Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.

This question is incomplete, the missing image is uploaded along this 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) × 30 × 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 - P₂A₂

m"( V₂ - V₁ ) = P₁A₁ - F - P₂A₂

F = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )

we substitute

F = ((80×)×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s

F =  565.632 - 0 - 200.032

F = 565.632 - 200.032

F = 365.6 lb

Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb

your answer would be fusion

true

explanation: