Answer to Question #77066 in Mechanics | Relativity for Tshanduko

Question

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

Answer 1

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