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Svetllana [295]
3 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]3 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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A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i
vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be d_{s} and d_{p} respectively

and head be  H_{s} and  H_{p} respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = \rho gQH                  (1)

From eqn (1):

P \propto QH

and

QH \propto \sqrt{H}D^{2}

So,

P \propto H^{\frac{3}{2}} D^{2}

Therefore,

\frac{P_{s}}{P_{p}} = \frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{5\sqrt{5}}{16}

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8 0
3 years ago
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa
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Answer:

Explanation:

The formula for critical stress is

\sigma_c=\frac{K}{Y\sqrt{\pi a} }

\sigma_c =\texttt{critical stress}

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

\sigma_c=\frac{54.8}{1\sqrt{\pi  \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

4 0
3 years ago
Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t
HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

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So force exerted in y-direction

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F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

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So the resultant force R

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{0.763^2+0.763^2}

R=1.079

So the force required to hold the pipe is 1.08 N.

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Answer:

See attached pictures.

Explanation:

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