A. I believe, lmk if I’m right
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is

put here value with
turbine is insulated so q = 0
so here

solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.
<h3>
What is power factor of a transformer?</h3>
Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).
PF = working power / apparent power
PF = kW/kVA
kVA = kW/PF
kVA = 50 kW/0.8
kVA = 62.5 kVA
Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.
Learn more about power factor here: brainly.com/question/7956945
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Answer:
no idea
Explanation:
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Answer:
a) 8kW
b) $128
Explanation:
Given the coefficient of performance of the heat pump cycle to be 2.5
Energy delivered by the heat pump = 20kW
a) net power required to operate the heat pump = Energy delivered / coefficient of performance
Net power required = 20/2.5
= 8kW
b) Given the cost of electricity is $0.08 for 1kWhour
Since net power required to operate heat pump = 8kW
If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour
since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128