Answer:
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Answer:
W=-940.36 KJ
Explanation:
Given that
Process follows pv=constant
So this is the isothermal process and work in isothermal process given as
Now by putting the values (1.4 bar =140 KPa)
W=-940.36 KJ
Negative sign indicates that this is a compression process and work will given to the system.
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:
Area of Circle = 78.5398
Surface Area of Sphere = 1.2566 x 10^3 = 1256.6 ft
Volume of Sphere = 33.5103 ft
Explanation:
Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.
r=5; % define r as 5
a=pi*r^2;% calculate the area of the circle
AreaOfCircle=a
r=10; % define r and 10 ft
sa=4*pi*r^2; %Calculate the surface area of the sphere
SphereSurfaceArea=sa
r=2;% define r as 2 ft
vs=(4/3)*pi*r^3;% Calculate the volume of the sphere
VolumeShere=vs
Uhh idk this can u mabye look it up
