Invention | Free full text | Mathematical model of the working process of the plunger pump gas cap in the discharge pipeline

$$\mathrm{of}=\mathrm{dQ}–\mathrm{dL}+I_{pHAd}\mathrm{d}{\mathrm{medium\; size}}_{pHAd}–I_{pH\mathrm{oh}}\mathrm{d}{\mathrm{medium\; size}}_{pH\mathrm{oh}}+I_{Gr}\mathrm{d}{\mathrm{medium\; size}}_{Gr}–I_{\mathrm{oh}}\mathrm{d}{\mathrm{medium\; size}}_{G\mathrm{oh}}$$

(4)

$$\mathrm{dQ}=A{F}_1\left({\stackrel{￣}{\mathrm{time}}}_{\mathrm{dimension}}–\mathrm{time}\right)\mathrm{d}t+A{F}_2\left({\mathrm{time}}_{\mathrm{w}}–\mathrm{time}\right)\mathrm{d}t$$

(5)

$${\mathrm{dM}}_{\mathrm{w}}=\Sigma {\mathrm{dM}}_{\mathrm{Well}I}–{\mathrm{dM}}_{\mathrm{oh}}–{\mathrm{dM}}_{\mathrm{Pad}}+{\mathrm{dM}}_{\mathrm{Pho}}–{\mathrm{dM}}_{\mathrm{grid}}+{\mathrm{dM}}_{\mathrm{go}}$$

(7)

$$v_w=\frac{{\mathrm{dM}}_{\mathrm{w}}}{F_2\mathrm{d}\mathsf{t}·{\mathsf{r}}_{\mathrm{w}}}$$

(8)

$$\mathrm{dV}=–{\mathrm{v}}_{\mathrm{w}}{\mathrm{F}}_2\mathrm{d}\mathsf{t}$$

(10)

$${\mathrm{d}\mathrm{medium\; size}}_{\mathrm{pH}}={\mathsf{Second}}_{\mathrm{pH}}{\mathrm{F}}_2\left({\mathrm{C}}_{\mathrm{w}}–{\mathrm{C}}_2\right)\mathrm{d}\mathsf{t}$$

(11)

$${\mathrm{C}}_{\mathrm{w}}={\mathrm{p}}_{eIst}/\left({\mathrm{right}}_{\mathrm{s}}{\mathrm{time}}_{\mathrm{w}}\right)$$

(12)

$${\mathrm{Second}}_{\mathrm{pH}}=\frac{\mathsf{A}}{{\mathrm{C}}_{\mathrm{p}}\mathsf{r}{\mathsf{A}}_{\mathrm{time}}/\mathrm{D}}$$

(14)

$$\mathrm{dM}={\mathrm{d}\mathrm{medium\; size}}_{\mathrm{Pad}}–{\mathrm{d}\mathrm{medium\; size}}_{\mathrm{Pho}}+{\mathrm{d}\mathrm{medium\; size}}_{\mathrm{grid}}–{\mathrm{d}\mathrm{medium\; size}}_{\mathrm{go}}$$

(15)

$$\mathrm{of}=\mathrm{d}\left({\mathrm{MC}}_{\mathrm{v}}\mathrm{time}\right)={\mathrm{C}}_{\mathrm{v}}\mathrm{time\; domain\; management}+{\mathrm{C}}_{\mathrm{v}}\mathrm{time\; domain\; management}$$

(17)

$$\left\{\begin{array}{c}\mathrm{d}\mathrm{U}=\mathrm{d}\mathrm{ask}–\mathrm{p}\mathrm{d}\mathrm{V}\\ dV=–\left({\mathsf{S}\mathrm{dM}}_{\mathrm{northwest}I}–{\mathrm{dM}}_{\mathrm{SW}}\right)/{\mathsf{r}}_{\mathrm{w}}\\ \mathrm{p}=\left(\mathrm{k}–1\right)\mathrm{U}/\mathrm{V}\\ \mathrm{time}=\mathrm{Voltage}/\left(\mathrm{gentlemen}\right)\end{array}\right.$$

(20)

$$Dp=\frac{{\mathrm{watt}}_{\mathrm{this}}}{{\mathrm{F}}_2}=\frac{{\mathrm{C}}_{\mathrm{Stef}}\left(I_{\mathrm{w}}–I_{\mathrm{w}0}\right)}{{\mathrm{F}}_2}$$

(twenty two)

$${\mathrm{of}}_{\mathrm{w}}={\mathrm{dQ}}_{\mathrm{w}}–{\mathrm{VAT}}_{\mathrm{bathroom}}+\Sigma I_{AdwI}{\mathrm{dM}}_{\mathrm{wireless}}–I_{\mathrm{oh}w}{\mathrm{dM}}_{0\mathrm{w}}+I_{pHAd}{\mathrm{dM}}_{\mathrm{Pad}}–I_{pH\mathrm{oh}}{\mathrm{dM}}_{\mathrm{Pho}}+I_{Gr}{\mathrm{dM}}_{\mathrm{grid}}–I_{\mathrm{oh}}{\mathrm{dM}}_{\mathrm{go}}$$

(twenty three)

$${\mathrm{of}}_{\mathrm{w}}=\mathrm{d}({\mathrm{you}}_{\mathrm{w}}·{\mathrm{medium\; size}}_{\mathrm{w}})={\mathrm{medium\; size}}_{\mathrm{w}}{\mathrm{of}}_{\mathrm{w}}+{\mathrm{you}}_{\mathrm{w}}{\mathrm{dM}}_{\mathrm{w}}$$

(25)

$${\mathrm{of}}_{\mathrm{w}}={\left(\frac{\partial {\mathrm{you}}_{\mathrm{w}}}{\partial v_{\mathrm{w}}}\right)}_{\mathrm{time}}\mathrm{d}{v}_{\mathrm{w}}+{\left(\frac{\partial {\mathrm{you}}_{\mathrm{w}}}{\partial {\mathrm{time}}_{\mathrm{w}}}\right)}_{{\mathrm{v}}_{\mathrm{w}}}{\mathrm{d}\mathrm{time}}_{\mathrm{w}}$$

(26)

$${\mathrm{of}}_{\mathrm{w}}={\mathrm{C}}_{\mathrm{w}}{\mathrm{medium\; size}}_{\mathrm{w}}{\mathrm{d}\mathrm{time}}_{\mathrm{w}}+{\mathrm{C}}_{\mathrm{w}}{\mathrm{time}}_{\mathrm{w}}{\mathrm{d}\mathrm{medium\; size}}_{\mathrm{w}}$$

(27)

$${\mathrm{d}\mathrm{time}}_{\mathrm{w}}=\frac{1}{{\mathrm{C}}_{\mathrm{w}}{\mathrm{medium\; size}}_{\mathrm{w}}}$$

dQ w + ∑ i adw i dM wi − i 0 w dM 0 w + i phad dM phad − i p h 0 dM ph 0 + i g r dM gr − i 0 dM p 0

(28)

$${\mathrm{d}\mathrm{T}}_{\mathrm{w}}=\frac{1}{{\mathrm{C}}_{\mathrm{w}}{\mathrm{M}}_{\mathrm{w}}}\left({\mathrm{dQ}}_{\mathrm{w}}+\sum i_{\mathrm{adw}i}{\mathrm{dM}}_{\mathrm{wi}}-i_{0w}{\mathrm{dM}}_{0\mathrm{w}}\right)$$

(29)

$$i_{adwi}={\mathrm{C}}_w{T}_{adi}; i_{0w}={\mathrm{C}}_w{T}_w$$

(30)

$${\mathrm{dQ}}_{\mathrm{w}}={\overline{\alpha }}_w{F}_w\left({\overline{T}}_{wl}-T_w\right)d\tau$$

(31)

$${\stackrel{￣}{A}}_w=\frac{A_w}{d_C}$$

0.33 R e w 0.3 P r w 0.43 P r w / P r w l 0.25

(32)

$${\overline{\alpha }}_w=\frac{{\alpha }_w}{d_c}\left[0.021R{e}_w^{0.8}P{r}_w^{0.43}{\left(P{r}_w/P{r}_{wl}\right)}^{0.25}\right]$$

(33)

$$F_w=l_w\pi d_c+\frac{\pi d_c^2}{2}$$

(34)

$${\mathrm{dM}}_{\mathrm{w}}=\sum {\mathrm{dM}}_{\mathrm{adw}i}-{\mathrm{dM}}_{o\mathrm{w}}$$

(35)

$${{\displaystyle \int }}_{I_{pp2}}^{}\frac{\partial }{\partial I}\left(z+\frac{p}{j}+\frac{v_{pp2}^2}{2G}\right)dI+\frac{1}{y}{{\displaystyle \int }}_{I_{pp2}}^{}\frac{\partial v_{pp2}}{\partial t}dI+\mathsf{S}D{H}_I=0$$

(36)

$$v_{pp2}=\sqrt{\frac{2G\left[\left(z_{1pp2}+\frac{p}{{\rho }_{w}g}\right)-\left(z_{2pp2}+\frac{P_d}{{\rho }_{w}g}\right)\right]}{\left({\lambda }_{pp2}\frac{l_{pp2}}{d_{pp2}}+\mathsf{\Sigma }{\xi }_i\right)}}$$

(37)

$$\begin{array}{c}{r}_w\frac{\partial {\mathrm{ask}}_{wpp2}}{\partial t}+F_{pp2}\frac{\partial p}{\partial X}+\frac{I_{pp2}{r}_w}{2d_{pp}{F}_{pp2}}{\mathrm{ask}}_{wpp\mathrm{time}\mathrm{phosphorus}2}\left|{\mathrm{ask}}_{w\mathrm{time}\mathrm{phosphorus}2}\right|=0\\ \end{array}$$

(38)

$$\frac{r_w{A}^2}{F_{pp2}}\frac{\partial p}{\partial X}+\frac{\partial p}{\partial t}=0$$

(39)

$${\mathrm{dM}}_{oh\mathrm{w}}=r_w{v}_{pp2}{F}_{pp2}dt={\mathrm{ask}}_{wpp2}{r}_{w}dt$$

(40)