A) The pressure (p)=400kPa=0.400MPa
Mass of mixture =M
quality (X)= 0.60
Volume of mixture (V)=10 m3
From steam table at P=0.400MPa
Specific volume of saturated water (vf)=0.00108355 m3/kg
Specific volume of saturated steam (vg)=0.46238 m3/kg
Therefore, the volume of steam having X=0.6 is given by
V=M[vf+X(vg-vf)]
10= M[0.00108355+0.6(0.46238-0.00108355)]
M=36.04748 kg
If pressure is lowered to 300kPa
p=0.300MPa
From steam table we get,
Specific volume of saturated water (vf)=0.00107317 m3/kg
Specific volume of saturated steam (vg)=0.60576 m3/kg
Therefore, the volume of steam having X=0.6 is given by
V=M[vf+X(vg-vf)]
10= M[0.001007317+0.6(0.60576-0.00107317)]
M= 26.4825 kg
Answer:
The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.
Explanation:
The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:
Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:
Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:
[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]
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