Respuesta :
Answer:
Please find the attached graph of temperature as a function of tube length
Explanation:
The given information are;
The temperature of the air = 310 K
The mass flow rate = 4 kg/s
k = 177 WmK1
The inner diameter of the tube = 0.20 m
The tube wall thickness = 0.02 m
The convection coefficients of the water = 150 W·m²/k
The convection coefficients of the water = 30 W·m²/k
The area of the tube = π×0.2^2/4 = 0.03142 m²
The density = 1000 kg.m³
The mass flow rate = 4 kg/s
[tex]U_m = \dfrac{\dot m}{\rho \cdot A_c} = \dfrac{4}{ 1000 \times 0.03142} = 0.127 \ m/s[/tex]
[tex]Re_{D} = \dfrac{\rho \times u_m \times D}{\mu} = \dfrac{1000 \ kg/m^3 \times 0.127 \ m/s \times 0.2 \ m}{8.01 \times 10^{-4}\ N \cdot S/m^2} = 31,791.25[/tex]
The Reynolds number is > 2300 for pipe therefore, we have turbulent flow, and the entry length is estimated at 10 pipe diameters
The Nusselt number, Nu = h*D/k = 0.023*31791.25^(0.8)*4.32^(0.4) = 165.11
The total resistance = [tex]R_{tot}=R_{conv, i} + R_{conv, i} + R_{tube}[/tex]
[tex]R_{conv, i} + R_{conv, i}[/tex] = (1/150*(1/(0.2)) + 1/30*(1/(0.22)))/π = 0.059 K/W
Resistance of tube, [tex]R_{tube} = \dfrac{ln(r_2/r_1}{2\cdot \pi \cdot k} = \dfrac{ln(0.12/0.1)}{2\times\pi \times 177} = 1.64 \times 10^{-4} \ K/W[/tex]
[tex]R_{tot}[/tex] = 0.059 + 1.64 × 10⁻⁴ = 0.059 K/W
The heat transfer [tex]\dot Q[/tex] = [tex]\dfrac{t_A - t_B}{R_{tot}}[/tex]
[tex]\dot Q[/tex] = (390 - 310)/0.059 = 1355.99 W ≈ 1356 W
Given that the water velocity = 0.127 m/s, we have;
Time to make one meter = 1/0.127 = 7.874 seconds
Mass of water that will have flowed in 7.874 seconds = 4×7.874 = 31.496 kg
The heat transferred in 7.874 seconds = 1356 × 7.874 = 10677.144 J
The specific heat capacity of water = 4,200 J/(kg·°C)
Therefore for one meter, we have;
10677.144 = 4,200 ×31.496 × ([tex]t_B[/tex] - 310)
([tex]t_B[/tex] - 310) = 10677.144 /(4,200 *31.496) =
[tex]t_B[/tex] =0.0807 + 310 = 310.0807 K
At two meters, we have;
2*10677.144 = 2*4,200 *31.496 × ([tex]t_B[/tex] - 310)
([tex]t_B[/tex] - 310.0807 ) = 2*10677.144 /( 2*4,200 *31.496 )
[tex]t_B[/tex] =0.0807 + 310.0807 = 310.1614
At three meters, we have
0.0807 + 310.0807 = 310.1614
The other values are;
m, T
1, 310.0807143
2, 310.1614286
3, 310.2421429
4, 310.3228571
5, 310.4035714
6, 310.4842857
7, 310.565
8, 310.6457143
Which gives the attached graph
