For a two-source network, if current produced by one source is in one direction, while the current produced by the other source
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Answer: Heat transfer (Q) is 521 kJ.
Explanation: In the piston-cilinder assembly, we can suppose that the oxygen act as an ideal gas, so, it can be used the General Gas Equation:
PV=RT, where:
P is pressure;
V is volume;
m is mass;
M is molar mass;
R is a constant: R = 8.314. m³bar.K⁻¹.mol⁻¹;
T is temperature;
Using this equation, find the intial temperature:
PV = RT
1.2 = .8.314.
.T
T = 1.924. K
To determine the final temperature, use Combined Gas Law:
=
, in which, the left side of the equality is related to the initial values and the right side, to the final values.
As pressure is constant:
T =
T =
T = 3.85. K
With the temperatures, calculate the heat transfer of the process:
Q = m.k.ΔT, where:
k is heat constant
ΔT = T -
Q = m.k.ΔT
Q = 2.1.35.(3.85 - 1.92).
Q = 521 kJ
The heat transfer in the process is 521 kJ.
Answer:
isobaric expansion = 281.09 Btu
isothermal compression= 72 Btu
Explanation:
The first law of thermodynamics is:
where:
Q=heat transferred
W= work
U=internal energy
P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.
In a isobaric process heat transferred is:
For an isothermal process (T2-T1 = 0) so
From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.
For the first process
we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)
we use this information in the appropriate unit to find the pressure in thermodynamic tables.
T1= 176°C
v1= 1.49 m^3/kg
P=1.37 bar
in the second state we have
P=1.37 bar =137000Pa
with thee properties we check in the thermodynamic tables
T2= 255°C
we usually find Cp on tables for water but from the Mayer relation we have:
Cp for water vapor is: 33.12 J/mol*K
R=8.314 J/mol*K
Cv= 41.434 J/mol*K
replacing in the equation for Q
296569J =281.09 Btu
Answer:
The roll force is 1.59 MN
The power required in this operation is 644.96 kW
Explanation:
Given;
width of the annealed copper, w = 228 m
thickness of the copper, h₀ = 25 mm
final thickness, hf = 20 mm
roll radius, R = 300 mm
The roll force is given by;
where;
w is the width of the annealed copper
is average true stress of the strip in the roll gap
L is length of arc in contact, and for frictionless situation it is given as;
Now, determine the average true stress, , for the annealed copper;
The absolute value of the true strain, ε = ln(25/20)
ε = 0.223
from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.
Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa
Finally determine the roll force;
The power required in this operation is given by;