Week 9 - Assignment 8 Fluid Properties.docx

Week 9 - Assignment 8 Fluid Properties.docx - Problem P6.1...

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Problem P6.1 1.5 × 10 3 kg /( m∙s )= 1.5 × 10 3 kg /( m∙s ) × 0.06852 slug kg 3.281 ft m = 3.132 × 10 5 slug /( ft ∙s ) 1.5 × 10 3 kg /( m∙s )= 1.5 × 10 3 kg /( m∙s ) × 10 3 cp 1 kg /( m∙s ) = 1.5 cp Problem P6.3 w = ρ×g×V →w = 1.32 × 32.2 × ( 14 gal× 0.1336 ft 3 1 gal ) = 79.5 lb Problem P6.4 ∆ p = ρ×g×h→∆ p = 1.32 slug / ft 3 × 32.2 ft s 2 × 18 ft = 765.072 slug / ft∙ s 2 = 765.072 psf ∆ p = 765.072 psf × 6.944 × 10 3 psi 1 psf = 5.31 psi Problem P6.5 ∆ p = ρ×g×h→∆ p = ( 13.54 × 10 3 ) kg m 3 × 9.81 m s 2 × [ 40 mm× 10 3 m 1 mm ] = 5313.096 kg / m∙s 2 = 5313.096 N / m 2 = 5313 Problem P6.6 M H = 0 ( 3 ) ( F water × 4 ) = 0
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F = F water × ( 4 3 ) →F = p avg × A× ( 4 3 ) →F = ∆ p 2 × A × ( 4 3 ) →F = ρ×g×h 2 × A× ( 4 3 ) F = 1000 kg / m 3 × 9.81 m / s 2 × 6 m 2 × ( 6 × 4 ) m 2 × ( 4 3 ) = 941760 kg∙ m s 2 = 941760 N = 941.76 kN Problem P6.7 Hinge is placed at the top of the gate: M H = 0 ( 3 ) ( F water × 4 ) = 0 → F = F water × ( 4 3 ) F = p avg × A× ( 4 3 ) → F = ∆ p 2 × A× ( 4 3 ) → F = ρ×g×h 2 × A × ( 4 3 ) F = 1000 kg / m 3 × 9.81 m ¿ s 2 × 6 m 2 × ( 6 × 4 ) m 2 × ( 4 3 ) = 941,760 kg∙m / s 2 = 941,760 N = 941.76 kN Hinge at the bottom of the gate: M H = 0 ( 3 ) ( F water × 2 ) = 0 → F = F water × ( 2 3 ) F = p avg × A× ( 2 3 ) → F = ∆ p 2 × A× ( 2 3 ) →F = ρ ×g×h 2 × A× ( 2 3 ) F = 1000 kg / m 3 × 9.81 m ¿ s 2 × 6 m 2 × ( 6 × 4 ) m 2 × ( 2 3 ) = 470,880 kg∙m / s 2 = 470,880 N = 470.88 kN
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If hinge is placed at the top of the gate, the holding force required is 941.76 kN .
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  • Fall '17
  • Dr. Xu
  • Physics, pH, Orders of magnitude, ηf, mgold msilver, 2 4 4 1 cm, 2 1000 kg, 470,880 kg

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