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Potential Flows Analyses - Lab Report Example

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Extract of sample "Potential Flows Analyses"

TЕNTIАL FLОWS АNАLYSЕS Student Course Date РОTЕNTIАL FLОWS АNАLYSЕS Any potential flow is assumed to be irrational and frictionless. This is because of the degree of their viscosities’ effects of certain fluids being small enough and thus, they can be neglected. Due to these characteristics of the potential flow, it is applied in several applications such as stationary and non-stationary flows. The irrotational nature of the velocity field is due to the curl effect on the gradient of a scalar quantity that is always equal to zero (Kundu, Cohen, Dowling, & Tryggvason 2016). Thus, in this case, the incompressible flow of the velocity potential will be considered. Also, it should be known that the knowledge of potential flows is important due to it its numerous applications it is employed in. For example, it is used in designing of outer fields of water waves, aerofoils, ground water flow and electro-osmotic flow. In this case, the derivation of the stream and potential functions of potential flow and their application is shown by using the digits of my ID number as the strength of the standard flows as follows. Since my ID number is 14074524, then the three standard flows chosen were as follows. 14, 74 and 52 as my uniform flow, sink and source strengths respectively. With respect to that, the derivation of the stream function was done as follows; Derivation of Stream function Since the stream function works only when the continuity equation is reduced to two terms, then from the potential flow equation of the cylindrical coordinates; By eliminating the unsteady state from the above equation, so that it becomes the plane incompressible flow, the equation above becomes; Thus, in order to satisfy the above equation, a function is defined and the simplified equation above becomes; By taking the curl of the momentum equation and applying it in the curl of the above equation, the following equation is obtained Thus, in order to solve the above equation, the boundary conditions are applied at the body and at infinity as follows, At the body; At infinity, Thus, the above equation reduces to; Therefore, At infinity; At the body; Therefore, the solution of the above equation under the uniform flow, sink and source in cylindrical coordinates with an assumption of the values of the constant being zero as follows; Uninform flow; Sink flow, Source flow Thus, the overall stream function of the From the combined flow function above, the velocity components in relation to the stream functions were as follows; And Derivation of the velocity potential functions From the above condition of the irrotationality of the velocity potential of the flow, it means that the curl of the velocity giving to zero shows that the velocity is a vector with a grad of a scalar that can be represented as follows; If This scalar quantity is similar and complementary to the stream function. Thus, from this knowledge of the velocity potential function, the velocity components are found as; Therefore, taking the divergence of the velocity potential flow, the equation of the continuity reduces to the following; Thus, solving the above equation by applying similar boundary equations as those applied in the stream function and then solving for the three flows, the following are the results that was obtained. Uninform flow; Sink flow, Source flow Thus the overall potential function of the From the combined flow function above, the distribution components in relation to the potential function are as follows; And Plots Based on the above velocity components and distribution lines, the plots were done as follows; In order to find the point of stagnation, , thus from Thus, But since then from , must produce a positive number that can give a result of zero in . Thus, . From this, considering the value of Therefore, the stagnant point will be at; Thus, By considering the stream function when it at zero, the SL is found as; Thus, the streamline is found by equating And then, From the above equations, then solution of the streamline is obtained as (θ) r1(+θ) r2 (-θ) 0.314159 1.5976 1.5976 0.628319 1.6798 1.6798 0.942478 1.8307 1.8307 1.256637 2.0763 2.0763 1.570796 2.4684 2.4684 1.884956 3.1145 3.1145 2.199115 4.2715 4.2715 2.513274 6.7192 6.7192 2.827433 14.3782 14.3782 3.141593 40295430613829800.0000 40295430613829800.0000 Also, considering the velocity distribution functions above in this case, then from the combined flow function above, the distribution components in relation to the potential function are as follows; And Thus, from the above velocity distribution functions obtained from the potentials, the solution of the distribution functions is obtained as; Thus, in order to obtained a zero in this functions, the But, in order to obtain a zero in the second equation, the must be negative since, the value of r cannot be negative. Thus, then the solution of the distribution is found by replacing the angle by as follows; Then Stagnant point is obtained as follows; Thus, from the potential function obtained above, the solution of the potential function is as shown below Differentiating the above equation and equating to zero, the following results is obtained, This implies that; By using the MS Excel in solving the above equations, the following table shows the results obtained; (θ) dr1(+θ) dr2(-θ) 0.314159 29.1246118 -29.1246118 0.628319 15.31171455 -15.31171455 0.942478 11.1246118 -11.1246118 1.256637 9.463160018 -9.463160018 1.570796 9 -9 1.884956 9.463160018 -9.463160018 2.199115 11.1246118 -11.1246118 2.513274 15.31171455 -15.31171455 2.827433 29.1246118 -29.1246118 3.141593 7.34605E+16 -7.34605E+16 From the above graphs, the combined stream function and the velocity potential graphs were obtained from the following data. (θ) r1(+θ) r2 (-θ) dr1(+θ) dr2(-θ) 0.314159 1.5976 1.5976 29.1246118 -29.1246118 0.628319 1.6798 1.6798 15.31171455 -15.31171455 0.942478 1.8307 1.8307 11.1246118 -11.1246118 1.256637 2.0763 2.0763 9.463160018 -9.463160018 1.570796 2.4684 2.4684 9 -9 1.884956 3.1145 3.1145 9.463160018 -9.463160018 2.199115 4.2715 4.2715 11.1246118 -11.1246118 2.513274 6.7192 6.7192 15.31171455 -15.31171455 2.827433 14.3782 14.3782 29.1246118 -29.1246118 3.141593 40295430613829800.0000 40295430613829800.0000 7.34605E+16 -7.34605E+16 From this data, the combined graph obtained was as follows; From the combined graphs above of the velocity potential and the stream functions, it was observed that the two different functions had different direction of motion at the point when the function were approaching at . But, it should be noted that the best results required is when the two functions are at 90 degrees to each other. Therefore, the function of the velocity functions at which the stream function was at 90 degrees was that when the negative of the angle was considered. Thus, changing the position of the stream lines will lead to the change in the distribution of the stream lines of the flow. By changing the position of the stream source and the sink in this case will lead to the change of the stream line distribution through changing the value of the angle through which the stream line function will be operating. Since, in this case, the source, and the sink were considered at the origin, then changing the position of the two will have depicted a different angle the s to one another that could have made it more complex to solve. For the case of the pressure distribution required to work out, the following formula was used; In this case, the fluid considered here is air with a density of 1.225 kg/m3 and 100, 000 Pascal. Thus, the pressure distribution was obtained at 2.7 radians. Therefore, the pressure obtained was 99967.66 Pa From the analysis, it was observed that the pressure distribution was directly affected by the source or sink strength and indirectly by the uniform flow and the position on the source and sink flow to the system. Thus, in this case, the minimum pressure was observed at the point when the angle of flow was approaching 180 degrees. In summary, knowledge of fluid flow analysis is important because it helps one to determine the various factors that affect the flow of fluids. Through this, one is able to design the right parameter of the object or equipment used in airlifts. References Kundu, P. K., Cohen, I. M., Dowling, D. R., & Tryggvason, G. 2016. Fluid mechanics. Massey, B. S., & Ward-Smith, J. 2012. Mechanics of fluids. London, Spon Press. Schlichting, H. 1949. Lecture series "boundary layer theory". Washington, DC, National Advisory Committee for Aeronautics. Vandyke, M. 2012. ˜An album of fluid motion. Stanford, Calif, Parabolic Press. White, F. M. 1986. Fluid mechanics. New York, McGraw-Hill. Read More
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