BPW Horizontal Centrifugal Pump

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<1>Technical data:

a)Flow:4-2400m3/h

b)Head:8-150m

c)Medium temperature:-20-120oC

d)Operation pressure:≤1.6Mpa

<2>Features:

a)Compact structure

b)Stable running, low noise

c)No leakage

d)Convenient operation

e)Viewing from the inlet of the pump, the outlet of it can
be mounted in one of the three ways, horizontally leftward,
vertically upward and horizontally rightward

<3>Application:

a)water supply and drainage in industries and cities

b)boost water-feeding for high buildings

c)garden irregation

d)fire-fighting water supply boost

e)remote distance transportation.

Q:All water pump have diferent caracteristics, but it is imposible to overcome a certain hight...what is that hight and why?: More height requires more push (pressure) to achieve it, but over time you will get to a point that the flow is no longer laminar but turbulent and then the motion is very chaotic and very little makes it to the other end, most is lost in friction

Q:i have a 1994 honda accord lx. i was wondering what the water pump is for and if it goes bad can i still drive around without it thanks: This Site Might Help You. RE: 1994 honda accord lx water pump what does it do? i have a 1994 honda accord lx. i was wondering what the water pump is for and if it goes bad can i still drive around without it thanks

Q:50 meters deep well with what water pump: home?: For deepwater submersible pumps, ordinary submersible pumps are inefficient.

Q:I cant get water to flow though the motor It seems like the pump isnt working any advise!: You should never ever start your engine without water to cool the impeller or it will burn the impeller vanes in the first place. As described, you have to split the lower from the upper case, there you find the pump, replace the impeller in side of it, get a manual first for the rebuild. And hope that the vanes are still there or you gonna have to find the pieces that came loose and entered the engines cooling circuit and clogged it up. Good luck!

Q:I have a 2001 chevy malibu and need to know how to check if the water pump needs replaced. The engine has been running much hotter than normal and is leaking coolant.: have a plumer come look at it every so often

Q:Cant get the bolt loose fromthe fan that is attached to the water pump in my ford ranger. need to remove bolt in order to remove water pump.: you're taking a wrench which will greater healthful at the back of the fan, and knock with a hammer on the wrench to interrupt the nut loose. you turn the nut to loosen it an identical way the fan turns while the engine is working

Q:I have an 1982 El Camino 305 small block and am having troubles with determining if it is the gasket or the pump. I can't quite locate a leak, but it seems to centralize from the water pump and drip down the hose to the radiator. After my engine gets hot, which mind you my temp light has never turned on, and I shut of my car steam rolls out from behind my pump. I think it is the pump and please be it so. (Easier and cheaper fix. Replacement with a inexpensive part XP.): Pull the oil dip stick if it has little bubbles on it and it is discolored you have a blown head gasket.Then pull your radiator cap when engine is cold, is it full or not? You can not trust what I call idiot lights because they usually come on after the damage is done.You can tell if the water pump is working by letting eng get to normal and grasp the top hose and squeeze it just a little,can you feel water running thru it,if not replace it.

Q:how do you change the water pump: When you change the timing belt you as well change the water pump as well. If you take it to anyone for timing belt change they will ask you if you want to change water pump because its convenient to change it out at that time. But if you dont have any knowledge how to work with the timing dont mess with it take it to someone who is knows what there doing because paying couple hundred dollars is better then paying thousands for engine rebuild or new engine.

Q:A rectangular tank that is 2 feet long, 3 feet wide and 6 feet deep is filled with a heavy liquid that weighs 80 pounds per cubic foot. How much work is done pumping all of the liquid out over the top of the tank?How much work is done pumping all of the liquid out of a spout 5 feet above the top of the tank?How much work is done pumping two-thirds of the liquid out over the top of the tank?How much work is done pumping two-thirds of the liquid out of a spout 5 feet above the top of the tank?I thought that math was hard enough and now they added physics..: Work is force x distance, which is conveniently pounds x feet. Since we are only interested in the height here (as that is the direction we are pumping), we only need to integrate along that path. The differential weight/force being pumped is the volume (LxWxH = 2 x 3 x dH) times the density (80) and the distance is the height that volume is pumped (6 - H). So W = integral( 6 * 80 * dH * (6-H)) from H=0 to 6 W = integral ((2880 - 480H) dH) = 2880H - 240*H^2 = 2880(6) - 240*(6)^2 = 8640 foot-pounds of work. If you are pumping from the spout, replace the 6-H with 11-H, and the answer becomes 23,040 foot-pounds, which makes sense since you have almost tripled the averge height you were lifting. For the 2/3, change the range of the integration to H = 2 - 6, since H = 0 - 2 represents the water at the bottom of the tank that will still be there later. This gives you answers of 3840 and 13,440 foot-pounds respectively, showing that the last couple of feet are the hardest to pump out (because they have the farthest to go...)

Q:Water is pumped out of a holding tank at a rate of 6-6e^(-.09t) liters/minute, where t is in minutes since the pump is started. If the holding tank contains 1000 liters of water when the pump is started, how much water does it hold one hour later??Round your answer to one decimal place.: Let V(t) be the volume of water (in Liters) in the tank after t minutes. We are given that V(0) = 1000, and we wish to find V(60). We are also given the rate of change of the volume: V'(t) = -[6 - 6e^(-0.09t)]. Note that a negative sign needs to be placed because the rate represents water leaving the tank, i.e., being pumped out of the tank. By the Fundamental Theorem of Calculus, Integral (0 to 60) V'(t) dt = V(60) - V(0), so that V(60) = V(0) + Integral (0 to 60) V'(t) dt V(60) = V(0) + Integral (0 to 60) -[6 - 6e^(-0.09t)] dt V(60) = V(0) + Integral (0 to 60) [-6 + 6e^(-0.09t)] dt V(60) = V(0) + [-6t + (6/-0.09)e^(-0.09t)] (Evaluated at t = 60 - Evaluated at t = 0] V(60) = V(0) + [-6(60) + (6/-0.09)e^(-0.09(60))] - [-6(0) + (6/-0.09)e^(-0.09(0))] V(60) = 1000 + [-360 - (200/3)e^(-5.4)] - [0 - (200/3)] V(60) = 1000 - 360 + 200/3[1 - e^(-5.4)] V(60) = 640 + 200/3[1 - e^(-5.4)] V(60) = approx. 706.4 Therefore, approximately 706.4 L of water occupy the holding tank 1 hour after the pump is started.

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