Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point
= 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point
= 0.1 m^3
The pressure at the second point
= 1 bar —> 1 x 102 = 102 kPa
The volume at the second point
= 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
![W_{a(1)}= 0\\P=10^{2}](https://tex.z-dn.net/?f=W_%7Ba%281%29%7D%3D%200%5C%5CP%3D10%5E%7B2%7D)
The work is defined by
![W_{a(2)}=\int\limits^V_V {P} \, dV](https://tex.z-dn.net/?f=W_%7Ba%282%29%7D%3D%5Cint%5Climits%5EV_V%20%7BP%7D%20%5C%2C%20dV)
║
V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
![W_{b}=\int\limits^V_V {P} \, dV\\P=10^{2} V^{-1}\\ W_{b}=\int\limits^V_V {10^{2} V^{-1}} \, dV\\\\](https://tex.z-dn.net/?f=W_%7Bb%7D%3D%5Cint%5Climits%5EV_V%20%7BP%7D%20%5C%2C%20dV%5C%5CP%3D10%5E%7B2%7D%20V%5E%7B-1%7D%5C%5C%20W_%7Bb%7D%3D%5Cint%5Climits%5EV_V%20%7B10%5E%7B2%7D%20V%5E%7B-1%7D%7D%20%5C%2C%20dV%5C%5C%5C%5C)
║
ln(V)║limit 1--0.1
=230.26 kJ
PV = 400 x 0.08 = 32 J
Hope this helps
Answer:
basic quantity. s.I unit. equipment
length M or Cm . metre rule
mass. Kg or G. spring/weight balance
time. Seconds. clock/stopwatch
Temperature. kelvin/celcius thermometer
volume. L/Ml/Cl. measuring cylinder
The answer is A) electric current ⚡️
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