Answer:
You so good is ué KKKKKK a gente não não é possível que o seu ponto de vista da minha irmã me ligou
Explanation:
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Answer:
A. Overall efficiency =83.95%
B. 87.45%
C.120.465kw
Explanation:
H=75m
Flow rate= 120L/s=1.02m^2/s
Po=Output power= 630kw
As power input Pi= Total power available
=gQH/1000
Where
g=9.81
Q=120L/s=1.02m^2/s
H=75m
= (9.81m/s^2 × 1.02m^2/s×75m)/1000
750.465kw
A. Overall efficiency of generator turbine unit (Po/Pi)
= Output power/ power input
= 630/750.465= 0.8395
Overall efficiency of the unit= 0.8395 × 100
=83.95%
B. Turbine efficiency if generator efficiency is 96%
no × nt= 0.96
nT= 87.45%
C. Power losses= Power input - Power output
Pi - Po
=750.465- 630
=120.465kw
Power losses due to inefficiencies in turbine of the generator is 120.465kw
Answer:
When a 10A electric heater is directly connected to 2A PLC output module for switching then PLC output module will most probably be damaged due to excess current drawn by the electric heater.
Explanation:
Usually, the high current rating loads such as electric heater are connected to the PLC via relays or contactors. A relay or contactor is a device which is used to make or break the electrical circuit.
These relays are capable of handling heavy currents and perform reliable switching. Contactor is preferred when the current is more than 10 A and relay is preferred when current is less than or equal 10 A.
