91Ó°ÊÓ

Firstly, let's compute the rolling resistance of the vehicle. We know the resistance is 6 lbf per 1000 lbf of vehicle weight, that is \(R_{r} = 6 \times \frac{{72,000}}{{1,000}} = 432 \, \mathrm{lbf}\). This is the force due to the rolling resistance of the vehicle.

Secondly, let's compute the aerodynamic drag. We know that the drag force \(D_{f}\) = \(0.5 \times C_{D} \times \rho_{a}\times A \times V^{2}\), where \(\rho_{a}\) is air density, \(V\) is velocity, \(C_{D}\) is the drag coefficient, and \(A\) is the frontal area of the vehicle. Given that \(\rho_{a}\) is about \(0.002378 slug/ft^{3}\) (at sea level and at 15.6°C) and \(V = 55mph = 80.67 ft/sec\), we get \(D_{f} = 0.5 \times 0.9 \times 0.002378 \times 102 \times (80.67)^{2} = 533.67 \, \mathrm{lbf}\).

The net force required to move the vehicle at constant speed is \(F_{n} = R_{r} + D_{f} = 432 + 533.67 =965.67 \, \mathrm{lbf}\).

Power \(P\) to overcome resistance is \(P = F_{n} \times V\). Substituting \(F_{n} = 965.67 \, \mathrm{lbf}\) and \(V = 80.67 \, \mathrm{ft/sec}\), we get \(P = 77872.85 \, lbf.ft/sec\). Since \(1 \, \mathrm{hp} = 550 \, \mathrm{lbf.ft/sec}\), we can convert the power to hp getting \(P =\frac{{77872.85}}{{550}} = 141.59 \, \mathrm{hp}\).

Given the specific fuel consumption of the engine to be 0.34 lbm/hp-hr, the fuel consumed \(F_{c}\) = \(0.34 \times 141.59 = 48.14 \, \mathrm{lbm/hour}\). Since the engine is 92% efficient, the actual fuel consumed should adjust for this. So, the actual fuel consumption per hour \(F_{ca} = \frac{{48.14}}{{0.92}} = 52.33 \, \mathrm{lbm/hour}\).

Since the density of diesel fuel is 6.9 lbm/gal, we can convert fuel consumption \(F_{ca}\) from lbm/hour to gal/hour, so the fuel consumption rate is \(\frac{{52.33}}{{6.9}} = 7.58 \, \mathrm{gal/hour}\). Then, to find the fuel economy, one can divide the speed by the fuel consumption rate, giving \( \frac{{55}}{{7.58}} = 7.26 \, \mathrm{mpg}\).

Now, let's calculate the fuel saved per year by the roof fairing. The air fairing system can reduce aerodynamic drag by 15 percent, that is a new drag coefficient \(C_{D,n} = 0.9 \times (1 - 0.15) = 0.765\). Using this new \(C_{D,n}\) can give a new aerodynamic drag \(D_{f,n} = 0.5 \times 0.765 \times 0.002378 \times 102 \times (80.67)^{2} = 453.12 \, \mathrm{lbf}\). Consequently, fuel economy with the fairing system is computed similar with Steps 3-6 but with \(D_{f,n}\). Suppose the fuel economy with the fairing system computed is \( \mathrm{mpg_{n}}\). Then, the fuel saved per year will be \((1 - \frac{{\mathrm{mpg}}}{\mathrm{mpg_{n}}}) \times \mathrm{Annual} \, \mathrm{Mileage} \, \mathrm{distance}/ \mathrm{mpg}\).