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professor190 [17]
3 years ago
11

5. (5 points) Select ALL statements that are TRUE A. For flows over a flat plate, in the laminar region, the heat transfer coeff

icient is decreasing in the flow direction B. For flows over a flat plate, in the turbulence region, the heat transfer coefficient is decreasing in the flow direction C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface D. For flows over a flat plate, if the length of the plate in the flow direction is long enough, the flows will inevitably become turbulent E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6. (5 points) Select ALL statements that are TRUE A. For the flow in a pipe with constant cross-section area, both hydrodynamic and thermal boundary layer thicknesses increasingly grow in the flow direction B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant C. In the thermally fully developed region, the mean temperature of the flow becomes constant D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed E. For internal flows, if Pr>1, the flows become thermally fully developed before becoming hydrodynamically fully developed
Engineering
1 answer:
finlep [7]3 years ago
7 0

Answer:

The following statements are true:

A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction

C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface

E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.

Select ALL statements that are TRUE

B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant

D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed

Explanation:

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BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of
Tanya [424]

(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ

<u>Explanation:</u>

<u />

Given-

Lattice parameter of Li  = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200

(a)

Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b)

Density is represented by ρ

ρ = n X (AM) / v X Nₐ

where,

Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

4 0
4 years ago
On some engines, after torquing cylinder head fasteners, you
djverab [1.8K]

Answer:

Re-torque the bolts as required while your engine is warm. But if you're using aluminum cylinder heads, you should wait until your engine is complete cooled until re-torquing

3 0
3 years ago
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0
4 years ago
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resis
Rufina [12.5K]

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

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b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

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The losses in the transmission line are given by

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Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

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Fudgin [204]

Answer:

6.4 m/s

Explanation:

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4 years ago
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