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Answer:
Enthalpy, hsteam = 2663.7 kJ/kg
Volume, Vsteam = 0.3598613 m^3 / kg
Density = 2.67 kg/ m^3
Explanation:
Mass of steam, m = 1 kg
Pressure of the steam, P = 0.5 MN/m^2
Dryness fraction, x = 0.96
At P = 0.5 MPa:
Tsat = 151.831°C
Vf = 0.00109255 m^3 / kg
Vg = 0.37481 m^3 / kg
hf = 640.09 kJ/kg
hg = 2748.1 kJ/kg
hfg = 2108 kJ/kg
The enthalpy can be given by the formula:
hsteam = hf + x * hfg
hsteam = 640.09 + ( 0.96 * 2108)
hsteam = 2663.7 kJ/kg
The volume of the steam can be given as:
Vsteam = Vf + x(Vg - Vf)
Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)
Vsteam = 0.3598613 m^3 / kg
From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3
Answer:
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Answer:
Explanation:
Given data in question
mean stress = 50 MPa
amplitude stress = 225 MPa
to find out
maximum stress, stress ratio, magnitude of the stress range.
solution
we will find first maximum stress and minimum stress
and stress will be sum of (maximum +minimum stress) / 2
so for stress 50 MPa and 225 MPa
=
+
/ 2
50 =
+
/ 2 ...........1
and
225 =
+
/ 2 ...........2
from eqution 1 and 2 we get maximum and minimum stress
= 275 MPa ............3
and
= -175 MPa ............4
In 2nd part we stress ratio is will compute by ratio of equation 3 and 4
we get ratio =
/
ratio = -175 / 227
ratio = -0.64
now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.
magnitude =
-
magnitude = 275 - (-175) = 450 MPa
Answer:
In the attached solution, the following calculations have already been calculated: Per phase , Maximum , RMS value of Internal Voltage in this sequence.
Explanation: