XBD-HY series Constant-pressure Fire-fighting Pump
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1. Summary
XBD-(HY) series constant-pressure fire-fighting pump comes as a new product developed in this Co. upon the market
de-mand and the performance and technical condition of it con-forms the requirements in the latest issued standard GB6245 Fire-fighting Pumps .
2. Operating Condition
Medium temperature: up to 80℃
Capacity range: 10 ~ 70 L/s
Pressure range: 0.2Mpa ~ 2.2Mpa
3. Application
Suitable for water supply for the fixed fire-fighting system (the fire extinguishing system with fire hydrants, spraying fire-
extinguishing system and smog-spraying with water fire-extinguishing system) in both industrial and civil buildings, especially suitable for the quartered fire-fighting water supply system of high buildings, also for the fire-fighting and living (production) mutually used water supply system and those in construction, municipal works, industry and mine, boiler etc. various places.
- Q: what water pump is on a 1978 mercruiser 170hp 224cc 3.7 liter
- Something in your description doesn't add up. Mercruiser did not have a motor branded as 170 hp in 1978, only 85-86. The 224s where a Mercury invention, all about the same for around 15 years. Do you you have a ser #, much more accurate identification.
- Q: my water pump has gone south on me and I want to repair it myself any ideas on how to will be greatly appreciated, do to space between the water pump of the engine and the chasi or body of the van that makes working on thi vehicle kind of difficult any knoledgeable instruction will be welcome. thanks for the help. I am in California,the name is Elmer
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- Q: im attaching a motor from another filter onto my homemade filter. but i dont want to have to run a cord all the way from the pond to a powerplug. does anyone know of another way to such water into a filter without using electricity? some kind of siphon?
- 1. I had an electrician put a buried cable in the ground coming up to a weatherproof post with outlets and an on-off switch. It was not as expensive as I thought it would be. 2. You can put the water pump and filters by the house or even inside and run buried water pipes out to the pond, or just put the pump inside and have the filter by the pond which would run by water pressure or gravity. 3. A windmill or solar panel. A siphon uses potential energy stored in the water to work. Once the water has dropped through the siphon, there is no more energy to draw from and the siphon stops. A windmill or solar panel could provide the power to lift the water back up and make the process continuous, or at least while it's windy or sunny. However when pricing these two options even though they would be very cool if you have the large yard to accommodate them, they turn out to cost much more than either 1. or 2. and have much higher maintenance costs associated with them. 4. If it's a very small pond, you can get a solar powered filter/fountain that looks like a floating plastic lily pad. My son had one of these, but it only worked on the sunniest of days, and lasted one summer before wearing out. If you add a large block of filter foam over the intake it may last a little longer. Rinse it out two or three times a week.
- Q: Is it cheaper to run an electric water pump or a furnace?
- heat water/pump. not sure what that is but I know what a heat pump is. If that whats your talking about thean the answer is well...yes and no. depending on cost of gas and electric typical a heat pump is better to run for heat as along as its above freezing outside. It will still work and put out heat all winter but it won't be much heat for what your paying for, at which point your better with gas heat.
- Q: Ok, so if i have an 80 watt solar panel and want to run a 12v dc fan and water pump ONLY when the sun is out or enough light to power them, will i need something like a voltage regulator? I would like to have the panels charge deep cycle 12 v batteries at the same time or when the fan and pump is not running. For instance, i have the panels out and my fan and pump on allowing them to run only on solar as well as charge the batteries. What diagram would i use? I would probably get a charge control for the batteries but more importantly what would i need to regulate the power from the panels? Inverter?
- You may have to do some math. The 80 watt panel is only 80 watts at peak sun on a 75°F day at the equator. Chances are you will have something less than 80 watts to work with. But you can add up all the hours of partial sun to get an equivalant number of peak sun hours. An example would be a few hours in the morning and evening at partial power and a couple of hours at solar noon at nearly full power may give you 5 peak sun hours worth of light. 5 psh x 80 w/ps = 400 wh Your supply may have 400 watt hours worth of power per day. You state that the fan is 12v dc but what is the wattage? It could be a little 12v dc fan out of a computer or it could be a huge 12v dc fan out of an RV. What is the power requirements of the water pump? Is it a little 12v dc one for a foot tall decorative fountain or an industial 3 phase pump for a well? To charge the battery you need a voltage 120% higher than the battery voltage. 12v x120%=14.4 v To add up your loads convert them all the use to dc watt hours per day Here is an example to give you an idea of how to play with your numbers: Fan 12vdc x 1.5 A = 18 watts, use this for 5 hours your load would be 18w x 5hr = 90 watt hours Pump 120 vac x 2.5 amps = 300 watts (The AC will need to come from an inverter. The inverter has a certain amount of loss. How good it does the job of converting dc to ac is know as it's efficency. Lets use 90% to be safe) 300 watts / 0.90 = 333.4 watts, use this for 1/2 hour per day 333.4w x 0.5hr = 166.7 watt hours The 90 watt hours + the 167 watt hours = 257 watt hours per day. This would leave about 140 watt hours to put into the battery. Yes it would be best to use a charge controller to protect the battery from overcharging if the pump and fan are off, or from draining the battery too much if the fan or pump stays on.
- Q: i installed a new pump don't know how to increase water pressure and to what what pressure it should be set
- This is very simple i would sugest the pressure to be set betwene 60 to 70 NO MORE THAN 70. 1. turn off power to pump (ether the circut breaker or a switch near the pressure switch) you dont want to weld the wrench to the switch. 2. remove the cover to the pressure switch its self (usly one #2 phillips screw) 3. now you will see 4 wires under some sloted screws nearest to the front of the switch, behint that is 1 or 2 bolts with a spring and a nut. 4. if you see 2 bolts you will want to turn the nut on the larger one to the right. you will want to tighten it about 1 to 2 full turns. (the same goes if there is only one bolt also) 5. now open the valve that is on the tank untill you here the switch click. then close the valve. 6. turn the power back on tio the pump and watch the pressure gauge on the tank. i DO NOT RECOMEND MORE THAN 70PSI. 7. if it goes over the pressure you want repete steps above exept loosen the nut in about 1/4 to 1/2 turns untill you reach the desired pressure 8. if it does not reach your desired pressure then you do the same thing exept tighten the nut in 1/4 to 1/2 turns untill you attain your desired pressure. 9. when the pressure is reached, replace cover so you dont risk rusting the contacts. if you need more help feel free to email me. and i will assist you in any way that i can.
- Q: replaced water pump and thermostat, flushed, opened bleeder valve had long life red colored original coolant, no visible leaks obstruction, when became hot on highway, accelerated quickly and temp back down to norm, eventually after 3hrs driving o.heated stopped and got towed home, new water pumps fins/pulley not turning as freely as old one, no leak out of inspection hole, hoses good shape.2001 2.2l ohv sunfire, air cond.
- get your vehicle up to operating temps and runs your hand along the core of the radiator from top to bottom. If it's cold at the bottom and hot at the top, it's plugged... Also I see where you bleed the system, might wanna check it again... air pockets are stubborn... Last but not least, have a C.O. test performed at a local radiator/automotive shop... This will tell you if you have a blown head gasket
- Q: Water is pumped through a pipe of diameter 13.0 cm from the Colorado River up to Grand Canyon Village, on the rim of the canyon. The river is at 564 m elevation and the village is at 2311 m.(a) At what minimum pressure must the water be pumped to arrive at the village? (in MPa)b) If 4300 m3 are pumped per day, what is the speed of the water in the pipe? (in m/s)(c) What additional pressure is necessary to deliver this flow? Note: You may assume that the free-fall acceleration and the density of air are constant over the given range of elevations.( in kPa)I have tried this over and over but i know B it is 3.75 m/s please help if you know to do this THANKS!!!!!
- a) Pressure is the weight of a 1 meter square column of water. The height of the pipe is 2311-564 = 1747 m volume is 1747 m? density of water is density of water at 20C = 0.998 g/cm? = 998 kg/m? mass of water is 1747 m? x 998 kg/m? = 1744000 kg weight of water is 1744000 x 9.8 = 17090000 N Pressure is 17090000 N/m? or 17.09 MP b) Cross sectional are of pipe is A = πr? A = π(.065)? = 0.0133 m? one meter of pipe has a volume of .0133 m? 4300 m? are pumped per day The water in that one meter of pipe has to be changed 4300 / 0133 = 324000 times per day to handle the flow That is 324000 / (24*3600) times per second or 3.75 times per second. Since the section of pipe is is 1 meter, that is 3.75 meters per second.
- Q: How do I check my well water pumping system to see if there is a leak? I have terrible rusty water.
- the only way to check it is to have a well pump service pull the pump and check the verticle lines. they tend to rust out. you can have them replaced by pvc lines now. when our pump went out we had all of the lines replaced because several of the galvanized lines had pin holes in them
- Q: A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 for pi . Round your answer to three significant digits.)The tank is a spherical shape has a radius r= 6m and a spout height of h=2m.I really dont know how to start this problem correctly. Any help will be appreciated. Thanks!
- The pump will not have to work until the water drains by gravity to the level of the spout. Then, Work = Weight of water * Height of lift Find the distance, d, of the center of mass for the remaining water in the bottom 2 m of the sphere. The height of lift will then be 2m - d. The hardest part is finding the distance d to the center of mass. I don't remember how to do that.
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XBD-HY series Constant-pressure Fire-fighting Pump
- Ref Price:
-
- Loading Port:
- China Main Port
- Payment Terms:
- TT OR LC
- Min Order Qty:
- -
- Supply Capability:
- -
OKorder Service Pledge
Quality Product, Order Online Tracking, Timely Delivery
OKorder Financial Service
Credit Rating, Credit Services, Credit Purchasing
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