Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate = 
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.

Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is

As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%
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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The LCA process is a systematic, phased approach and consists of four components: goal definition and scoping, inventory analysis, impact assessment, and interpretation. The standards are provided by the International Organisation for Standardisation (ISO) in ISO 14040 and 14044, and describe the four main phases of an LCA: Goal and scope definition. Inventory analysis. Impact assessment.
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Answer:
>>pounds=13.2
>>kilos=pounds/2.2
Explanation:
Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.