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: *changed*
Explanation: Because you peed
Answer:
98°C
Explanation:
Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7
22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²
Temperature change, t = (50 - 25)°C = 25°C = 298K
Hence, Temperature = 150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =
∴ Temperature change = 2.00K
But temperature, T= (373 - 2)K = 371 K
In °C = (371 - 273)K = 98°C
If i would visit one of my friends it’s c