This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
