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:
Explanation:
, integrated circuit packaging is the final stage of semiconductor device fabrication, in which the block of semiconductor material is encapsulated in a supporting case that prevents physical damage and corrosion.
Answer:
Explanation:
Hello,
In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:
Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:
Thus, the liquid and liquid-vapor entropies are included to compute the quality:
Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:
Then, by using the first law of thermodynamics, the maximum specific work is computed via:
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I believe A is the answer: creat materials for electrical cables
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