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aliya0001 [1]
3 years ago
10

Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work

that can be delivered by thisturbine.
Engineering
1 answer:
Naily [24]3 years ago
3 0

Answer:

w_{out}=319.1\frac{BTU}{lbm}

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

s_2=s_1

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}

Best regards.

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In Uniform motion distance traveled is equal in equal interval of time .

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3 years ago
Impedance is defined as the total opposition to current in an AC circuit. Question 17 options: True False
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2 years ago
An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa
maxonik [38]

Answer:

\dot m_{2} = 0.199\,\frac{kg}{s}

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

1 + y = z

Then, the expression is simplified afterwards:

\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0

\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0

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h = 2994.3\,\frac{kJ}{kg}

State 2 (Saturated liquid)

h = 1008.3\,\frac{kJ}{kg}

State 3 (Liquid-Vapor mixture)

h = 2444.22\,\frac{kJ}{kg}

The ratio of the stream at state 2 to the stream at state 1 is:

y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}

y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }

y = 0.397

The mass flow rate of the saturated liquid is:

\dot m_{2} = y\cdot \dot m_{1}

\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )

\dot m_{2} = 0.199\,\frac{kg}{s}

5 0
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erastova [34]

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Explanation:

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Plz refer to the image below

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