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Jet001 [13]
3 years ago
14

A ship tows a submerged cylinder, 1.5 m in diameter and 22 m long, at U = 5 m/s in fresh water at 20°C. Estimate the towing powe

r in kW if the cylinder is (a) parallel, and (b) normal to the tow direction.
Engineering
1 answer:
shusha [124]3 years ago
7 0

Answer:

(a) 120 kW

(b) 800 kW

Explanation:

Given:

diameter: 1.5 m

length = 22 m

U = 5 m/s

temperature = 20°C

For water at 20°C, take ρ = 998 kg/m³  and µ = 0.001 kg/m⋅s

To find:

power in kW

(a) if the cylinder is parallel

Length / Diameter = L / D = 22 / 1.5 = 14.6 = 15

Re(L) = ρ*U*L / µ = 998 * 5 * 22 / 0.001 = 109780000 = 1.1E8

C(D.Frontal) ≈ 1.1

  Force = F = 1.1 * ρ/2 * U² * π / 4 * D

            =  1.1 (998 / 2) (5)²(π / 4)(1.5)²  

            = 1.1 * 499 * 25 * 0.785 * 2.25

            = 24000 N

Power = Force * Displacement / time

           = F * U

           = 24000 * 5

           = 120000

Power = 120 kW

b) if the cylinder is normal to the tow direction.

Re(L) = ρ*U*D / µ = 998 * 5 * 1.5 / 0.001 = 7485000 = 7.5E6

C(D.Frontal) ≈ 0.4

Force = F = 0.4 * ρ/2 * U² * D * L

            =  0.4 (998 / 2) (5)²(1.5)(22)  

            = 164670 ≈ 165000

Power = Force * Displacement / time

           = F * U

           = 165000 * 5 = 825000

           ≈ 800 kW

Power = 800 kW

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In Joule:

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T_2=50 C= 122 F

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A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

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A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

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At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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