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dangina [55]
2 years ago
11

Air at 80 kPa and 400 K enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the

airstream is decreased from 230 to 30 m/s as it passes through the diffuser. Find (a) the exit temperature of the air and (b) the exit area of the diffuser.
Physics
1 answer:
PilotLPTM [1.2K]2 years ago
3 0

Answer:

a)  425.6 K = 152.6 degree Celsius

b)  678.6 square centimeter

Explanation:

Initial enthalpy = Final enthalpy

m(h1 + v1^2/2) = m(h2 + v2^2/2)

(h1 + v1^2/2) = (h2 + v2^2/2)

h 2 = (h1 + v1^2/2) - (v2^2/2)

h 2 = 400.98 + (230^2 – 30^2)/2 * 10^-3

= 426.98 KJ/Kg

T2 = T1’ + (T2’ -T1’)(h2-h1’)/( h2’-h1’)

T2 = 420 + (430-420)(431.43-421.26)(426.98-421.26)/(431.43 – 421.26)

T2 = 425.6 K = 152.6 degree Celsius

Area = mRT2/P2V2

Area = (600/60*60*1) *0.287*425.6/(100*30) *10^4

Area = 678.6 square centimeter

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andriy [413]

Answer:

-929.5Joules

Explanation:

To get the work done by sam, we will calculate the kinetic energy of sam expressed as;

KE = 1/2mv²

m is the mass = 1100kg

v is the velocity = 1.3m/s

KE = 1/2(1100)(1.3)²

KE = 550(1.69)

KE = 929.5Joules

Since Sam is opposing the direction of movement, work done by him will be a negative work i.e -929.5Joules

5 0
3 years ago
At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end
SpyIntel [72]

Answer:

5.590278\ m/s^2

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Explanation:

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s = Displacement

a = Acceleration

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{161}{3.6}-0}{8}\\\Rightarrow a=5.590278\ m/s^2

The acceleration is 5.590278\ m/s^2

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{1610}{3.6}-\dfrac{161}{3.6}}{60-8}\\\Rightarrow a=7.74038461538\ m/s^2

The acceleration is 7.74038461538\ m/s^2

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 5.590278\times 8^2\\\Rightarrow s=178.888896\ m

Distance traveled in the first 8 seconds is 178.888896 m

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=\dfrac{161}{3.6}\times 52+\dfrac{1}{2}\times 7.74038461538\times 52^2\\\Rightarrow s=12790.56\ m

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Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.

We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.

Mathematically\ energy\ E=\frac{hc}{\lambda}

Here h is the Planck's constant,c is the velocity of light.

Here we have been given red light and blue light.

In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.

Hence, the blue light will activate the most and red the least.

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