Answer:
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The exit temperature is 586.18K and compressor input power is 14973.53kW
Data;
- Mass = 50kg/s
- T = 288.2K
- P1 = 1atm
- P2 = 12 atm
<h3>Exit Temperature </h3>
The exit temperature of the gas can be calculated isentropically as
Let's substitute the values into the formula
The exit temperature is 586.18K
<h3>The Compressor input power</h3>
The compressor input power is calculated as
The compressor input power is 14973.53kW
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Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
