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mezya [45]
3 years ago
9

A closed rigid tank contains water initially at 10,000 kPa and 520ºC and is cooled to a final temperature of 270° C. Determine t

he final pressure of the water.
Engineering
1 answer:
jasenka [17]3 years ago
5 0

Answer:

final pressure is 6847.41 kPa

Explanation:

given data:

P_1 =  10,000 kPa

T_1 =520\ degree\ celcius = 793 K

T_2  = 270 degree celcius = 543 K

as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream

for ischoric process we know that

\frac{P_1}{T_1} =\frac{P_2}{T_2}

\frac{10*10^6}{793} = \frac{P_2}{543}

P_2 = 6.84741*10^6 Pa

final pressure is 6847.41 kPa

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A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu
suter [353]

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

emf= N\frac{d \phi}{dt}

where;

ΔФ is change in magnetic flux

ΔФ  = BA = B x πr²

ΔФ  = 0.573 x π(0.0395)² = 0.002809 T.m²

emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0
3 years ago
Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is
professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

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3 0
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Solid Isomorphous alloys strength
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Answer:

Explanation:

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3 0
3 years ago
I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going to tr
Alina [70]

Answer:

Δx = 25 ft.

Explanation:

Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:

1 ) Egg falling:

If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:

yf-y₀ = 1/2* g* t²

If we take the up direction as positive, we can solve for t as follows:

0-100 ft = 1/2* (-32.15 ft/s²)* t²

⇒ t = \sqrt{(100*2)/32.15} = 2.5 sec.

2) Person on the ground running away:

In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.

We can get the distance at which he can reach, applying the definition of velocity:

v = (xf-x₀) / (tfi-t₀)

If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:

xf = v*t = 20 ft/sec*1.25 sec = 25 ft.

8 0
3 years ago
Find the rate of heat transfer through a 6 mm thick glass window with a cross-sectional area of 0.8 m2 if the inside temperature
kiruha [24]

Answer:

6.9

Explanation:

I had the same question lol your welcomr if itd not right in sorry

3 0
3 years ago
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