__RECIPROCATING PUMP TEST RIG__

Reciprocating pump is a positive displacement plunger pump. It is often used where relatively small quantity of water is to be handled and delivery pressure is quite large. Reciprocating pump is widely used as Automobile Service Stations, Chemical Industries, or as metering and dosing pumps.

The “

**UNICOOL**” apparatus consists of a single cylinder, double acting reciprocating pump mounted over the sump tank. The pump is driven by Dimmer Control DC motor with stepped cone pulley. An energy meter measure electrical input to motor. Measuring tank is provided to measure discharge of the pump. The pressure and vacuum gauges provided to measure the delivery pressure and suction vacuum respectively.__SPECIFICATIONS__

Ø Reciprocating pump – 37.5mm bore, stroke length 49mm, double acting with air vessel on discharge side, suction 28mm, discharge 22mm.

Ø D.C motor, Dimmer Control Type.

Ø Measuring tank – ___mm x ___mm x ___mm height.

Ø Sump tank – 600mm x 900mm x 600mm height.

Ø Measurements –

a) Pressure gauge –0-4 Kg/cm

^{2}for discharge pressure.b) Vacuum gau00ge –0-760mm Hg for suction vacuum.

c) 1 Ph. Energy meter for motor input measurement.

__EXPERIMENTAL PROCEDURE__

Ø Fill up sufficient water in sump tank.

Ø Fill up the air vessel for about 2/3

^{rd}capacity.Ø Open the gate valve in the discharge pipe of the pump fully.

Ø Close the Ball valve and drain valve of the measuring tank.

Ø Check nut bolts & the driving belt for proper tightening.

Ø Driven the outlet pipe into funnel and slowly increase the pump speed, slightly close the discharge valve. Note down the various reading in the observations table. Repeat the procedure for different gate valve openings. Take care that discharge pressure does not rise above 4 Kg/cm

^{2}.Ø Change the speed and tank readings for different Ball valve openings. Repeat the procedure for different speeds and complete the observation table.

__OBSERVATIONS__

Sr No. | Pump speedN_{p}RpM | DischargeVacuumKgs/cm^{2} | Suction VacuumMm of Hg | Time for 10 Lts.In measuring tank t sec | Time for 10 rev of Energy meter te sec |

__CALCULATIONS__

**1)**

**Suction head**

Hs = suction vacuum of H

_{p}x (p_{Hg}- p_{w}) Where,

p

_{Hg}= Sp gravity of mercury 13.6 p

_{w}= Sp gravity of water 1 Hs = 12.6 x suction vacuum mtrs.

**2)**

**Delivery head**

H

_{a}= Discharge pressure, Kg / cm^{2}x 10 (as 10m of water = 1 Kg / cm

^{2})**3)**

**Total head**

H

_{q}= H_{s}+ H_{p}+ 3mtr Where,

Loss of head in piping and fittings is assumed to be 3 mtrs.

**4)**

**Discharge**

L x B x .01

Actual discharge, Q

_{a}= --------- m^{3}/ sec T

**5)**

**Out put power of pump**

W. Q

_{a}. H_{p}P

_{w}= -------------------- Kw 1000

Where,

W = Specific weight of water = 9810 N/m

^{3} Q

_{a}= Discharge m^{3}/sec H

_{p}= total head. Mtr.**6)**

**Input power to pump**

Let time required for 10 rev of energy meter is to sec.

Then,

10 3600

IP = --------- x -------------- Kw

te 1200

where,

Energy meter constant is 1200 Rev/ Kwh.

Taking motor efficiency 60%, we have input shaft power

S.P = I.P. x 0.6

**7)**

**Over all efficiency of pump**

P

_{w}ŋ

_{o}= -------- x 100% SP

__PRECAUTIONS__

Ø Operate all the controls gently.

Ø Sever allow to rise the discharge the pressure above 4 Kg/cm

^{2}.Ø Always use clean water for experiment.

Ø Before starting the pump ensure that discharge valve is opened fully.

**SAMPLE CALCULATION****OBSERVATIONS**

SrNo. | Pump speedNpRPM | DischargePressure(Kg/cm^{2}) | Suction Vacuummm of Hg | Times for 10 lit rise in measuring tank t sec | Time for 10 rev of Energy meter te Sec |

1 | 445 | 0.5 | 452 | 16.12 | 354.1 |

2 | 437 | 1 | 445 | 16.75 | 317.6 |

3 | 434 | 2 | 440 | 17.16 | 269.6 |

4 | 425 | 3 | 422 | 17.37 | 225.0 |

**Suction head –**

Hs = 12.6 x 0.452 mtrs

= 5.695 mtrs

**Delivery head –**

H

_{a}= 0.5 x 10 = 5 mtrs

**Total head –**

H

_{q}= 5.695 + 5 + 3 mtr = 13.695 mtrs.

**Where,**

Loss of head in piping and fitting is assumed to be 3 mtrs.

**Discharge**

0.01

Actual discharge, Q

_{a}= ------------ m^{3}/sec 16.12

= 6.20 x 10

^{-4}m^{3}/sec**Output power of pump**

9810 x 6.20 x 10

^{-4}x 13.695 P

_{w}= ----------------------------------------- kw 1000

= 0.083 kw

**Input power to pump –**

Let time required for 10 rev. of energy meter be to sec.

Then, 10 3600

IP = ------------ x ---------------- kw

354.1 240

= 0.42 kw

Where, energy meter constant 240 rev/kwh

Taking motor efficiency 80% we have input shaft power

Sp = 0.42 x 0.8

= 0.34 kw

Overall efficiency of pump –

0.083

η

_{O}= ------------ x 100% 0.34

= 24.41%

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