A pulley in the shape of a solid cylinder of mass 1.50 kg and radius 0.240 m is free to rotate around a horizontal shaft along t
1 answer:
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Answer:
The density of the fluid is 1100 kg/m³.
Explanation:
Given that,
Height = 5.00 cm
Pressure at top =594 Pa
Pressure at bottom = 1133 Pa
We need to calculate the change in pressure
Using formula of change in pressure
Where, = Pressure at bottom
= Pressure at top
put the value into the formula
Using formula of pressure for density
Where, = density
P = pressure
h = height
Put the value in to the formula
Hence, The density of the fluid is 1100 kg/m³.
Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99