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Answer:
the estimated time of submersion is 17.7 years
Explanation:
Given the data in the question;
estimate the time of submersion in years.
we write down the relation between time of submersion and corrosion penetration as follows;
CPR(mpy) = K × W(mg) / [ A(in²) × p(g/cm³) × t(hr) ]
we solve for t
t = (K × W) / ( AP × CPR )
given that;
Area A = 5 in²
W = 2.3 kg = 2.3 × 10⁶ mg
density of steel p = 7.9 g/cm³
CPR = 200
we know that K is 534
so we substitute
t = (534 × 2.3 × 10⁶ mg) / ( 5 in² × 7.9 g/cm³ × 200 mpy )
t = 1,228,200,000 / 7900
t = 155468.3544 hr
t = 155468.3544 hr × ( 1 yrs / ( 365 × 24 hrs )
t = 17.7 years
Therefore, the estimated time of submersion is 17.7 years
Answer:
a) 229.4281 hp.
b) 262.15 ft3/min.
Explanation:
Given data:
P1 = 14.2 psi
T1 = 60°F = 520° R
P2 = 120 psi
T2 = 500°F = 960° R
volumetric flow rate ( Av1 ) = 1200 ft^3 /min = 20 ft^3 / sec
attached below is the detailed solution
Answer:
I have 72 baes just kidding I have 73 but they all don't love me
Answer: 1.98 × 10^4 N
Explanation:
Form similar triangle ADE and ABC
a/x= 2/3, a=2/3x
Width of the strip w= 2(4+a) = 8+2a
W= 8 +2 (2/3x)= 8+4/3x
Area of the strip = w Δx
(8 +4/3x) Δx
Pressure on the strip p= pgx= 10^3 ×9.81x= 9810x
But,
Force= Pressure × area= 9810x × (8+4/3x)Δx
Adding the forces and taking lim n to infinity
F total= lim n--> infinity E 9810x × (8+4/3x)Δx
Ftotal= Integral 2,0 9810x × (8+4/3x)Δx
F total= 9810 integral 2, 0 (8+4/3x)dx
= 9810(8+x^2/2 + 4/3x^3/3)2,0
=9810(4x^2 + 4/9x^3)
=9810(4x2^2 + 4/9×2^3-0)
=9810(16 + 32/9)
Hydrostatic force as an integral
Ft= 19.18 ×10^4N