A straight fin fabricated from 2024 Aluminum alloy (k=185 W/mK) has a base thickness of t=3 mm and a length of L=15 mm. Its base
temperature is Tb=100oC, and it is exposed to a fluid for which T[infinity] =20oC and h=50 W/m2K. For the foregoing conditions and a fin of unit width, compare the fin heat rate, efficiency, and volume for rectangular, triangular, and parabolic profiles.
1 answer:
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Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
Answer:
Pound is the product of slug and foot/square second.
Explanation:
We are given that
Force=1 N
1N=
We have to find the units comprise the pound.
Force=1 Pound
Mass=Slug
Acceleration=
Therefore,
1 pound=
Therefore, we can write as 1 pound is equal to the product of slug and ft/square second.
Hence, pound is the product of slug and foot/square second.