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Svetllana [295]
2 years ago
6

A classroom that normally contains 40 people is to be air-conditioned with window air-conditioning units of 5 kW cooling capacit

y. A person at rest may be assumed to dissi- pate heat at a rate of about 80 Watts. There are 10 light bulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h.
a. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window air-conditioning units required.
b. What is the source of heat being generated for the light bulbs and for the students?
Engineering
1 answer:
kap26 [50]2 years ago
4 0

Answer:

Explanation:

a ) Heat energy given out by 40 people

= 40 x 80 = 3200 J in one second

Heat energy given out by 10 light bulbs

= 10 x 100 = 1000 J in one second

Heat transfer by wall = 15000 x 1000 J in one hour

= 15000 x 1000 / 60 x 60

= 4166.67 J in one second

Total energy

= 3200 + 1000 + 4166 .67 J in one second

= 8366.67 J

To drive this much of heat , we need 2 air -conditioner of 5000 W each.

b ) Source of heat for bulb is electrical energy of bulb

For the students , source of heat by student is the bio- chemical reactions taking place in the body of the student which is exothermic.

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Suppose the working pressure for a boiler is 10 psig, then what is the corresponding absolute pressure?
yanalaym [24]

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

From Fluid Mechanics, we remember that absolute pressure (p_{abs}), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:

p_{abs} = p_{atm}+p_{g} (1)

Where:

p_{atm} - Atmospheric pressure, measured in pounds per square inch.

p_{g} - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

If we suppose that p_{atm} = 14.696\,psi and p_{g} = 10\,psi, then the absolute pressure is:

p_{abs} = 14.696\,psi+10\,psi

p_{abs} = 24.696\,psi

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

8 0
2 years ago
Miller Indices:
svetlana [45]

Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with  lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

<u>In this case, for [1 2 1]:</u>

l=1=1/a_0 \rightarrow a_0=1

m=2=2/b_0 \rightarrow b_0=0.5

n=1=1/c_0 \rightarrow c_0=1

<u>for </u>[1 2 \overline{4}]<u>:</u>

l=1=1/a_0 \rightarrow a_0=1

m=2=2/b_0 \rightarrow b_0=0.5

n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25

B) The closest distance between planes with the same Miller indices can be calculated as:

With \pi:[l m n] ,the distance is d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}} with lattice constant a.

<u>In this case, for [1 2 1]:</u>

<u />d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a<u />

<u>for </u>[1 2 \overline{4}]<u>:</u>

d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

5 0
3 years ago
A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 37.0 m long piece of such wire.
Alinara [238K]

Answer:

R=1923Ω

Explanation:

Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.

Coil length(L) of the wire=37.0m

Cross-sectional area of the conductor or wire (A) = πr^2

A= π * (2.053/1000)/2=3.31*10^-6

To calculate for the resistance (R):

R=ρ*L/A

R=(1.72*10^8)*(37.0)/(3.31*10^-6)

R=1922.65Ω

Approximately, R=1923Ω

5 0
2 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
A fixed mass of saturated water vapor at 400 kpa is isothermally cooled until it is a saturated liquid. Calculate the amount of
Kazeer [188]
This is the explanation

6 0
3 years ago
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