Which is better for outdoor use wicker or resin wicker?
Check inside of each unit to see what the setting is for the alarm to go off, they might be set low so you are having these problems. Another thing to do is call your power company and tell them what is happening with the smoke alarms. They can check by putting a recording device on your power lines to your house. You might be lucky to have someone from the power comany come out and check the alarms and settings for you. Also check with your neighbors around you to see if they have the same problem with that type of smoke alarm. The only time you need to be concerned about batteries in smoke alarms is if you lost power for a long time, or if you had the type that wasn't connect to the main power supply. What would happen with the battery type only, when the battery power starts to get low it beeps often to indicate the battery power is low. The battery in your alarms are to maintain the power to the alarm in case you lose power all together. Save your money on the batteries for now, it has to be your power supply or the setting in the alarms.
one likely cause for this is are you running your A/C if so do you constently open the doors? if so that is the reason for the alarms sounding.
Believe it or not, I have also seen smoke alarms trip because of steam. W ehad one right outside our bathroom and it would go off all the time after taking a shower. What a pain.
The minimum power is calculated by assuming no friction (so no power is lost on that) and aiming straight up. Assuming no friction allows for conservation of mechanical energy to be invoked. For a water particle of mass m, equating its kinetic energy at the nozzle to the gravitational potential energy at height h gives 1/2 m v^2 m g h So given the height h the speed of the flow at the nozzle must be v sqrt(2gh) Given the diameter of the nozzle, we therefore can calculated with this speed the amount of energy given to the flow in t seconds: The mass flowing out through area A in t seconds is rho A v t, so the energy given to it in that time is E 1/2 (rho A v t) v^2 1/2 rho A v^3 t The power applied therefore is P E/t 1/2 rho A v^3 1/2 rho (pi d^2 / 4) v^3 1/8 pi rho d^2 v^3 But we saw from energy conservation that v (2 g h)^(1/2) So P pi rho d^2 (gh/2)^(3/2) P 3.1415* 1000 kg/m^3 *(0.030 m)^2 * (9.8 m/s^2 * 35m / 2)^(3/2) 6.4*10^3 kg m^2 / s^3 6.4 kW
