| Literature DB >> 25332866 |
Caroline W Kanyiri1, Mathew Kinyanjui1, Kang'ethe Giterere1.
Abstract
In this research study, fluid flow through a cylindrical collapsible tube has been investigated. Of particular interest is the effect of flow parameters on the cross sectional area of a collapsible tube, flow velocity and internal pressure of the fluid. The flow parameters considered are longitudinal tension and volumetric flow rate. The tube is considered collapsible in the transverse direction, taken to be perpendicular to the main flow direction. Collapse happens when external pressure exceeds internal pressure and hence the tube results to a highly noncircular cross sectional area. The fluid flow in consideration is steady and incompressible. Equations governing the flow are non-linear and cannot be solved analytically. Therefore an approximate solution to the equations has been determined numerically. In this case, finite difference method has been used. A computer program has then been used to generate the results which are presented in form of graphs. The results show that the longitudinal tension is directly proportional to both the cross sectional area and internal pressure and inversely proportional to the flow velocity and that change in volumetric flow rate has no effect on the cross sectional area but it is directly proportional to the flow velocity and inversely proportional to the internal pressure.Entities:
Keywords: Collapsible tube; Flow parameters; Newtonian fluid; Steady flow
Year: 2014 PMID: 25332866 PMCID: PMC4193970 DOI: 10.1186/2193-1801-3-566
Source DB: PubMed Journal: Springerplus ISSN: 2193-1801
Figure 1Cross sectional area versus distance with K = 1.21 × 10 ρ = 1.0 × 10 Pe = 4.00 × 10 r = 4.3 × 10 Q = 5 × 10 .
Figure 2Flow Velocity versus distance with K = 1.21 × 10 ρ = 1.0 × 10 Pe = 4.00 × 10 r = 4.3 × 10 Q = 5 × 10 .
Figure 3Internal Pressure versus distance with K = 1.21 × 10 ρ = 1.0 × 10 Pe = 4.00 × 10 r = 4.3 × 10 Q = 5 × 10 .
Figure 4Cross sectional area versus distance for T = 4.0 × 10 K = 1.21 × 10 ρ = 1.0 × 10 Pe = 4.00 × 10 r = 4.3 × 10 .
Figure 5Flow velocity versus distance for T = 4.0 × 10 Kp = 1.21 × 10 ρ = 1.0 × 10 Pe = 4.00 × 10 r = 4.3 × 10 .
Figure 6Internal pressure versus distance for T = 4.0 × 10 Kp = 1.21 × 10 ρ = 1.0 × 10 Pe = 4.00 × 10 r = 4.3 × 10 .