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3 years ago

Helpppppp

Engineering

1 answer:

matrenka [14]3 years ago

5 0

Answer:

Among us is the answer

Explanation:

If you play among is everyday then Among us is always the answer

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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:

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

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

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

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

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:

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

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

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

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for...?

thenks for the points :))

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