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kobusy [5.1K]
2 years ago
13

Under steady state, a compressor is used to increase the pressure of an ideal gas air from 100 kPa to 1 MPa. Meanwhile the tempe

rature is also raised from 300 K to 500 K. To be safe, a separate stream of water is used to cool the air down to 320 K while the pressure of air remains unchanged during the cooling process. The temperature of the cooling water increases from 15°C to 30°C, while the pressure of the cooling water remains the same at 1 bar approximately. Assume the kinetic and potential energy change is negligible, there is no stray heat transfer from the heat exchanger to the surroundings.Draw a p-v diagram, one for air identify state 1 to 3 on air p-v diagram. Please show work for how to obtain relative pressures and temperatures for this p-v diagram.
Physics
1 answer:
Art [367]2 years ago
4 0

Answer:

‍♀️

Explanation:

...

You might be interested in
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
2 years ago
A particle has a charge of -4.25 nC.
SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m

The distance from the electric field is 1.71436 m

4 0
3 years ago
Read 2 more answers
Describe the relationship between kinetic energy and speed and give an example of how changing and object speed would affect its
Andreas93 [3]

Answer:

I'm not sure

Explanation:

I have had that question to Uchida c r go crew in to go be

6 0
3 years ago
A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s
Andrej [43]

Answer:

1300 m

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 20 s to 70 s is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)

=200+1000+100

=1300

As velocity is in m/s and time is in s so the unit of area is m

Hence, the total distance is  1300m.

8 0
3 years ago
What to do if vehicle struck by lightning
agasfer [191]
Lighting flows around the outside of a truck, and the majority of the current flows from the cars metal cage into the ground below. It's not very safe to be in a car or truck during bad weather.
4 0
3 years ago
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