A corn ethanol production plant receives 500,000.0 kg/day corn feedstock at a moisture content of 15.5% (wet basis). If all of t
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Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1
For pipe 2
Q=2.8 l/s
We know that Q=AV
head loss (h)
Now putting the all values
So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:
It is necessary to get the Reynold's number:
Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:
The overall heat transfer coefficient:
Here
Substituting values: