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avanturin [10]
3 years ago
11

Air at 80 kPa and 10°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

ssure of 100 kPa. The exit area of the diffuser is 5 times the inlet area. Determine: i. the exit temperature of the air, and ii. the exit velocity of air.
Engineering
1 answer:
il63 [147K]3 years ago
7 0

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

A_2=5A_1

If consider that density of air is not changing from inlet to exit then by using continuity equation

A_1V_1=A_2V_2

So   A_1\times 150=5A_1V_2

V_2=30m/s

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

Here Q=0 and w=0

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

When air is treating as ideal gas  

h=C_pT

Noe by putting the values

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}

T_2=293.74K

So the exit temperature is 293.74 K.

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Answer:

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Answer:

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Solution:

As per the question:

System Load = 96000 Btuh

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Temperature rise, T' = 15^{\circ}

Now,

The system load is taken to be at constant pressure, then:

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Now, for a rise of 20^{\circ} in temeprature:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s

Now, for 15^{\circ}:

\dot{m}C_{p}\Delta T = 96000

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The tensile stress is a type of normal stress, in which a perpendicular force creates the stress to an object’s surface.

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