \text{R}

Since process 4s to 1 is isochoric, the ideal gas equation of state gives

p_\text{max} = p_1 = p_{4s} (T_1/T_{4s})

and, since the process 3 to 4s is isentropic,

T_{4s}/T_3 (p_{4s}/p_3)^{(k−1)/k}

or

p_{4s} = p_3 (T_{4s}/T_3)^{k/(k-1)} = (8.00  \text{psia}) (\frac{1250  \text{R}}{520  \text{R}})^{1.40/0.40} = 172  \text{psia}

then,

p_\text{max} = (172  \text{psia}) [(8520  \text{R})/1250  \text{R}] = 1170  \text{psia}

c. Equation (13.34) gives the indicated power as

(\dot{W}_1)_\text{out} = \frac{(η_T)_\text{ASC} (\dot{Q}/\dot{m})_\text{fuel} (DNp_1/C)}{(A/F) (RT_1) (CR – 1)}               (13.34)

\left|\dot{W}_1\right|_\text{out} = \frac{(0.585)(20.0 × 10³  \text{Btu/lbm})(260.  \text{ in³/rev})(4000.  \text{rev/min})(1170  \text{lbf /in²})/2}{(16.0)[0.0685  \text{Btu/(lbm.R)}](8520  \text{R})(9.00 − 1)(12  \text{in/ft})(60  \text{s/min})}

=(132,00  \text{ft.lbf/s}) (\frac{1  \text{hp}}{550  \text{ ft.lbf/s}} ) = 241  \text{hp}

d. Equation (13.33) gives the mechanical efficiency of the engine as

η_m = \frac{(\dot{W}_B)_\text{out}}{(\dot{W}_1)_\text{out}} = \frac{85.0  \text{hp}}{241  \text{hp}} = 0.353 = 35.3\%

e. Finally, the actual thermal efficiency of the engine can be determined from 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_T)_{\substack{\text{Otto}\\\text{actual}\\}} = \frac{(\dot{W}_B)_\text{out}}{\dot{Q}_\text{fuel}} = \frac{(η_m)(\dot{W}_1)_\text{out}}{\dot{Q}_\text{fuel}} = (η_m) (η_T)_{\substack{\text{Otto}\\\text{cold ASC }\\}}

= (0.353)(0.585) = 0.207 = 20.7\%