Conservation of energy in explosion

I think when you say “shell” you might mean “gun”.

When a bullet is fired from a gun, the first stage is that the chemical energy in an explosive in the bullet’s cartridge is released to create hot gas under pressure. This compressed gas has potential energy, which is the energy of the explosion $E$.

As the hot gas expands in the barrel of the gun it propels the bullet forward and exerts an equal force on the gun, pushing it backwards - this is known as “recoil”. Most of the gas’s potential energy is converted into kinetic energy of the gun and kinetic energy of the bullet. However, some energy is lost as the gas escapes from the open barrel of the gun and expands into the surrounding air. So it would be more accurate to say that $E$ is slightly greater than the kinetic energy of the gun plus the kinetic energy of the bullet.

If we neglect the momentum of the escaping gas, then the momentum of the gun and the bullet will be approximately equal and opposite, due to conservation of momentum. However, the bullet’s mass is much smaller than the gun’s, so it has a much higher velocity.