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%
Answer:
#Initialise a tuple
team_names = ('Rockets','Raptors','Warriors','Celtics')
print(team_names[0])
print(team_names[1])
print(team_names[2])
print(team_names[3])
Explanation:
The Python code illustrates or printed out the tuple team names at the end of a season.
The code displayed is a function that will display these teams as an output from the program.
Answer:
Change in entropy S = 0.061
Second law of thermodynamics is satisfied since there is an increase in entropy
Explanation:
Heat Q = 300 kW
T2 = 24°C = 297 K
T1 = 7°C = 280 K
Change in entropy =
S = Q(1/T1 - 1/T2)
= 300(1/280 - 1/297) = 0.061
There is a positive increase in entropy so the second law is satisfied.
Answer:
True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.
Option: A
<u>Explanation:
</u>
To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. <em>Regeneration</em> technique <em>is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine. </em>
<em>Thermal efficiency</em> of a turbine is increased as <em>the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor.
</em>
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
