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expeople1 [14]
3 years ago
3

Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -12oC with a volumetric flow rat

e of 0.18 m3/s. Refrigerant exits at 7 bar, 70oC. Changes in kinetic and potential energy from inlet to exit can be ignored. Determine the volumetric flow rate at the exit, in m3/s, and the compressor power, in kW.
Engineering
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

\dot V_{out} = 0.061\,\frac{m^{3}}{s}, \dot W = 108.875\,kW

Explanation:

The process in the compressor is modelled after the First Law of Thermodynamics:

\dot W + \dot m \cdot (h_{1}-h_{2}) = 0

The specific enthalpies of the refrigerant at inlet and outlet are, respectively:

Inlet (Saturated vapor)

\nu = 0.10744\,\frac{m^{3}}{kg}

h = 243.34\,\frac{kJ}{kg}

Outlet (Superheated Vapor)

\nu = 0.036373\,\frac{m^{3}}{kg}

h = 308.34\,\frac{kJ}{kg}

The mass flow is:

\dot m = \frac{\left(0.18\,\frac{m^{3}}{s} \right)}{0.10744\,\frac{m^{3}}{kg} } \co

\dot m = 1.675\,\frac{kg}{s}

The volumetric flow rate at the exit is:

\dot V_{out} = \left(1.675\,\frac{kg}{s} \right)\cdot \left(0.036373\,\frac{m^{3}}{kg} \right)

\dot V_{out} = 0.061\,\frac{m^{3}}{s}

The power needed to make the compressor work is:

\dot W = \dot m \cdot (h_{2}-h_{1})

\dot W = \left(1.675\,\frac{kg}{s} \right)\cdot \left(308.34\,\frac{kJ}{kg}-243.34\,\frac{kJ}{kg} \right)

\dot W = 108.875\,kW

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Answer:

1) 63.66 ohm

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Explanation:

Data provided in the question:

Part 1

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Part 2

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