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Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.
Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:
Without any extra adjustments or corrections, either 125% of the continuous load, OR
When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).
This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.
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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:
