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Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogy
Following are property of air
ρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/ms
- a)82
- b)94
- c)100
- d)85
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Air flows over a heated plate at a velocity of 40 m/s. The local skin...
Here for air ρ=0.88 kg/m3, Cp = 1001 J/kg°K , k = 0.035 W/mK
μ = 2.286×10-5 Kg/ms
local skin friction coefficient at a point Cf = 0.004
Reynolds Colburn analogy with
Stanton no St
As
h = 93.62 = 94 W/m2K
Most Upvoted Answer
Air flows over a heated plate at a velocity of 40 m/s. The local skin...
Given data:
Velocity of air, V = 40 m/s
Skin friction coefficient, Cf = 0.004
Density of air, ρ = 0.88 kg/m3
Specific heat capacity of air, Cp = 1001 J/kg K
Thermal conductivity of air, k = 0.035 W/m K
Dynamic viscosity of air, μ = 2.286 x 10^-5 kg/ms
Calculations:
Reynolds number, Re = ρVl/μ, where l is the characteristic length of the plate
Using Reynolds number, we can determine the Nusselt number, Nu, using the Reynolds-Colburn analogy:
Nu = (Cf/2) * (Re)^0.5 * (Pr)^0.33, where Pr = Cpμ/k is the Prandtl number
Local heat transfer coefficient, h, can be determined using Nu = h*l/k
Assuming the plate is flat, we can take the length of the plate as the characteristic length: l = 1 m
Re = 0.88*40*1/2.286 x 10^-5 = 153363.6
Pr = 0.88*1001*2.286 x 10^-5/0.035 = 0.057
Nu = (0.004/2) * (153363.6)^0.5 * (0.057)^0.33 = 93.7
h = 93.7 * 0.035/1 = 3.28 W/m2 K
Therefore, the local heat transfer coefficient at the point on the plate is 94 W/m2 K (approx). The correct answer is option B.
Velocity of air, V = 40 m/s
Skin friction coefficient, Cf = 0.004
Density of air, ρ = 0.88 kg/m3
Specific heat capacity of air, Cp = 1001 J/kg K
Thermal conductivity of air, k = 0.035 W/m K
Dynamic viscosity of air, μ = 2.286 x 10^-5 kg/ms
Calculations:
Reynolds number, Re = ρVl/μ, where l is the characteristic length of the plate
Using Reynolds number, we can determine the Nusselt number, Nu, using the Reynolds-Colburn analogy:
Nu = (Cf/2) * (Re)^0.5 * (Pr)^0.33, where Pr = Cpμ/k is the Prandtl number
Local heat transfer coefficient, h, can be determined using Nu = h*l/k
Assuming the plate is flat, we can take the length of the plate as the characteristic length: l = 1 m
Re = 0.88*40*1/2.286 x 10^-5 = 153363.6
Pr = 0.88*1001*2.286 x 10^-5/0.035 = 0.057
Nu = (0.004/2) * (153363.6)^0.5 * (0.057)^0.33 = 93.7
h = 93.7 * 0.035/1 = 3.28 W/m2 K
Therefore, the local heat transfer coefficient at the point on the plate is 94 W/m2 K (approx). The correct answer is option B.
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Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer?
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Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer?.
Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogyFollowing are property of airρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/msa)82b)94c)100d)85Correct answer is option 'B'. Can you explain this answer?.
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