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:

Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number (
) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

where
is the density of the fluid,
is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

Answer:
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Answer:
if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society
Answer: B Excretory
Explanation: Hope this helps :)