Answer:
Explanation:
Hello!
To solve this problem you must follow the following steps, which are fully registered in the attached image.
1. Draw the complete outline of the problem.
2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties.
3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.
3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.
4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values and finds the work done.
5. draw the pV diagram using the 300F isothermal line
Answer:
••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.
Answer:
A) 282.34 - j 12.08 Ω
B) 0.0266 + j 0.621 / unit
C)
A = 0.812 < 1.09° per unit
B = 164.6 < 85.42°Ω
C = 2.061 * 10^-3 < 90.32° s
D = 0.812 < 1.09° per unit
Explanation:
Given data :
Z ( impedance ) = 0.03 i + j 0.35 Ω/km
positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km
A) calculate Zc
Zc =
=
=
= 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl =
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = 
= 
= 0.622 < 87.55 °
gl = 0.0266 + j 0.621 / unit
C) exact ABCD parameters for this line
A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )
C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )
D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
where : cos h (gl) = 
sin h (gl) = 
Answer:
Looks like mold that got frosted over
Explanation: