# Research on the Principle of a New Flexible Screw Conveyor and Its Power Consumption
## 3 The Conveying Power Consumption Model
### Power Consumption Calculation  
The friction consumption of fibers is \(P_{1}\).  
As shown in the Figure 6, the frictional force can be represented as:  
\[df_{1}=\tan\delta\rho_{(z)}hbgd\cos\alpha \tag{14}\]  
Figure 5: Diagram of the velocity of a single particle.  
where \(\rho_{(z)}\) is the density of the powder material, \(h\) is the thread height, \(b\) is the spiral groove width, \(\mathrm{d}l\) is the length along the spiral element, and \(\delta\) is the friction angle.  
The corresponding torque infinitesimal \(\mathrm{d}M_{1}\) can be represented as:  
\[\mathrm{d}M_{1}=\frac{1}{2}(D-d_{Z})\rho_{(z)}hbg\tan\delta\cos\alpha\lambda \frac{\pi D}{p}\mathrm{d}z \tag{15}\]  
where \(D\) is the diameter of the inner wall, and \(d_{\mathrm{z}}\) is the screw shaft outer diameter.  
By the formula \(P=M\omega\), we have:  
\[P_{1}=M_{1}\omega=\omega\int_{0}^{z}\frac{(D-d_{Z})\rho_{(z)}hbg\tan\delta \cos\alpha\lambda\pi D}{2p}\mathrm{d}z \tag{16}\]  
In Figure 7, the frictional force \(df_{2}\) is:  
\[df_{2}=\tan\delta_{1}\omega^{2}\frac{D-d_{Z}}{2}dm=\frac{\rho_{(z)}hb\omega^{2 }(D-d_{Z})\pi D\tan\delta_{1}}{2p}\mathrm{d}z \tag{17}\]  
Figure 6: Diagram of the stress of the material along the spiral surface.  
Figure 7: Vertical plane stress analysis.  
The corresponding torque infinitesimal d\(M_{2}\) can be represented as:  
\[\mathrm{d}M_{2}=\frac{D-d_{Z}}{2}df_{2}=\frac{(D-d_{Z})^{2}}{4}\frac{\rho_{(z)}hb \omega^{2}\pi D\tan\delta_{1}}{p}\mathrm{d}z\]  
The consumption of the material due to the interaction with the spiral wall is \(P_{2}\).  
By the formula \(P=M\omega\), we have:  
\[P_{2}=M_{2}\omega=\frac{\omega(D-d_{Z})^{2}hb\omega^{2}\pi D\tan\delta_{1}}{4p} \int_{0}^{z}\rho_{(z)}\mathrm{d}z \tag{18}\]  
Power is required to overcome the elastic deformation of the flexible helical fibers [13; 14]. When the flexible screw rotates with the helical axis, the centrifugal force of single fiber will produce a force \(f_{0}\) on tangent direction.  
\(f_{4}\) is the difference between d\(N_{1}\) and \(f_{0}\). It is the elastic potential energy, as shown in Figure 8.  
The centrifugal force\(f_{0}\) and the friction force\(f_{4}\) are:  
\[f_{0}=m\omega l^{2} \tag{19}\]  
\[f_{4}=\mu m\omega l^{2}+f_{3} \tag{20}\]  
where \(m\) is the quality of a single fiber, and \(\omega\) is the rotating speed of the screw shaft.  
We assumed the elastic potential energy of the fibers as the elastic potential energy of a spring:  
\[E_{P}=\frac{1}{2}k\delta^{2} \tag{21}\]  
where \(\delta\) is the deformation of the fiber bundle, and \(k\) is the spring stiffness which depends on the material itself:  
\[\delta=\frac{f_{4}l}{\mathrm{E}A} \tag{22}\]  
The total elastic energy of the nylon fiber is:  
\[E_{z}=NE_{p}=\frac{1}{2}\mathrm{N}k\bigg{[}\frac{(\mu m\omega l^{2}+\mu dN_{1} )l}{\mathrm{E}\pi{d_{s}}^{2}}\bigg{]}^{2} \tag{23}\]  
Figure 8: Stress analysis of the elastic deformation of the flexible spiral strand.  
We can obtain the total power consumption of elastic energy of the nylon fiber:  
\[P_{3}=\frac{E_{Z}}{t}=\frac{1}{2t}\text{Nk}\Big{[}\frac{(\mu m\omega l^{2}+\mu dN _{1})l}{\text{E}\tau{d_{s}}^{2}}\Big{]}^{2} \tag{24}\]  
where \(t\) is the working time.  
To sum up, the total power consumption \(P\) is:  
\[\left\{\begin{array}{l}P_{1}=M_{1}\omega=\omega\int_{0}^{z}\frac{(D-d_{Z}) \rho_{(z)}h{p}\tan\delta\cos\alpha\Lambda\pi D}{2p}\text{d}z\\ P_{2}=M_{2}\omega=\frac{\omega(D-d_{Z})^{2}h\omega^{2}\pi D\tan\delta_{1}}{2 \pi{d_{s}}^{2}}\int_{0}^{z}\rho_{(z)}\text{d}z\\ P_{3}=\frac{E_{Z}}{t}=\frac{1}{2t}\text{Nk}\Big{[}\frac{(\mu m\omega l^{2}+\mu dN _{1})l}{\text{E}\pi{d_{s}}^{2}}\Big{]}^{2}\\ \\ P=P_{1}+P_{2}+P_{3}\end{array}\right. \tag{25}\]