Sea Water Cooling Pump

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Loading Port:: Shanghai

Payment Terms:: TT OR LC

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Supply Capability:: 5000 unit/month

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Quick Details

Place of Origin:Henan, China (Mainland)
Model Number:200QJ80-140/7
Theory:Centrifugal Pump
Structure:Multistage Pump
Usage:Water
Power:Electric
Standard or Nonstandard:Standard
Fuel:Electric
Pressure:High Pressure
Application:Submersible
Material:Cast Iron / Stainless Steel
Color:Blue
Flow:80m3/h
Head Range:140m
Power:55kW
Outlet:4 inch
OEM,Design Service:Accept

Packaging & Delivery

Packaging Details:	wooden box packing
Delivery Detail:	5-7days after payment confirmation or as customers' requirements

Specifications

sea water cooling pump
1.Flow range:10~500 m3/h
2.Lift range:10~600m
3.no leakage
4.high head

sea water cooling pump

1. Operation parameters

Flow range	10~500 m3/h
Lift range	10~600m
Applicable well Borehole diameter	φ175~φ400mm
Diameter scope	φ32~φ203mm
Temperature range	-15~80 centidegree

2.Product overview

QJ submersible pumps are widely used in

agriculture irrigation, sprinkler irrigation well water lifting
water supply, drain system of industrial and civil building
the hill, mountain areas' water diversion
drainage of low-lying areas
water supply, drainage of industrial and mining enterprises in urban and rural area
all the other water facilities

agriculture irrigation, sprinkler irrigation well water lifting

water supply, drain system of industrial and civil building

the hill, mountain areas' water diversion

drainage of low-lying areas

water supply, drainage of industrial and mining enterprises in urban and rural area

all the other water facilities

3. Technical parameters

Several type parameters, other type, please contact with manufacturer.

Model	Flow range	Head range	Rotation speed	Pump efficiency	Rated power of motor	Maximum outer diameter of unit	Weight
Model	(m3/h)	(m)	(r/min)	(%)	(kw)	(mm)	(kg)
175QJ10-40/3	9-11	42- 38	2850	53	2.2		48
175QJ15-65/5	13-17	68- 62	2850	58	5.5	168	73
200QJ15-26/2	13-17	28- 24	2850	62	2.2	184	48
200QJ20-30/2	18-22	31- 29	2850	66	3	184	59
250QJ140-105/7	130-150	91-120	2875	76	63	233	430
300QJ300-160	279-321	167-153	2900	76	220		620

4. QJ using condition

(1). 380V voltage deviation is no more than ±5%, frequency is 50HZ, three -phase power deviation is no more than ±1%.

(2). Motor cavity must be filled with water.

(3). The first stage impeller of the pump should be more than 2mm below the moving water surface, pump unit should be less than 70m below the static water level.

(4). The distance from the motor bottom to well bottom must be more than 3m.

(5). Water temperature should be no more than 20 centidegree.

(6). Sand content of the water ( by mass) should be no more than 0.01%.

(7). PH value of water 6.5~8.5.

(8). Hydrogen sulfide content of the water should be no more than 1.5mg/L.

(9). Chloride ion content of the water should be no more than 400mg/L.

(10). Water yield of water source should ensure the requirements of pump continuous working.

5. Characteristics

(1). Motors and pumps assembled into one.

(2). Simple structure.

(3). Easy installation and removal, small area.

(4). Simple using and maintenance.

(5). Safe and reliable operation, no pollution to water, etc.

Q:About 2 months ago I started hearing a noise coming from my bathroom that sounded like loud humming every 3 or 4 seconds after I would turn on the water or flush the toilet. I noticed that when this noise happened the water pressure would increase and then it would go back to normal. Then the noise started to get louder. Last night while my husband was taking a shower the water lost pressure and then completely turned off. Now I have no water at all whatsoever except for maybe a few drops when I turn the faucet on. I have a private well and when I bought the house I was told that the well is underneath one of my bedrooms because the house was built on to and they built over the well. I know I can get to the pump from the crawl space. Has anyone had this problem or does anyone know what it might be or maybe even how much it could cost to fix?: Hi. You could have a pressure tank that is in need of replacement or adding air to. The air cushion in the tank has either become compromised or has disappeared completely leaving no air cushion to let the pump build up pressure against. The pump is basically running every time you use even the smallest amount of water and this will basically cause the pump to self destruct. The next thing you may have is a bad pressure switch that is both causing the noise and the pressure variations from the nature of the points being bad. This is not expensive to fix. Under forty bucks. The other fix may run into the several thousands. Sorry. Hope this helps.

Q:Where is the water pump?How do i take it out, and replace it?: Another thing that's good to do is disconnect the battery , plus be sure to check the quality and mixture of the coolant with a anti-freeze tester , also have a container handy so nothing falls on the ground bec anti-freeze is poison to animals if they drink it etc. Hope that helps and best of luck.

Q:Why can't sprinkler system and hydrant system share water pump in fire fighting system?: The design code of sprinkler system GB 50084 - 2001 (2005 Edition) stipulates that the sprinkler system shall be provided with an independent water supply pump, that is, the water pump shall not be shared with the hydrant system. The aim is to ensure the reliability of the water supply and prevent interference.See the code for design of automatic sprinkler system, GB 50084 - 2001 (2005 Edition), 10.2.1 and the corresponding provisions of the following interpretation.

Q:What is the reason for the overload of the water pump indicator?: If the power is normal:1, the pump is indeed overloaded (at work to see how much the ammeter indicates).2 、 the motor is out of order.3 、 the thermal relay is not adjusted properly or aging.

Q:1978 Ranchero 500. Pumps leaking like crazy. I have a belt to remove, which i assume is the serpentine belt? Its tight as can be on the pulleys. I have no clue how to remove this and any other advice on this do-it-yourself project would be much appreciated. Thanks: being a 78 it probably has a V belt instead of a serpentine belt. you loosen the adjustment bolt on the alternator and the long through bolt, push the alternator and the belt should come off. you will need to take the fan off, it should have 4 bolts to take out to remove the fan. then you take the hose off after you drain the radiator and take out the water pump bolts and pry off the pump. now before installing the new pump make sure the gasket surface is clean and dry. put some good sealer on both sides of the gasket. to keep the gasket from moving i use string like thread and tie the gasket on. put the pump in place, put the bolts back in and tighten them, put the hose back on, put the fan back on and tighten the bolts and refill the radiator. if you want good sealer use what i use. i use indian head shellac. once that stuff gets set up it won,t leak. auto zone sells it in a small brown bottle.

Q:my basement flooded and the water came in contact with the wiring on my water pump.Now the pump tries to turn on but shuts off without pumping any water.How do i get my pump pumping water again?: sounds like not only did it come in contact withe wiring but also with the pump,when your working with electricity you need to be very careful and have the right tools to do the job with. to start this job you need a multi testing meter,this will let you check things like amp,volts,ohms,you know things that comes in contact with the pump and wiring,you didn't say what the voltage is on this pump,so first i'd find out what that is and make sure you have that going to the pump,then if so make sure the pump isn't froze or locked up,next go to the breaker box and locate the breaker for the pump,and test to see how many amps the pump is pulling when turned on.then find out if there is anything restricting the pump from doing its job,it may be something as simple as a run capacitor or even a shorted or burnt wire,but first make sure you have the correct voltage and amperage going to and coming out of the motor,once you get that taken care of you may need to prime the pump in order to get it pumping again

Q:guys i have a mini-water pump.....i wanna pump kerosene instead of water,to make a mini stove,on my own..my doubt is ...if the water pump is used to pump kerosene,will the kerosene ignite inside the pump itself or will cause any problem..i just want the pump to pump kerosene....will it ignite inside of the pump,because the pump runs on electricity....can u clear my doubt: Is the motor of the pump built-in so that the pumped fluid flows through the motor also? If the answer to the question is yes then do not use the pump for pumping kerosene. However, if the pumped liquid does not come in contact with motor then you can use the pump for pumping kerosene. But ensure that the pump shaft seal is of mechanical type (so that kerosene does not leak from the shaft sealing area) with parts (specially any rubber parts) compatible with hydrocarbons (use viton rubber parts instead of EPDM)

Q:i have a well that i am trying to pump and i need to have one pump to bring the water up from the well, and one pump to take it out to our lawn. i just wanted to know if i could hook the irrigation pump directly into the well pump, or if that would be damaging to the pump?thanks!: You can always use demand pumps to do the job. This way they just sit idle and wait on the call for water before they go to work.

Q:A water pump that consume 2kW of electric power when operating is claimedto take in water from a lake and pump it to a pool whose free surface is 30 m above the free surface of the lake at a rate of 50 L/s. Determine if this claim is reasonable.You have to show the minimum power required is 14.7 kWand i can't get it.: A liter of water mas a mass of 1Kg. How much work must be done to raise 50 Kg of water through a height of 30 m? You can say: Work done (W) = potential energy gained (mgh), Or you can say that Work done is the weight force (mg) * the distance moved against the force (h) Either way, W = m * g * h where m is the mass; g is the acceleration due to gravity {9.8 m/s?}; h is the height W = 50 * 9.8 * 30 = 14700 J If you need to do that amount of work in one second, that's a rate of working of 14700 J/s A joule/second is a watt. Minimum power required = 14700 W = 14.7 kW

Q:All the water to my house comes from the river close by - via a 1500kw surface pump, into a large pressurised tank. If there is ever any kind of blockage in the pipe - or if it freezes ...this causes the pump to run - 'dry' - which eventually burns out the motor. I have lost 2 pumps so far because of running dry - and they are very expensive.I would like to fit some kind of automatic cut-out - that will ALWAYS stop the pump running after 15 minutes or perhaps less - as that is ALL the time it needs - if the water is flowing freely - to fill the pressurised tank. I feel sure such a thing must exist - but I do not know what it is called or where to buy one. Does anyone know of a 'switch' of this kind? And where to look for one? What could I put into google to do a search - for suppliers and prices, please?: the best thing is buy a submersable pump it is cheap and never burn

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