Answer:32.4m/
Explanation:
Given data
=0.4m
Intial angular velocity
=4rad/s
angular acceleration
=5rad/
angular velocity after 1 sec
=
+
=4+5
=9rad/s
Velocity of point on the outer surface of disc
=
v=
m/s=3.6m/s
Normal component of acceleration
=
=
=32.4m/
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 what do you want i know a project but what is it on?
Explanation:
