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erica [24]
3 years ago
15

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

Engineering
1 answer:
levacccp [35]3 years ago
8 0

Answer:

\dot W_{out} = 3863.98\,kW

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

h_{in} = 3353.1\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mixture)

h_{out} = 890.1\,\frac{kJ}{kg}

The power produced by the turbine is:

\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})

\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )

\dot W_{out} = 3863.98\,kW

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Yuri [45]

Answer:

D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted

Explanation:

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3 years ago
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

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3 years ago
An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface
aivan3 [116]

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

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Substituting the given values

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The maximum allowable surface crack is 1.44 mm

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3 years ago
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Answer: Thx for points...

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