M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h = 0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
Answer:
Meteorology
Explanation:
That is indeed the study of weather and climate.
Answer:
Potential energy of the ball would increase
Explanation:
The formula for potential energy of an object is:
Potential Energy (object) = mass x gravity x height
In your question, the value for height is increased when the object (ball) is placed in a higher position. This creates a net positive change in the right-hand side of the formula and since the value of mass and gravity doesn't change because gravity is a constant (meaning it's always 9.8N/kg) and the mass is said to be kept the same, the ball's potential energy would increase.