Question

Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?

Answer 1

Answer:

The velocity is  $v_2= 0.45 \ m/s$

Explanation:

From the question we are told that

      The initial speed of the hot water is  $v_1 = 0.85 \ m/s$

     The pressure from the heater  $P_1 = 450 \ KPa = 450 *10^{3} \ Pa$

      The height of the hot water before flowing is  $h_1 = 0 \ m$

      The height of bathtub above the heater is $h_2 = 3.70 \ m$

       The pressure in the pipe is $P_2 = 414 KPa = 414 *10^{3} \ Pa$

       The density of water is $\rho = 1000 \ kg/m^3$

Apply Bernoulli equation

      $P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2$

Substituting values

     $(450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )$

=>   $v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}$

=>   $v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}$

=>    $v_2= 0.45 \ m/s$