Answer:
[tex]250\ \text{lbm/min}[/tex]
[tex]625\ \text{ft/min}[/tex]
Explanation:
[tex]A_1[/tex] = Area of section 1 = [tex]10\ \text{ft}^2[/tex]
[tex]V_1[/tex] = Velocity of water at section 1 = 100 ft/min
[tex]v_1[/tex] = Specific volume at section 1 = [tex]4\ \text{ft}^3/\text{lbm}[/tex]
[tex]\rho[/tex] = Density of fluid = [tex]0.2\ \text{lb/ft}^3[/tex]
[tex]A_2[/tex] = Area of section 2 = [tex]2\ \text{ft}^2[/tex]
Mass flow rate is given by
[tex]m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}[/tex]
The mass flow rate through the pipe is [tex]250\ \text{lbm/min}[/tex]
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
[tex]m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}[/tex]
The speed at section 2 is [tex]625\ \text{ft/min}[/tex].
(a) The mass flow rate will be "250 lbm/min".
(b) At section 2, the speed will be "625 ft/min".
According to the question,
Section 1 area, A₁ = 10 ft²
Section 2 area, A₂ = 2 ft²
Water's velocity, V₁ = 100 ft/min
Volume at section 1, v₁ = 4 ft³/lbm
(a) We know the formula,
Mass flow rate, m = ρA₁V₁
= [tex]\frac{A_1 V_1}{v_1}[/tex]
By substituting the values,
= [tex]\frac{10\times 100}{4}[/tex]
= [tex]\frac{1000}{4}[/tex]
= 250 lbm/min
(2) The speed will be:
→ m = ρA₂V₂
or,
V₂ = [tex]\frac{m}{\rho A_2}[/tex]
By substituting the values,
= [tex]\frac{250}{0.2\times 2}[/tex]
= [tex]\frac{200}{0.4}[/tex]
= 625 ft/min
Thus the responses above are correct.
Find out more information about speed here:
https://brainly.com/question/4931057
