In a motorcycle engine, a piston is forced down toward the crankshaft when the fuel in the top of the piston’s cylinder undergoes combustion. The mixture of gaseous combustion products then expands adiabatically as the piston descends. Find the average power in (a) watts and (b) horsepower that is involved in this expansion when the engine is running at 4000rpm, assuming that the gauge pressure immediately after combustion is 15atm, the initial volume is 50cm³, and the volume of the mixture at the bottom of the stroke is 250cm³. Assume that the gases are diatomic and that the time involved in the expansion is one-half that of the total cycle.

1. The engine is running at 4000rpm.
2. The gauge pressure immediately after combustion is p_i=15atm.
3. The gases are diatomic.
4. The initial volume of the gas,$V_i=50c{m}^{3}or50\times {10}^{-6}{m}^3.$
5. The final volume of the gas, $V_f=250c{m}^{3}or250\times {10}^{-6}{m}^3$.
6. The time involved in the expansion is one-half that of the total cycle.

When a gas undergoes adiabatic expansion, the system undergoes a no heat interaction thermodynamic process with having no connection to the heat of the surroundings, where the work done by the system is because of its internal energy change. The compression of a gas adiabatically will cause a rise in the temperature of the gas, which will result in a decrease in the volume of the gas. Similarly, Adiabatic expansion results in a drop in temperature and increase in volume.

The pressure-volume relation during an adiabatic expansion,

$p{V}^{\gamma }=constant$ (i)

where, $\gamma$is the ratio of molar specific heat at constant pressure to molar specific heat at constant volume. The work done by the system in a adiabatic expansion,

localid="1662691461102" $W=\frac{(p_f{V}_f-p_i{V}_i)}{(1-\gamma )}$ (ii)

where, p_f is the final pressure on the gas, p_i is the initial pressure on the gas, V_f is the final volume of the gas, V_i is the initial volume of the gas.

The average power for the thermodynamic expansion,

$P=\frac{W}{\Delta t}$ (iii)

where, W is the total work done by the system, $\Delta t$is the time taken for the process.

The conversion factor of power from watt to horsepower,

$1W=1.34\times {10}^{-3}hp$ (iv)

For diatomic gas, $\gamma =\frac{7}{5}$.

From equation (i), the final pressure of an adiabatically expanded as can be given using the given data as follows:

(where, p_fis the final pressure on the gas, p_i is the initial pressure on the gas, V_f is the final volume of the gas, V_i is the initial volume of the gas.)

$$\begin{gathered} p_f={\left(\frac{V_i}{V_f}\right)}^{\gamma }{p}_i \\ ={\left(\frac{50\times {10}^{-6}{m}^3}{250\times {10}^{-6}{m}^3}\right)}^{\frac{7}{5}}\times 15atm \\ =1.58atm \end{gathered}$$

The work done by the system can be given in joules using the data in equation (ii) as follows:

$$\begin{gathered} W=\frac{(p_f{V}_f-p_i{V}_i)}{(1-\gamma )} \\ =\frac{1.01\times {10}^{5}Pa/atm\times (1.58atm\times 250\times {10}^{-6}{m}^3-15atm\times 50\times {10}^{-6}{m}^{3)}}{\left(1-{\displaystyle \frac{7}{5}}\right)} \\ =89.64J \end{gathered}$$

Now, the period of each cycle is given as follows:

(Since, it revolves 4000 times in one minute)

$\begin{array}{l}\zeta =\frac{60s}{4000}\\ =0.015s\end{array}$

Since the time involved in the expansion is one-half of the total cycle, we get the required time of the expansion process as follows:

$\begin{array}{l}\Delta t=\frac{\zeta }{2}=\frac{0.015}{2}\\ =7.5\times {10}^{-3}s\end{array}$

Thus, the average power for expansion is given using the above values in equation (iii) as follows:

$$\begin{gathered} P_{avg}=\frac{89.64J}{7.5\times {10}^{-3}s} \\ =12\times {10}^{4}W \end{gathered}$$

Hence, the value of the average power in watts is 12$\times {10}^{4}W$.

From equation (iv), the value of one watt as horsepower can b given as:

$$\begin{gathered} 1W=\frac{1}{746}hp \\ =1.34\times {10}^{-3}hp \end{gathered}$$

The average power for the expansion in horsepower using the above value and power value of part (a) is given as follows:

$$\begin{gathered} p_{avg}=(1.2\times {10}^4\times 1.34\times {10}^{-3})hp \\ =16.1hp \\ =16hp \end{gathered}$$

Hence, the value of average power is 16hp.