Pressure Drop Rate - Math Problem Example

Pressure drop rate The first step is to generate a solid friction factor model λs by us of the provided data. The process involves the following steps Calculation of m* from the given values of ma and ms with the calculated values being as shown in table 1 column 4. From the given values of ma and the relating to the pipe cross section the was calculation of volume per unit time and velocity as can be seen in column 6 and 7 respectively. Fr values shown in column 8 were calculated from the mean velocities The solid friction factor λs was calculated for various pressure drop using the relation λs = where the D is the pipe diameter v is velocity of air and L is length of pipe with the calculated values being as in column 11 Finding the a and b The process of finding constants a and b involved plotting Fr (x-axis) verses λsm*τ (y-axis).Various values of τ were chosen with the respective graphs being generated. When the value τ was taken as 0.2 the result was as in figure 1 where the R2=0.954 The data used to generate the graph is shown in Table column 4 and column 13 Figure 1: solid friction factor graph for τ =0.2 When the value τ was taken as 0.3 the result was as in figure 1 where the R2=0.955 The data used to generate the graph is shown in Table column 4 and column 15 Figure 2 : Solid friction factor graph for τ =0.2 When the value τ was taken as 0.5 the result was as in figure 1 where the R2=0.511 Figure 3 : Solid friction factor graph for τ =0.7 When the value τ was taken as 0.9 the result was as in figure 1 where R2=0.807 The data used to generate the graph is shown in Table column 4 and column 17 Figure 4 : Solid friction factor graph for τ =0.9 When the value τ was taken as 0.7 the result was as in figure 1 where R2=0.994 The data used to generate the graph is shown in Table column 4 and column 19 Figure 4 : Solid friction factor graph for τ =0.9 From the graphs it can be seen that the value of τ giving the optimum value of R2 was 0.7 where R2 was found to be 0.994 which is the highest compared to the other R2 values. From the graph the relation for y and x is = Now through rearranging So that the equation Is in the form y = Kxz Results to λs = 17.17m*0.7Fr -1.76 Thus a = 17.17 and b = -1.76 Table 1 Product mass flow rate for system = 40000/3600 = 11.1kg/s From the data the relationship between product mass flow rate ms and pressure drop is given by the graph figure. The equation for the relationship is given by y = 61.92x+112.5 Substituting for x with 11.1 y=61.92x11.1+112.5= 799.8kPa Equivalent length Le = h+2v+Nb Test pipe line equivalent length = 173+2x0+14x20 = 453m Plant pipe line equivalent length = 455+5 + 2x29 + 5x20 = 618 (for silo 1) or 455+11 + 2x29 + 5x20= 624m (for silo 2) Scaling pressure drop for length (silo 1) =799.8x= 1091 Scaling pressure drop for diameter in silo 1 = 1091x = 578.23 Scaling pressure drop for length (silo 2) = 799.8x =1101kPa Scaling pressure drop for diameter 1101x= 583.53kPa Calculating air flow rate silo 1 Taking conveying inlet air velocity C1 =17m/s and using equation T1 is the temperature at inlet =288K and =578.23kPa = 768 = = 0.76m3/s Calculating air flow rate silo 2 Taking conveying inlet air velocity C1 =17m/s and using equation T1 is the temperature at inlet =288K and =583.53kPa = = = 0.77m3/s Feeder type and specification In order to ensure that the ratio of mean flow rate to the steady state flow rate is brought close to unity, the blow tank will be set in parallel. This will be achieved by ensuring that while one of the blow tanks is in the discharging process, the other blow tank will be in the process being depressurized, filled and pressurizing in readiness to discharge when the other tank will be empty. This will ensure that there is almost continuous conveying of material throughout the common pipeline. For this to be achieved there will be need of a full set of discharge vent and isolating valves. Also required will be level switches for each tank in addition to having automatic control system that will facilitate correct timing. There is need for the system to have an automatic control facility because the material flow rate will be changing due to the difference in distances of the discharge points. Air supply being the controlling parameter of system, there is a requirement for feedback control on the supply for the blow tank so as the conveying line can always work at the maximum capacity that is allowable at the supply pressure. As away of achieving effective control of the blow tank rate of discharge, a modulating valve will be fitted on one of the air supply lines. Through this there will be automatic proportioning of the total air supply between the tank and the supplementary line. Power required The power required to drive the compressor drive is given by P = 203VIn The power required for line =203X0.76In = 270.66Kw =203X0.77In = 275.5Kw It can be seen that the power will vary for the two lines and as mentioned earlier the feeder need to be designed to take care of the varied power requirement for the two lines. The line requiring more power may need to have a lower discharge so as to reduce power required. Read More