water at the rate 10kg/s is compressed adiabatically from 5Pa to 50Pa in a steady flow process. Find the power required. the specific volume is 0.03
Answers
Explanation:
In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.
Answer:
The energy required to compress water from 5 Pa to 50 Pa at a charge of 10 kg/s in an adiabatic regular glide technique is 35.3 kW.
Explanation:
From the above question,
To discover the electricity required to compress water from 5 Pa to 50 Pa at a charge of 10 kg/s in an adiabatic constant glide process, we can use the following equation:
Power = m * (h2 - h1)
Where:
m = mass waft price of water = 10 kg/s
h2 = unique enthalpy of water at the outlet stress of 50 Pa
h1 = particular enthalpy of water at the inlet stress of 5 Pa
Since the procedure is adiabatic, there is no warmness switch involved, and we can anticipate that the unique enthalpy of water stays regular at some point of the process. Therefore, we can locate the unique enthalpy of water at each inlet and outlet pressures the use of the steam tables or software:
h1 = 497.42 kJ/kg
h2 = 500.95 kJ/kg
Substituting the values into the equation, we get:
Power = 10 kg/s * (500.95 kJ/kg - 497.42 kJ/kg) = 35.3 kW
Therefore, the energy required to compress water from 5 Pa to 50 Pa at a charge of 10 kg/s in an adiabatic regular glide technique is 35.3 kW.
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