DESIGN PROJECT-2 CENTRIFUGAL PUMP

ME 5427 By

INDU SHEKHAR KUMAR

1.)

The following equation was derived to find the relation between the head of the system and the volumetric flow rate through the pipes.

The various values of the parameters were put in EES and an equation was found with Hsystem and Q2 as the variables. The EES code and results can be seen in Appendix-2. While calculating head losses there were a total of eight loss terms, four concentrated and four distributed. The relation was run for various values of flow rate and the following graph was obtained.

The resulting equation which was derived in the graph is as follows:

= 7.5 + 0.3771 2

2.)

The graph displayed above was merged on to the pump map figure to obtain values of torque at different operating speeds of the pump.

The different possible operating point of the pump are shown in the table below.

Percentage Speed Speed (RPM) Power (kW) 60 2100 0.2 70 2450 0.38 80 2800 0.60 90 3150 0.88

100 3500 1.19

The hydraulic point were found at the intersection of the hydraulic circuit and pump-map graphs. Thus points were found for the pump-characteristics graph.

3.)

The mechanical characteristics of the pump was plotted using Matlab using the above points as reference and converting the Power values into torque values using the following equation.

The above graph however did not account for the torque and speed conversion of the gearbox. Thus, new values of corrected speed were plotted and the corresponding values of required Torque were calculated. In other words the Pump characteristics were plotted in terms of engine speed. The resulting looked as follows.

This graph was matched with the engine map to find the intersection point of the graphs, which translated to the operating point of the graph. This can be seen in the following figure.

The Matlab code for the entire process can be seen in the Appendix-1.

4.)

The operating point on the engine characteristic curve is obtained at torque of 2.1 N-m and at speed of 2380 RPM. The same point pump characteristic curve is obtained at a flow rate of 8.29 m3/hr. and a corresponding pump speed of 2713 RPM. This corresponds to the 68%.

Pump Head 33.4 m Flow Rate 8.29 m3/hr.

Pump Speed 2713 RPM Power Consumption 0.5234 kW Hydraulic Efficiency 0.5632

Engine Speed 2380 RPM Engine Torque 2.1 N-m

All the above values were calculated with EES. The EES code can be seen in Appendix-2.

5.)

The Net Positive Suction Head available at the operating point was found to be 17.9 m. From the pump characteristic graph it can be seen that the Net Positive Suction Head required is around 3 m. Hence, as NPSHa > NPSHr cavitation does not occur.

Appendices Appendix-1

clc clear format compact % Plotting mechacnical characteristics of the pump % Power values P = [0.2 0.38 0.60 0.88 1.19]; RPM = [0.6*3500 0.7*3500 0.8*3500 0.9*3500 1*3500]; %Torque values T = (30/pi)*((P*100)./(RPM*pi/30)); figure(1) plot(RPM,T) title('Pump Characteristics') xlabel('Speed (r/min)') ylabel('Torque (N-m)') grid on % Corrected Speed Transmission values RPM_cor = RPM/(0.95*1.2); % where 0.95*1.2 specifies the velocity ratio T_cor = (30/pi)*((P*100)./(RPM_cor*pi/30)); figure(2) plot(RPM_cor,T_cor) title('Corrected Pump Characteristics') xlabel('Speed (r/min)') ylabel('Torque (N-m)') grid on

Appendix -2

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{Indu Shekhar Kumar}{ME 5427}{Design Project 2}

{Given}//Pipe Dimensionsa = 2 {m} {Length}b = 15 {m}d_b = 0.05 {m} {diameter}c = 7.5d_c = 0.035d = 65d_d = 0.035e = 1d_e = 0.035 f = 1lambda = 0.04 {Assumed}g = 9.81*3600^2 {m/hr^2}

//Major and Minor Loss coefficientsxi_in = 1xi_out = 1xi_elbow = 0.3

{Calculations}//Cross-sectional area of pipesomega_b = pi*(d_b^2)/4omega_c = pi*(d_c^2)/4omega_d = omega_comega_e = omega_d

// Finding elevation changez_1 = az_4 = f+e+c

//Concentrated lossesC_L = xi_in/(omega_b)^2 + xi_elbow/(omega_c)^2 + xi_elbow/(omega_d)^2 + xi_out/(omega_e)^2

//Distributed lossesD_L = lambda*(((b/d_b)*(1/omega_b)^2) + ((c/d_c)*(1/omega_c)^2) + ((d/d_d)*(1/omega_d)^2) + ((e/d_e)*(1/omega_e)^2) )

//Head loss relationsH_system = z_4 - z_1 + (Q^2/(2*g))*(C_L + D_L)H_fAC = (Q^2/(2*g))*(C_L + D_L)

//Following the analysis Q = 8.29 {m^3/hr}

//Hydraulic efficiencyeta_hyd = H_system /(H_system+H_fAC )

//NPSHap_A = 101.325 {kPa}p_nu = 2.338 {kPa}rho_water = 998.2 {kg/m^3}g_new = 9.81 {m/s^2}NPSHa = (((p_A - p_nu)*convert(kPa,Pa)/(rho_water*g_new))- (z_1 + H_fAC))*(-1)

SOLUTIONUnit Settings: SI C kPa kJ mass deg

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a = 2 [m] b = 15 [m]c = 7.5 [m] CL = 1.988E+06 [m-4]d = 65 [m] db = 0.05 [m]dc = 0.035 [m] dd = 0.035 [m]de = 0.035 [m] DL = 9.386E+07 [m-4]e = 1 [m] hyd = 0.5632 f = 1 [m] g = 1.271E+08 [m/hr2]gnew = 9.81 [m/s2] HfAC = 25.9 [m]Hsystem = 33.4 [m] = 0.04 NPSHa = 17.8 [m] b = 0.001963 [m2]c = 0.0009621 [m2] d = 0.0009621 [m2]e = 0.0009621 [m2] pA = 101.3 [kPa]p = 2.338 [kPa] Q = 8.29 [m3/hr]water = 998.2 [kg/m3] elbow = 0.3 in = 1 out = 1 z1 = 2 [m] z4 = 9.5 [m]

No unit problems were detected.

Parametric Table: Table 2Q Hsystem

[m3/hr] [m]

Run 1 0 7.5 Run 2 0.2041 7.516 Run 3 0.4082 7.563 Run 4 0.6122 7.641 Run 5 0.8163 7.751 Run 6 1.02 7.892 Run 7 1.224 8.065 Run 8 1.429 8.269 Run 9 1.633 8.505 Run 10 1.837 8.772 Run 11 2.041 9.07 Run 12 2.245 9.4 Run 13 2.449 9.761 Run 14 2.653 10.15 Run 15 2.857 10.58 Run 16 3.061 11.03 Run 17 3.265 11.52 Run 18 3.469 12.04 Run 19 3.673 12.59 Run 20 3.878 13.17 Run 21 4.082 13.78 Run 22 4.286 14.42 Run 23 4.49 15.1 Run 24 4.694 15.8 Run 25 4.898 16.54 Run 26 5.102 17.31 Run 27 5.306 18.11 Run 28 5.51 18.94 Run 29 5.714 19.81 Run 30 5.918 20.7 Run 31 6.122 21.63

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Parametric Table: Table 2Q Hsystem

[m3/hr] [m]

Run 32 6.327 22.59 Run 33 6.531 23.58 Run 34 6.735 24.6 Run 35 6.939 25.65 Run 36 7.143 26.73 Run 37 7.347 27.85 Run 38 7.551 28.99 Run 39 7.755 30.17 Run 40 7.959 31.38 Run 41 8.163 32.62 Run 42 8.367 33.89 Run 43 8.571 35.19 Run 44 8.776 36.53 Run 45 8.98 37.89 Run 46 9.184 39.29 Run 47 9.388 40.72 Run 48 9.592 42.18 Run 49 9.796 43.67 Run 50 10 45.19

0 1 2 3 4 5 6 7 80

5

10

15

20

25

30

Q

Hsy

stem

(m^3/hr)

(m)

Deisgn Project 2Deisgn Project 2m