Answer:
Explanation:
Height h = 1.03m
Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3
Time t = 13.5 mins = 13.5 * 60 = 810 seconds
Length of pool L = 14 inch = 14 * 2.54 = 35.56cm
width of pool b = 48 inch = 48 * 2.54 = 121.92 cm
a.) Consider the bernoulli's equation is given as:
[tex]P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)[/tex]
consider the equation of bernoulli at the top of the pool
[tex]P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)[/tex]
where [tex]P_1=P_0[/tex] atm pressure
At the top of the pool [tex]v_1=0m/s[/tex], substitute in V_1 in equation (2)
[tex]P_0+\rho gh_1 =constant ...(3)[/tex]
Hence equation (3) serves as the bernoullis equation at the top.
b.) Consider the equation of bernoulli's at the opening of the pool
[tex]P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)[/tex]
where [tex]P_2=P_0[/tex] atm pressure and [tex]h_2=0m[/tex]
[tex]P_0+\rho v_1^2 =constant ...(6)[/tex]
Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.
c.) Consider the equation (3) and (4)
[tex]P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)[/tex]
Hence velocity is [tex]v_2=(\sqrt{2gh_1})m/s[/tex]
d.) consider (7)
[tex]v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)[/tex]
This is the norminal value of velocity
e.) consider the equation of flow rate interval of v and t
flow(t)=[tex]\frac{dv}{dt}(m^3/s)[/tex] hence this is the flow rate
f.) Consider the equation cross sectional area in terms of V,v2 and t
[tex]AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8[/tex]
hence this serves as the cross sectional area.
g.) Consider the equation of area from equation (8)
[tex]A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2[/tex]
