Rolling of the eyes. The person is either vexed or frustrated.
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
The work is defined by
║ 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
║ ln(V)║limit 1--0.1
=230.26 kJ
The force applied to raise the block would be 250N
One side will be uneven and fall farther than the other side.
Answer:
With a slower speed-perhaps 5 cm/s answer is c