Answer:
U = 5.53 kW/m² °C
Explanation:
Given:
N = 24
L = 3m
Di = 1.2 cm = 0.012m
For water:
m = 3 kg/s; Twin = 20°C; Twout= 55°C
Cp = 4180 J/kg·K
For oil:
To_in= 120°C; To_out = 45°C;
For change in temperature
ΔT1 = To_in - Twout
120° - 55° = 65°C
ΔT2 = To_out - Twin
= 45 - 20 = 25°C
For the rate of heat transfer from water to oil:
Q = m*Cp(Twout - Twin)
Q = (3kg/s)*(4180J/kg.k)(55°C -20°C)
Q = 439800 ≈ 439.8 KW
The total surface area, A, of the tube =
A = N*pi*Di*L
A = 24*3.142*0.012*3
A = 2.714m²
Let's find the overall transfer coefficient, U, using the formula :
Q = U * A * F * ΔTlm
Where F= correction factor.
ΔTlm = logarithmic mean temperature
Average temperature difference =
[tex] \delta T_lm = \frac{\delta T_1 - \delta T_2}{ln [\frac{\delta T_1}{\delta T_2}]} [/tex]
[tex] \delta T_lm = \frac{65 - 25}{ln [\frac{65}{25}]} = 41.862 [/tex]
[tex] P = \frac{Tw_out - Tw_in}{To_in - Tw_in} [/tex]
[tex] P = \frac{55 - 20}{120 - 20} = 0.35[/tex]
[tex] R = \frac{To_in - To_out}{Tw_out - Tw_in} [/tex]
[tex] P = \frac{120 - 45}{55 - 20} = 2.15[/tex]
Using the correction factor chart, at P and R = 0.35 and 2.15 respectively, F = 0.7
From the equation, let's make U the subject,
[tex] U = \frac{Q}{A*F* \delta T_lm} [/tex]
[tex] U = \frac{439.8}{2.714*0.7* 41.862} [/tex]
U = 5.53 kW/m² °C
