Solved Problem on Thermodynamics

A cannon launches a projectile with a mass equal to 1225 kg with a speed of 762 m/s. The projectile is propelled by a mass of 300 kg of gunpowder, it is known that 1 g of gunpowder burns by producing 1500 cal. What is the thermal efficiency of this cannon as a heat engine? Assume 1 cal = 4.2 J.

Problem data:

Mass of the projectile: m_P = 1225 kg;
Velocity of the projectile: v = 762 m/s;
Mass of the gunpowder: m_G = 300 kg;
Heat produced by the gunpowder: q = 1500 cal/g.

Problem diagram:

The mass of gunpowder m_G when it is detonated produces a quantity of heat Q, this energy does a work W on the projectile m_P. This work will be the kinetic energy K of the projectile being triggered with speed v (Figure 1).

Figure 1

Solution

First, we must convert the quantity of heat produced by the gunpowder from calories per gram (cal/g) to joules per kilogram (J/kg)

\[ q=1500\;\frac{\cancel{\text{cal}}}{\cancel{\text{g}}}\times \frac{4.2\;\text{J}}{1\;\cancel{\text{cal}}}\times \frac{1000\;\cancel{\text{g}}}{1\;\text{kg}}=6300000\;\frac{\text{J}}{\text{kg}}=6.3\times 10^{6}\;\frac{\text{J}}{\text{kg}} \]

The total energy produced by the powder will be

\[ \begin{gather} Q=m_{B}q\\ Q=300\times 6.3\times 10^{6}\\ Q=3\times 10^{2}\times 6.3\times 10^{6}\\ Q=18.9\times 10^{2}\times 10^{6}\\ Q\simeq 1.9\times 10^{9}\;\text{J} \tag{I} \end{gather} \]

The kinetic energy is given by

\[ \bbox[#99CCFF,10px] {K=\frac{m v^{2}}{2}} \]

the energy acquired by the projectile

\[ \begin{gather} K=\frac{m_{P} v^{2}}{2}\\ K=\frac{1225\times 762^{2}}{2}\\ K=\frac{1225\times 580664}{2}\\ K=\frac{711288900}{2}\\ K=355644450\\ K\simeq 3.6\times 10^{8}\;\text{J} \tag{II} \end{gather} \]

The thermal efficiency is given by

\[ \bbox[#99CCFF,10px] {\eta =\frac{W}{Q}} \]

As the work done is equal to kinetic energy W = K, the thermal efficiency will be

\[ \begin{gather} \eta =\frac{3.6\times 10^{8}}{1.9\times 10^{9}}\\ \eta=\frac{3.6\times 10^{8}}{1.9\times 10^{9}}\\ \eta \simeq 1.89\times 10^{8}.10^{-9}\\ \eta\simeq 1.89\times 10^{-1}\\ \eta \simeq 0.189=\frac{18.9}{100} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\eta \simeq 18.9\;\text{%}} \]

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