water is forced out of a fire extinguisher by air pressure.how much gauge air pressure in the tank is required for the water jet to have a speed of 30m/s when the water level is 0,500m below the nozzle
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Expert's answer
2018-05-09T09:30:08-0400
Use Bernoulli's equation z1 + P1/rho*g + V1^2/2g = z2 + P2/rho*g + V2^2/2g Lets call point 1 the surface on the water inside, and point 2 the water as it comes out of the nozzle. z1 = 0 m z2 = 0.5 m V1 = 0 m/s V2 = 30 m/s P1 = ?? P2 = 0 Pa guage Cancel out everything that is zero in the equation P1/rho*g = z2 + V2^2/2g Solve for P1 P1 = rho*g *(z2 + V2^2/2g) P1 = 1000 kg/m^3 * 9.8 m/s^2 * (0.5 m + (30 m/s)^2/(2*9.8 m/s^2) P1 = 450.25 kPa
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