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zhenek [66]
3 years ago
8

Air at 27°C, 1 atm flows parallel to a flat plate, which is electronically heated. The plate is 0.5 m long in the direction of f

low. The uniform heat flux from the plate to the air is 3,000 W/m². A trip wire placed at the leading edge of the plate ensures that the boundary layer is turbulent over the entire plate. A thermocouple mounted at the trailing edge of the plate records a temperature of 127°C. Consider the following:
a) Determine the local Nusselt number Nuy at the trailing edge of the plate.
b) Calculate the Reynolds number Rex at the trailing edge of the plate.
c) Evaluate the air free stream velocity (m/s).

Engineering
1 answer:
dmitriy555 [2]3 years ago
8 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 106 GPa and an original cross-sectional diameter of 3.9
kiruha [24]

Answer:

L= 312.75 mm

Explanation:

given data

elastic modulus E = 106 GPa

cross-sectional diameter d = 3.9 mm

tensile load F = 1660 N

maximum allowable elongation ΔL = 0.41 mm

to find out

maximum length of the specimen before deformation

solution

we will apply here allowable elongation equation that is express as

ΔL =     \dfrac{FL}{AE}     ....................1

put here value and we get L

L   =    \dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}

solve it we get

L = 0.312752 m

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4 years ago
Can the United States defeat Iranian forces
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5 0
3 years ago
Read 2 more answers
If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical
lubasha [3.4K]

Answer:critical stress= 20.23 MPa

Explanation:

Since there was an internal crack, we will divide the length of the internal crack by 2

Length of internal crack, a = 0.7mm,

Half length = 0.7mm/2= 0.35mm  changing to meters becomes

0.35/ 1000= 0.35 x 10 ^-3m

The formulae for critical stress is calculated using

σC = (2Eγs /πa) ¹/₂

σC = critical stress=?

Given

E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²

γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m

a= Half Length of crack=0.35 x 10 ^-3m

σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂

=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂

=(4.0920 x10 ^14)¹/₂

σC=20.23 x10^6 N/m² = 20.23 MPa

​  

​

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3 years ago
You read a research study that concludes that the higher a student's self-esteem, the better he performs in school. This sort of
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