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2 answers:
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Answer:
The total skin friction drag on the hull under these conditions is 276N
Explanation:
In this question, we are asked to determine the total skin friction drag on the hull under the given conditions.
Please check attachment for complete solution and step by step explanation
Answer:
the rate of heat loss is 2.037152 W
Explanation:
Given data
stainless steel K = 16 W
diameter (d1) = 10 cm
so radius (r1) = 10 /2 = 5 cm = 5 ×
radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 ×
temperature = 25°C
surface heat transfer coefficient = 6 6 W
outside air temperature = 15°C
To find out
the rate of heat loss
Solution
we know current is pass in series from temperature = 25°C to 15°C
first pass through through resistance R1 i.e.
R1 = ( r2 - r1 ) / 4 × r1 × r2 × K
R1 = ( 5.2 - 5 ) / 4
× 5 × 5.2 × 16 ×
R1 = 3.825 ×
same like we calculate for resistance R2 we know i.e.
R2 = 1 / ( h × area )
here area = 4 r2²
area = 4 (5.2 ×
)² = 0.033979
so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )
R2 = 4.90499
now we calculate the heat flex rate by the initial and final temp and R1 and R2
i.e.
heat loss = T1 -T2 / R1 + R2
heat loss = 25 -15 / 3.825 × + 4.90499
heat loss = 2.037152 W