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Phoenix [80]
2 years ago
6

A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

ined at 100 bar, 520 Celsius. The outlet is maintained at 10 bar, 280 Celsius. A rate of heat loss of 60 kW is measured. Determine the rate of work output from the turbine, in kW.
Engineering
1 answer:
strojnjashka [21]2 years ago
5 0

Answer:

\dot W_{out} = 399.47\,kW

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

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

The work done by the turbine is:

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

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

P = 10\,MPa

T = 520\,^{\textdegree}C

h = 3425.9\,\frac{kJ}{kg}

Outlet (Superheated Steam)

P = 1\,MPa

T = 280\,^{\textdegree}C

h = 3008.2\,\frac{kJ}{kg}

The work output is:

\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW

\dot W_{out} = 399.47\,kW

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Answer:n=0.973

Explanation:

Given

When True strain\left ( \epsilon _T_1\right )=0.171

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When True stress\left ( \sigma _2\right )=346.2 MPa

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We know

\sigma =k\epsilon ^n

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Substituting value

263.8=k\left ( 0.171\right )^n------1

346.2=k\left ( 0.226\right )^n-----2

Divide 1 & 2 to get

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Taking Log both side

ln\left ( 1.312\right )=nln\left ( 1.3216\right )

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2 years ago
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(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
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Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

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Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

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