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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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Direction: List down or enumerate the type of outlets you want to install on your dream house. Also indicate the quantity (in pi
Lorico [155]

Answer:

Apartment outlet 4pcs

Explanation:

6 0
2 years ago
A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po
Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

V(r)=\frac{kq}{r}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

q_1=+2.00\mu C=+2.00\cdot 10^{-6}C

and is located at the origin (x=0, y=0)

Charge 2 is

q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

r_{1A}=0.40 m

So the potential due to charge 2 is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V

The distance of point A from charge 2 is

r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

So the potential due to charge 1 at point B is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V

The distance of charge 2 from point B is

r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m

So the potential due to charge 2 at point B is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V

Therefore, the net potential at point B is

V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V

So the potential difference is

V_B-V_A=-29,500 V-(-2700 V)=-26,800 V

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

W=q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

In this problem, we have:

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

4 0
3 years ago
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15
Alenkasestr [34]

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

3 0
3 years ago
Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w
shusha [124]

Answer:

Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.

5 0
3 years ago
The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find
Nataly_w [17]

The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25

<h3>How to find the Instantaneous rate of change?</h3>

The annual inventory cost C for a manufacturer is given as;

C = (1012000/Q) + 7.5Q

where Q is the order size when the inventory is replenished.

Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341

Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23

Instantaneous rate of change in C is first order derivative C':

C'(Q) = -1,012,000/(Q²) + 7.5

C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25

Read more about Instantaneous rate of change at; brainly.com/question/14666106

#SPJ1

8 0
1 year ago
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