the energy of the m-th meson per unit rapidity can be calculated as : $$\frac{dE_m}{dy}=\frac{g_mV\lambda_m}{(2{\pi}^2)}\;\int \frac{m_T^{3}\;{\cosh^2}y\;dm_{T}}{\displaystyle\Big[\exp\left(\frac{m_T\;{\cosh}y}{T}\right)\Big]}.$$ Here, $g_m$ and $\lambda_m$ are the degeneracy factor and fugacity of the m-th meson. The above equations give the transverse energy of the particles in a stationary thermal source.

Transverse Energy of Hadrons in a Thermal Model with Flow
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The invariant yield with the inclusions of collective flow can be written as [@Schnedermann:1993] :

=(-)r drd.

Here, $\rho$ is the parameter given by $\rho={\tanh}^{-1}\beta_r$, with the profile chosen as $\beta_r=\displaystyle\beta_s\;\Big(\xi\Big)^n$ [@Schnedermann:1993; @Braun:1996]. $\beta_s$, the maximum surface velocity appears as a free parameter in our model calculation and $\xi=\displaystyle\Big(r/R_0\Big)$, $R_0$ is the maximum radial position of the fireball at freeze-out. The average of the radial flow can be calculated by the following formula [@Adcox:2004] : $$\begin{aligned}
<\beta_r> =\frac{\int \beta_s\xi^n\xi\;d\xi}{\int \xi\;d\xi}=\Big(\frac{2}{2+n}\Big)\beta_s.\end{aligned}$$ For the sake of simplicity, we take n=1 in our model calculation. After incorporating the collective flow in our thermal model, we get the expressions for energy of baryons per unit rapidity as follows :

=(-)m\_[T]{}\^[3]{}yrdrdddm\_[T]{},

where $r$ lies between $0$ and $R_0$ i.e. the radius of the fireball at freeze-out, the azimuthal angle $\phi$ lies between $0$ and $2\pi$, and the longitudinal space-time 