Question 11.12: A diesel engine has a compression ratio of 19.2 to 1. Air at...

First, draw a sketch of the system (Figure 11.2).

The unknowns here are the final temperature and pressure of the air at the end of the compression stroke and the work required per lbm of air present.

The piston-cylinder arrangement of a diesel engine forms a closed system for the air being compressed. The unknowns are T_2, p_2, and _1W_2/m. The energy balance for this system (neglecting any changes in the potential and kinetic energies of the air) is

_1Q_2 − _1W_2 = m(u_2 − u_1)

Now, _1Q_2 = 0 (isentropic processes are also adiabatic), so

_1W_2/m = u_1 − u_2

The gas tables are to be used for the thermodynamic properties of air here, because they are more accurate than the standard constant specific heat ideal gas equations. From Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at 60.0°\text{F} = 520.  \text{R},

u_1 = 88.62  \text{Btu/lbm}

p_{r1} = 1.2147

and

v_{r1} = 158.58

For a compression ratio of  19.2 to 1, v_2/v_1 = 1/19.2. Then, from Eq.  (11.37),

v_2/v_1 = v_{r2}/v_{r1}                     (11.37)

v_{r2} = v_{r1}(v_2/v_1) = 158.58/19.2 = 8.26

scanning down the v_r column in Table C.16a, we find that v_r = 8.26 at about

T_2 = 1600  \text{R}= 1140°\text{F}

u_2 = 286.06  \text{Btu/lbm}

and

p_{r2} = 71.73

Then, from Eq. (11.36),

p_2/p_1 = p_{r2}/p_{r1}                      (11.36)

p_2 = p_1(p_{r2}/p_{r1}) = (14.7  \text{psia})(71.73/1.2147) 

= 868.1  \text{psia}

Finally, from the preceding energy balance,

_1W_2/m = u_1 − u_2 = 88.62 − 286.06 = −197.44  \text{Btu/lbm}