## 2 Theoretical Analysis
Here, \(q_{\text{max}}\) is maximum flow of the pipeline connected with the accumulator, in m\({}^{3}\)/s; \(q_{a}\) is current flow, in m\({}^{3}\)/s. \[n=\frac{0.4}{q_{\text{max}}}q_{a}+1 \tag{1}\]  
In consideration of the heat dissipation loss, the corresponding energy conversion \(E_{h}\) along with the change of pressure of the hydraulic oil in the hydraulic accumulator can be calculated, as shown in Equation (2). According to the working principle of the hydraulic accumulator, pressure adjustment of the hydraulic oil is realized by controlling pressure of the pre-stored nitrogen in it. Here, \(V_{1}\) and \(V_{2}\) are volume of the nitrogen in the hydraulic accumulator before and after change, in m\({}^{3}\); \(P_{0}\) is pre-charge pressure of the hydraulic accumulator, in Pa; \(V_{0}\) is the corresponding gas volume in the hydraulic accumulator for the pre-charge pressure, in m\({}^{3}\). \[E_{h}=\int\limits_{V_{1}}^{V_{2}}{{P}dV}=\int\limits_{V_{1}}^{V_{2}}{{P_{0}}( \frac{V_{0}}{V})}^{n}{dV}=\int\limits_{V_{1}}^{V_{2}}{{P_{0}}(\frac{V_{0}}{V} )}^{\frac{0.4}{q_{a}+1}}{dV} \tag{2}\]  
Supposing that temperature in the hydraulic accumulator does not change in energy conversion process, the corresponding energy conversion can be calculated by Equations (3) and (4) [28]. Here, \(E\) is the energy conversion in the condition of the isothermal process, in J; \(E_{h}\) is the energy conversion in the condition of the adiabatic process, in J; \(E_{r}\) is the heat dissipation loss, in J; \(P_{1}\) and \(P_{2}\) are the pressure of the accumulator before and after the change, in Pa.  
\[E=\int\limits_{V_{1}}^{V_{2}}{{P}dV}=\frac{{{P_{0}}{V_{0}}}}{0.4}\left[{{{{ \left({\frac{{{P_{1}}}}{{{P_{0}}}}}\right)}^{\frac{0.4}{4}}}}-{{{\left({\frac {{{P_{2}}}}{{{P_{0}}}}}\right)}^{\frac{0.4}{4}}}}}\right] \tag{3}\]  
\[E_{r}=E-E_{h} \tag{4}\]  
Besides the heat dissipation loss, there are also on-way pressure loss in the pipeline and local pressure loss in the joint and elbow [29]. In order to simplify the calculation, the local pressure loss is replaced by the equivalent on-way pressure loss. Therefore, the pressure loss \(\Delta P_{\text{L}}\) (in Pa) between the two hydraulic components can be calculated by the Equation (5), and the energy loss \(W_{\text{L}}\) (in J) can be calculated by the Equation (6). Here, \(\xi\) is the on-way resistance coefficient, which can be consulted in the handbook for the mechanical design or handbook for the hydraulic transmission design; \(L_{\text{e}}\) is total equivalent effective length of the pipeline, in m; \(Q_{\text{L}}\) is the flow passed through the pipeline, in m\({}^{3}\)/s; \(g\) is the gravitational acceleration, in m/s\({}^{2}\); \(D\) is diameter of the pipeline, in m; \(A\) is cross-sectional area of the pipeline, in m\({}^{2}\). \[\Delta P_{\text{L}}=\frac{{\xi L_{\text{e}}{Q_{\text{L}}}^{2}}}{2 gDA^{2}} \tag{5}\]  
\[W_{\text{L}}=\int\limits_{0}^{\text{t}}{\Delta P_{\text{L}}Q_{\text{L}}dt}= \int\limits_{0}^{\text{t}}{\frac{{\xi L_{\text{e}}{Q_{\text{L}}}^{3}}}{2 gDA^{2 }}dt} \tag{6}\]  
When the vehicle is braked, the mechanical energies of the wheels are transferred to hydraulic energies of the hydraulic accumulator, and the pressure is increased until the maximum pressure, which indicates that the hydraulic accumulator is full filled. Afterwards, the residual braking energy flowsback to the low-pressure accumulator through the overflow valve, and these energies cannot be regenerated. The energy recovery rate is defined as the ratio between the recycled energy by the hydraulic accumulator and the kinetic energy change of the vehicle before and after braking, and it can be calculated by the Equation (7). Here, \(m\) is weight of the vehicle, in \(Kg\); \(v_{0}\) and \(v_{1}\) are the velocity of the vehicle before and after braking, in m/s. In this way, the theoretical energy recovery rate \(\eta_{\rm acc}\) can be obtained.