the total volume of the fireball formed at chemical freeze-out and $N_i$ is the total number of the i-th baryon. We assume that the volume of the fireball, V is the same for all types of hadrons at the time of the homogeneous emissions.

By using $E_i=m_T{\cosh}y$, eq. (2) can be written as [@Tiwari:2013]:

&=&.

Transverse Energy of Hadrons in a Thermal Model
-----------------------------------------------

The transverse energy, $E_T$ in an event is defined as: $$E_T = \sum_i E_i \sin \theta_i,$$ with $\theta_i$ as the polar angle made by the i-th particle in an event with the detector. The sum is taken over all the particles emitted into a fixed solid angle within the detector acceptance. Taking into account the calorimetry measurement of $E_T$, one redefines the energy of the individual secondaries as [@Adams:2004cb; @Adam:2016thv; @Adler:2004zn]

$$\label{Eq:ET_Cal}
       E_{\mathrm i} = \left\{
    \begin{array}{ll}
                       E_{\mathrm {total}}-m & \mbox{for baryons} \\
                       E_{\mathrm {total}} + m & \mbox{for anti-baryons} \\
                       E_{\mathrm {total}} & \mbox{for all other particles.}
    \end{array}
    \right.$$

Considering the above experimental formulae, we proceed with the formulation of the transverse energy in our excluded-volume model. Using eq. (3), we write the energy of the i-th baryon per unit rapidity as :

&=&.

Similarly, 