## Calculate Motor Pump Size

• Calculate Size of Pump having following Details
• Static Discharge Head (h1)=50 Meter.
• Required Amount of Water (Q1)=300 Liter/Min.
• Density of Liquid (D) =1000 Kg/M3
• Pump Efficiency (pe)=80%
• Motor Efficiency(me)= 90%
• Friction Losses in Pipes (f)=30%

## Calculations:

• Flow Rate (Q) =Q1x1.66/100000 =300×1.66/100000= 0.005 M3/Sec
• Actual Total Head (After Friction Losses) (H) = (h1+h2)+((h1+h2)xf)
• Actual Total Head (After Friction Losses) (H)=50+(50×30%)= 65 Meter.
• Pump Hydraulic Power (ph) = (D x Q x H x9.87)/1000
• Pump Hydraulic Power (ph) = (1000 x 0.005 x 65 x9.87)/1000 =3KW
• Motor/ Pump Shaft Power (ps)= ph / pe = 3 / 80% = 4KW
• Required Motor Size: ps / me =4 / 90% = 4.5 KW

## NEC Code 430.22 (Size of Cable for Single Motor):

• Size of Cable for Branch circuit which has Single Motor connection is 125% of Motor Full Load Current Capacity.
• Example: what is the minimum rating in amperes for Cables supplying 1 No of 5 hp, 415-volt, 3-phase motor at 0.8 Power Factor. Full-load currents for 5 hp = 7Amp.
• Min Capacity of Cable= (7X125%) =8.75 Amp.

## NEC Code 430.6(A) (Size of Cable for Group of Motors or Elect. Load).

• Cables or Feeder which is supplying more than one motors other load(s), shall have an ampacity not less than 125 % of the full-load current rating of the highest rated motor plus the sum of the full-load current ratings of all the other motors in the group, as determined by 430.6(A).
• For Calculating minimum Ampere Capacity of Main feeder and Cable is 125% of Highest Full Load Current + Sum of Full Load Current of remaining Motors.
• Example:what is the minimum rating in amperes for Cables supplying 1 No of 5 hp, 415-volt, 3-phase motor at 0.8 Power Factor, 1 No of 10 hp, 415-volt, 3-phase motor at 0.8 Power Factor, 1 No of 15 hp, 415-volt, 3-phase motor at 0.8 Power Factor and 1 No of 5hp, 230-volt, single-phase motor at 0.8 Power Factor?
• Full-load currents for 5 hp = 7Amp.
• Full-load currents for 10 hp = 13Amp.
• Full-load currents for 15 hp = 19Amp.
• Full-load currents for 10 hp (1 Ph) = 21Amp.
• Here Capacity wise Large Motor is 15 Hp but Highest Full Load current is 21Amp of 5hp Single Phase Motor so 125% of Highest Full Load current is 21X125%=26.25Amp
• Min Capacity of Cable= (26.25+7+13+19) =65.25 Amp.

## NEC Code 430.24 (Size of Cable for Group of Motors or Electrical Load).

• As specified in 430.24, conductors supplying two or more motors must have an ampacity not less than 125 % of the full-load current rating of the highest rated motor +  the sum of the full-load current ratings of all the other motors in the group or on the same phase.
•  It may not be necessary to include all the motors into the calculation. It is permissible to balance the motors as evenly as possible between phases before performing motor-load calculations.
• Example:what is the minimum rating in amperes for conductors supplying 1No of 10 hp, 415-volt, 3-phase motor at 0.8 P.F and 3 No of 3 hp, 230-volt, single-phase motors at 0.8 P.F.
• The full-load current for a 10 hp, 415-volt, 3-phase motor is 13 amperes.
• The Full-load current for single-phase 3 hp motors is 12 amperes.
• Here for Load Balancing one Single Phase Motor is connected on R Phase Second in B Phase and third is in Y Phase.Because the motors are balanced between phases, the full-load current on each phase is 25 amperes (13 + 12 = 25).
• Here multiply 13 amperes by 125 %=(13 × 125% = 16.25 Amp). Add to this value the full-load currents of the other motor on the same phase (16.25 + 12 = 28.25 Amp).
• The minimum rating in amperes for conductors supplying these motors is 28 amperes.

## NEC 430/32 Size of Overload Protection for Motor:

• Overload protection (Heater or Thermal cut out protection) would be a device that thermally protects a given motor from damage due to heat when loaded too heavy with work.
• All continuous duty motors rated more than 1HP must have some type of an approved overload device.
• An overload shall be installed on each conductor that controls the running of the motor rated more than one horsepower. NEC 430/37 plus the grounded leg of a three phase grounded system must contain an overload also. This Grounded leg of a three phase system is the only time you may install an overload or over – current device on a grounded conductor that is supplying a motor.
• To Find the motor running overload protection size that is required, you must multiply the F.L.C. (full load current) with the minimum or the maximum percentage ratings as follows;