Question

You carry a fire hose up a ladder to a height of 10 m above ground level and aim the nozzle at a burning roof that is 9 m high. You hold the hose horizontally and notice that the water strikes the roof at a horizontal distance of 7 m from where it exits the nozzle. The hose is connected to a large pressurized chamber in the fire truck 0.5 m above ground level. What is the pressure in the chamber

Answer 1

Answer:

The value is $P_1 = 314645 \ Pa$

Explanation:

From the question we are told that

The height is $h_2 = 10 m$

The height of the burning roof is $k = 9 m$

The horizontal distance is $d = 7 \ m$

The height of the truck is $h_1 = 0.5 \ m$

Generally the time for the water to hit the roof from the hose is mathematically represented as

$t = \sqrt{\frac{2 * (h_2 - k)}{g} }$

=> $t = \sqrt{\frac{2 * (10 - 9)}{9.8} }$

=> $t = 0.4518 \ s$

Generally the velocity of the water is mathematically evaluated as

$v_2 = \frac{d}{t}$

$v_2 = \frac{ 7}{0.4518}$

$v_2 = 15.5 \ m/s$

Generally from Bernoulli's Equation we have that

$P_1 + \frac{1}{2} v_1^2 * \rho + \rho *g *h_1 = P_2 + \frac{1}{2} v_2^2 * \rho + \rho *g *h_2$

Here $P_1 [\tex] is pressure in the chamber which we are to calculate , [tex]P_2 [\tex] is the atmospheric pressure with value [tex]P_2 = 101325 \ Pa [\tex] , [tex]v_1 [\tex] is the velocity of the water before it starts flowing with value [tex]v_1 = 0 m/s [\tex] , [tex]\rho [\tex] is the density of water with value [tex]\rho = 1000 \ kg/m^3 [\tex] So [tex]P_1 + \frac{1}{2} 0^2 * 1000 + 1000 *9.81 *0.5 = 101325 + \frac{1}{2}* 15.5^2* 1000 + 1000 *9.81 *10$  

         $P_1 = 314645 \ Pa$