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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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Answer:

Mercury and Venus lie closer to Sun than position of Earth

Explanation:

As we know that all planets around the sun in the range of their distance can be arranged as following:

1). Mercury.

2). Venus.

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4). Mars.

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5 0
3 years ago
An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a
Zolol [24]

Answer:

a) \omega = 0.421\,\frac{rev}{s}, b) \Delta \theta = 0.066\,rev, c) v = 0.966\,\frac{mm}{s}, d) a = 3.293\,\frac{mm}{s^{2}}

Explanation:

a) The angular velocity of the turntable after 0.200\,s.

\omega = \omega_{o} + \alpha\cdot \Delta t

\omega = 0.240\,\frac{rev}{s}  + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)

\omega = 0.421\,\frac{rev}{s}

b) The change in angular position is:

\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot  \alpha \cdot t^{2}

\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}

\Delta \theta = 0.066\,rev

c) The tangential speed of a point on the rim of the turn-table:

v = r\cdot \omega

v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

v = 9.655\times 10^{-4}\,\frac{m}{s}

v = 0.966\,\frac{mm}{s}

d) The tangential and normal components of the acceleration of the turn-table:

a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )

a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}

a_{t} = 2.078\,\frac{mm}{s}

a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}

a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

a_{n} = 2.554\,\frac{mm}{s^{2}}

The magnitude of the resultant acceleration is:

a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}

a = 3.293\,\frac{mm}{s^{2}}

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