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mojhsa [17]
2 years ago
9

Air enters a turbine with a stagnation pressure of 900 kPa and a stagnation temperature of 658K, and it is expanded to a stagnat

ion pressure of 100 kPa. Assuming the expansion process is isentropic, determine the turbine power for a mass flow rate of 0.04 kg/s. What does the sign of the calculated power indicate
Engineering
1 answer:
bezimeni [28]2 years ago
6 0

Answer:

12.332 KW

The positive sign indicates work done by the system ( Turbine )

Explanation:

Stagnation pressure( P1 ) = 900 kPa

Stagnation temperature ( T1 ) = 658K

Expanded stagnation pressure ( P2 ) = 100 kPa

Expansion process is  Isentropic, also assume steady state condition

mass flow rate ( m ) = 0.04 kg/s

<u>Calculate the Turbine power </u>

Assuming a steady state condition

( p1 / p2 )^(r-1/r)  = ( T1 / T2 )

= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )

=  ( 9 )^0.285 = 658 / T2

∴ T2 = 351.22 K

Finally Turbine Power / power developed can be calculated as

Wt = mCp ( T1 - T2 )

    = 0.04 * 1.005 ( 658 - 351.22 )

    = 12.332 KW

The positive sign indicates work done by the system ( Turbine )

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The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
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Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

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The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

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6 0
3 years ago
What is the connection between the air fuel ratio and an engine running rich/poor? please give clear examples and full sentances
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Explanation:

Air fuel ratio:

 Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

Air\ fuel\ ratio=\dfrac{mass\ of\ air}{mass\ of\ fuel}

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.

When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.

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In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
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Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

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P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

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⇒ n=\frac{PV}{RT}

By substituting the values, we get

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       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

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                   =28.12 \ g/mol

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