Answer:
(b) 56%
Explanation:
the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature
here we have given T₁ (Higher temperature)= 600+273=873
lower temperature T₂=110+273=383
Efficiency of power cycle is given by =1-
=1-
=1-0.43871
=.56
=56%
Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.
Answer:a
a) Vo/Vi = - 3.4
b) Vo/Vi = - 14.8
c) Vo/Vi = - 1000
Explanation:
a)
R1 = 17kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0
sin we know Va≈Vb=0
so
-Vi/5kΩ + -Vo/17kΩ = 0
Vo/Vi = - 17k/5k
Vo/Vi = -3.4
║Vo/Vi ║ = 3.4 ( negative sign phase inversion)
b)
R2 = 74kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
so
(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0
-Vi/5kΩ + -Vo/74kΩ = 0
Vo/Vi = - 74kΩ/5kΩ
Vo/Vi = - 14.8
║Vo/Vi ║ = 14.8 ( negative sign phase inversion)
c)
Also for ideal op-amp
Va≈Vb=0 so Va=0
Now for position 3 we apply nodal analysis we got at position 1
(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0 ( 5MΩ = 5000kΩ )
so
-Vi/5kΩ + -Vo/5000kΩ = 0
Vo/Vi = - 5000kΩ/5kΩ
Vo/Vi = - 1000
║Vo/Vi ║ = 1000 ( negative sign phase inversion)
Answer:
h = 287.1 m
Explanation:
the density of mercury \rho =13570 kg/m3
the atmospheric pressure at the top of the building is
the atmospheric pressure at bottom
we have also
1.18*9.81*h = (100.4 -97.08)*10^3
h = 287.1 m
