Answer:
The velocity is [tex]v_2= 0.45 \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed of the hot water is [tex]v_1 = 0.85 \ m/s[/tex]
The pressure from the heater [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]
The height of the hot water before flowing is [tex]h_1 = 0 \ m[/tex]
The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]
The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Apply Bernoulli equation
[tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]
Substituting values
[tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]
=> [tex]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}[/tex]
=> [tex]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}}[/tex]
=> [tex]v_2= 0.45 \ m/s[/tex]
