Answer:
Explanation:
Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:
The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:
The properties of the steam are obtained from steam tables:
Intial State
Final State
The entropy change of the steam during the process is:
Answer:
0.0297M^3/s
W=68.48kW
Explanation:
Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
state 1
X=quality=1
T=-26C
density 1=α1=5.27kg/m^3
entalpy1=h1=234.7KJ/kg
state 2
T2=70
P2=8bar=800kPa
density 2=α2=31.91kg/m^3
entalpy2=h2=306.9KJ/kg
Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.
m1=m2
(Q1)(α1)=(Q2)(α2)
the volumetric flow rate at the exit is 0.0297M^3/s
To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation
W=m(h2-h1)
m=Qα
W=(0.18)(5.27)(306.9-234.7)
W=68.48kW
the compressor power is 68.48kW