3. Briefly explain the conduction mechanism in metals?

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approxim

diamong [38]

Answer:

Explanation:

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:

[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )

where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.

3 0

3 years ago

A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an

vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0

2 years ago

A 20-mm thick draw batch furnace front is subjected to uniform heat flux on the inside surface, while the outside surface is sub

lara [203]

Answer:

hello your question is incomplete attached below is the complete question

answer :

To ( inside temperature ) = 598 K

TL ( outside temperature ) = 594 k

Explanation:

a) Determine the surface temperature To and TL based on the known conditions provided in the drawing

To ( inside temperature ) = 598 K

TL ( outside temperature ) = 594 k

attached below is the detailed solution

5 0

3 years ago

-0-1" -0 -20 -15 -10 0 -5

kari74 [83]

Answer:

what

Explanation:

what is that

3 0

3 years ago

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

3. Briefly explain the conduction mechanism in metals?​

3. Briefly explain the conduction mechanism in metals?