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tatiyna
3 years ago
14

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o

f the refrigerant is 1.2 kg/s. Determine: (a) The power input to the compressor, and (b) The volume flow rate of the refrigerant at the compressor inlet.
Engineering
1 answer:
Leni [432]3 years ago
6 0

Answer:

(a) The power input to the compressor: \dot{W}=73.07 kJ/s = 73.07 kW

(b) The volume flow rate of the refrigerant at the compressor inlet: \dot{v}=0.209 m^{3}/s

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg

Now we proceed to know the work with the following expression:

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

Now we replace values and our result is:

\dot{W}=73.07 kJ/s = 73.07 kW

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

\nu=0.1739m^{3}/kg

With our specific volume and the mass rate, we can calculate the volume rate:

\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s

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3 years ago
A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage
Kobotan [32]

Answer:

Following are the solution to the given question:

Explanation:

Line voltage:

V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v

Power supplied to the load:

P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi

10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A

Check wye-connection, for the phase current:

I_{ph}=I_L= 32.68\ A

Therefore,

Phasor currents: 32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A  

Magnitude of the per-phase load impedance:

Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega

Phase angle:

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Please find the phasor diagram in the attached file.

8 0
3 years ago
16. A circular swimming pool has a diameter of 1818m. The sides are 33m high and the depth of the water is 2.52.5m. How much wor
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Answer:

Explanation:

Radius ( r ) = 9 m

Volume = πr²h

= 3.14 x 9² x 2.5

= 635.85 m³

mass m = 635.85 x 10³

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work done

= mgh

=  635.85 x 10³x 9.8 x 31.75

= 197.84 x 10⁶J

4 0
3 years ago
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