Matlab Software to Determine the Saving in Parallel Pumps Optimal Operation Systems, by Using Variable Speed.

Transcription

1 IEEE Energy00 Atlanta, GA USA November, 008 Matlab Software to Determine the Saving in Parallel Pumps Optimal Operation Systems, by Using Variable Speed. Dr. Mariano David Zerquera Izquierdo. Dr. Juan José Sánchez Jiménez. Dr. Alexis Martinez del Sol Electrical mechanic deparment, Guadalajara University, Ave. Revolucion 1500 CP 44840, Phone: () , Abstract In this paper a software is described, which was made with the purpose to determines the systems behaviour in which many centrifugal pumps are involved operating in parallel. The software permits to determinate the flux distribution to give for each pump, for a total flow to supply, in order to obtain a minimum of total losses. For this purpose it is taken in consideration that each pump is driven for an induction motor, fed for a variable frequency voltage driver. INTRODUCTION As it is known there are many industry systems of pumping, in which centrifugal pumps intervene operating in parallel, giving a certain flow to a common System. (Fig. No.1). For a certain total flow to give, infinite combinations of individual flows exist of each one of the pumps that intervenes in the system. With the program in computer that is described in this paper, it is possible to determinate the optimal flow distribution among the pumps, for a total flux to give, so a minimum operating cost is obtained. This includes the losses of the hydraulic system and the electric motors which drives the pumps. In the present work a brief description of a program made in language Matlab making use of the visual programming, is carried out. This software allow to consider the effect of the motors, and in automatic form, to determine the best operation point, from the point of view of energetic demand of the supply source. That is to say that the best result, means that the operation of the pumps and motors, throws the electric minimum energy consumption. DEVELOPMENT To carry out the outlined work, it will be considered a system with n pumps (See figure No.1.), discharging to a common system by means of which liquid is pumped to a tank. It is considered that the pumps b1, b,..bn, are driven for motors m1,m, mn fed with voltages and frequencies V1,f1,V,f. Each combined motor pumps forms systems S1,S,.. Sn, which feed liquid 1,,.n to a common system which demand a total flux c. For a total flow c to pump, there are infinite flows combinations to be distributed among each pump. Each one of these distributions throws a different economic technical behaviour, existing only one combination that gives as a result a minimum operating cost. For each combination of flow distribution, each pump is drive to a certain speed and therefore each motor should be fed with the voltage and the required frequency. Example case study without taking in consideration the motor losses To start up it is considered a simple system formed by two pumps, with the behaviours parameters shown in the table No.1. It is known that the pump behaviour correspond to a speed of 1800 (r/min) and the liquid density is ρ =1000 kg : m It is wanted: a) Determine the system behaviour if it is required to pump a total flow c=100 liters / s when the speeds of the motors are adjusted so the pumps No.1 and N o. gives flows: 1=60 litros/s =40 liters/s. respectively. b) To carry out a study to determine which should be the flow to give for each pump, to give the common total flow c=100 liters/s, so the system operation gives as a result the smallest pumps electric demand cost. Determination of the system behaviour. Pumps and systems curve fitting. With the data given in the table No.1, was carried out a curve fitting, by means of the library corresponding of the language Matlab, what threw the coefficients shown in the table No.. In agreement with these coefficients as a result, the following equations are obtained: Pump No1: H = (m) (1) Pump No.: H = (m) () Systems No.1 y No.: H = 1, , , (m) ()

2 Common System: considered the point A as the in the figure No., when H = 6, , applying the corresponding affinity equation: 10,099 (m) (4)PumpNo1: = 0, ,48 1,7.10 0, Pb b b b + 1 (7) H 1 = H H 1 1 = 51,6 = 0,0 (8) 1 40 (kw) (5) If it is considered as the condition (1), the corresponding to a PumpNo.: point of the characteristic of the pump to the manufacturer s P b = 0,14 0,6077 1,66.10 b b b 1, speed, the equations (8) and (1) can be equalled, being obtained the following equation: (kw) (6) To calculate the mechanical power it demands for the pumps 0,0 = shown in the equations (5) and (6), the following equation is (9) 1 used: From the equation (9) it is obtained: 9,8ρH 100 Pb =. (Kw) Efb , ,1 100 = 0 1 1= 50,7441 liters/s Determination of the system load The value of the flow found 1 is pointed out in the figure No. Since the characteristic of the pump belongs to the manufacturer s speed, and in accordance with the outlined nomenclature, can be wrote, by applying the corresponding affinity equation: To determine the pump load it is necessary firstly to know the total system load H. For a total flux given c, the pump No.1 will give a flow 1 and the pump No. a flow. The total head seen by the pump No.1, is fixed by the sum of the head corresponding to the system No.1 with the head of the system common. Equally the head seen by the pump No., is the sum of the head corresponding to the system No. added with the common one. In the figure No. are shown the curves of H vs of the systems No1 and No. and that of the common one, based on the data of the table No.1, as well as the operation points for the given flows. In accordance with these flows, the heads that each one of the pump see are given for: 50, = So: N = 1419 r/min N In accordance with the previous results, so that the pump impels a flow of 40 liters/s to the head of 51,6 meters must be rotated to 1419 r/min. The pump power, if the same was moved to the speed of 1800 r/min, can be determine by means of the equation (5), by substituting the flow for their value: 1=50,7441 liters / s Pb=47946 W Pump No.1: Hs1c=Hs1+Hc=11,6+40=51,6 (m) Pump No.: Hsc=,6+40=6,6 ( m). That is to say that the pump No.1 should pump a flow of 40 liters/s against a head of 51,6 m; the pump No. a flow of 60 liters/s against a head of 6,6 m and for this, should be rotated to a speed that we will denominate operation speed. However the values of H and introduced as data, belong to 1800 r/min speed, for each one of the pumps. This outlines the necessity to calculate the operation speeds to which the pumps should be drive by their primary motors. In the figure No. are shown the characteristics of H vs of the pump No.1 to the speed o n=1800 r/min and a point of the characteristic (point A) that belongs together with the operation point which should go by the couple: 1=40 liters/s H1=51,6 m To determine the pump speed, corresponding to the characteristic that goes by the point A of the figure No., it is necessary to apply the centrifugal pumps affinity law. If it is The pump power input to the operation speed N1=1419 r/min, is obtained applying the power affinity equation 47, = P =,464 kw P 1419 For a similar procedure to the previous, for the pump No. is obtained: 1=50,556 liters/s Pb=1,997 kw P=5,485 kw If the previous procedure is repeated, the results shown in the table No. are obtained, and their graphic representation in showed in the figure No.4 For the condition of minimum cost it is obtained: PTb=75,75 kw Pb1=,7 kw Pb=75,55 1==50 liters/s, Nb1=154 (r/min), Nb=009 (r/min), (it is indicated shadowed in the table No.) Optimal Operation Motors Pumps

3 The function objective to optimize corresponds with the total losses that is to say: P T = P sb1 + P sb +...P sbn + P m1 + P m +...P mn + Psc To use this function objective it is necessary to fulfil the following inequality restrictions for the powers of the pumps and the electric current of the motors respectively: Pbn PNbn 0 In INn 0. Also it is necessary fulfil the following equality equation for fluxes: n = c Software description optimization of this. Of the obtained results, it can be proven that making use of the program, a considerable saving can be obtained with a optimal operation, just as it can be seen of the tables No.9 and No.10 When the software was run, were obtained the results shown in the tables No.5 and No.6, corresponding to the non optimal operation. The results corresponding to the optimal operation are shown in the tables No.7 and No8 The program presents two options: a) Non Optimal Operation of pumps in parallel b) Optimal Operation of pumps in parallel. By means of the first option the user should introduce among other data, the flow that is wanted each one of the pumps gives, and as a result the point of operation of the system is obtained. With the second option, the total flow should be given to pump for the group of pumps and as a result the program obtains the optimal operating point. Also it is necessary to introduce to the software another data, such as: Parameters of the equivalent circuit of the motor, data of the characteristics of the pumps and of the hydraulic systems, costs schedules, etc. To determine the optimal operation of the system of pumping, the corresponding optimization library of the language Matlab of was used. The curve fitting library was also used, to represent mathematically the characteristics of the pumps and of the hydraulic systems. So that the program is friendly to the user, the graphic interfaces (guide) were used, as much in the data entry as in the results. Example case study taking in consideration the motors losses To validate the elaborated program, three pumps and the hydraulic systems were analyzed with the data shown in the table No.4. It was wanted: a) To determine the behaviour of the system if it is required to pump a total flow c=15 liters / s when the speeds of the motors are adjusted so the pumps No.1, N o. and No. gives flows: 1=50 liters/s, =5 liters/s and =50 liters/s (Non optimal operation) b) To determine which should be the flow to surrender for each pump to give the common total flow c=15 liters/s so the operation of the system gives as a result the smallest cost in the electric demand. (Optimal operation) CONCLUSIONS A program has been made in language Matlab, which allows to determine the optimal operation of pumps in parallel, driven by induction motors fed by adjustable voltagefequency drivers. The elaborated program allows to determine the system behaviour with a non optimal distribution of the flow among the pumps and with the

4 Table 1 : Systems data : Flux (l/s) H: Head (m) Efb: Pump efficiency (%) Bomba 1 Hb , , ,9 Bomba Hb 48 47,8 46, ,1 7,4,88 Sistema 1 y Hs1,Hs 4,4 11,6,6 40,4 6 88,4 Sistema común Hsc 10 11, 14,8 0,8 9, 40 5 Bomba1 Efb Bomba Efb Table Polynomials Coefficients obtained by the curve fitting. from H vs of the pumps and the systems. Coeficientes Orden Orden Orden1 Independiente Bomba1 (b1) Bomba (b) Sistema1 (S1) ,006 7, Sistema (S) 1, ,006 7, Sistema Común 7 (Sc) ,001-0, Bomba1 (b1) 0, ,7.10 0, Bomba (b) 0,14-0,5077 1, , Table. Pumps input power for a total flow c=100 litros/s Pb1, Pb: Each pump input mechanical power PTb: Total power demanded for the pumps 1,: Each pump flux. Nb1,Nb: Each pump speed. Bomba No.1 Bomba No. PTb(kW) 1(l/s) Pb1(kW) Nb1(r/min) (l/s) Pb(kW) Nb(r/min) 10 5, , , , ,8 49 9,5 0 16, , ,11 40, , ,96 50, , , , , , , , , , , , , , ,85

5 c Sc Tank S1 1 S S Sn n b1 b b bn m1 V1,f1 m m mn V,f V,f Vn,fn c1 c c cn V, f Fig.1. Pumping system with several pumps in parallel Fig.. Characteristic of H vs. Fig.. Characteristic of H vs corresponding to the pump No.1

6 Fig. 4. Pumps mechanical input power for different fluxes. Table 4. Points of the pumps and the systems Pumps 1 y (75hp, 1800 H , , ,9 r/min) Pump(50hp,1800r/min) H 48 47,8 46, ,1,74,88 Systems 1 y H System H 4,4 11,6,6 40,4 6 88,4 Common System H 10 11, 14,8 0,8 9, 40 5 Pumps 1 y Ef Pump Ef

7 Table 5. Each motor behaviour (Non optimal operation) Motor I(A) f(hz) n(r/min) P1(kW) Ef(%) V1(V) Pm(kW) C ($/año) 1 60,55 57, ,85 91,4 47.6, ,4 40,96 71,78 18,1 8,9 91,49 550,09 141,4 60,55 57, ,85 91,4 47.6, ,4 Table 6. Each pump behaviour (Non optimal operation) Pump (l/s) H (m) Pb (kw) Ef (%) Peb (kw) C($/año) ,6 41,8 86,07 5, ,9 5, , ,6 41,8 86,07 5, Table 7. Each motor behaviour (Optimal operation) Motor I(A) F(Hz) n/r/min) P1(kW) Ef (%) V1 (V) Pm(kW) C$/año) 1 7,8 60,8 181,5 59,46 9,14 455,1 4, ,8 1,16 67,9 0,6 11,87 79,6 50,5,4 1098,8 7,8 60,8 181,5 59,46 9,14 455,1 4, ,8 Table 8. Each pump behaviour (Optimal operation) Pump (l/s) H (m) Pb (kw) Ef (%) PT(kW) C ($/año) , 54,74 86,11 7,6 776,6 5 61,15 9,4 1,7 6, , , 54,74 86,11 7,6 776,6 Motor- Pump Table 9. Motor Pump combination total losses and costs Non optimal Optimal PT(kW) C($/año) PT(kW) C($/año) 1 9, , ,, ,7 8, ,8 9, , , Table 10. Whole system losses and costs Non optimal Optimal PT (kw) 4,04,4 C ($/año) Total save =6966 ($/año)

8 NOMENCLATURE n- Angular speed of the motors and pumps (r/min) f - Frequency of the applied voltage to the motors. (Hz) In- Motor n Electric current (A) IN- Motor nominal electric current (A) P1 Electric power consumed by each motor (kw) Pmn: Losses of the motor n (kw) Pm- Motors losses (kw) Peb- Pumps losses (kw) PT Total losses (Motor pump- Hidraulic system - Convertor) (kw) PTb- Total mechanics power demands for the pump (kw). Pb- Mechanics power demands for the pump (kw o W) Psbn: Losses of the pump and system n (kw) Psc: Losses of the common system (kw) C Losses total cost ($/ year) - Flow (liters/s). H - Head (m) Ef: Efficiency (%) ρ - Liquid specific weigh to pump kg m BIBLIOGRAPHY [1] Jimmie J: Cathey, Electric machines: Analysis and design applying Matlab. Mc Garw Hill, 001 [] Stephan J. Chapman: Electric machines, Mc Graw Hill, 000 [] Ron Carlson: The correct Method of Calculating Energy Savings to Justify Adjustale_Frecuency Drives Pumps., IEEE Transactions on Industry Aplications, Vol. 6, No.6, Nov/Dec 000. [4] Manual of the User of the Matlab 7.1 [5] Karachi, Manual on centrifugal pumps, Alfaomega, 00. [6] Kenneath Mc Naughton, Pumps, selection, use and maintenance. Mc. Graw Hill, [7] M. Zerquera and others, Economic operation of transformers, Memoirs of the XIII meeting of summer of the IEEE, Acapulco, Mexico, 00. [8] Delores M. Etter:, Troubleshooting of Engineering with Matlab, Prentice Hall, 199 [9] McPerson G., Laramore R: "An introduction to Electrical Machines and Transforms", Second Edition, Prentice Hall, 1994.