Question 12.8: Water Flow Inside Tubes* The internal convective heat transf......
Water Flow Inside Tubes*
The internal convective heat transfer coefficient is to be computed for conditions when water at a temperature of 10°C with a velocity of 2.5 m/s flows through a tube with 8 mm internal diameter. These conditions are common in the water side of the evaporator of a refrigeration system.
Given: v = 2.5 m/s, characteristic length D = 8 mm
Find: h_{con} using Equation 12.20 for cooling condition
Lookup values: From property tables of water, \rho = 1000 kg/m³, \mu = 0.00131 Pa · s, c_{p} = 4190 J/(kg · K), k = 0.573 W/(m·K)
N uD_{D} = 0.023 Re_{D}^{0.8}Pr^{n} (12.20)
n = 0.4 heating
n = 0.3 cooling
* Same inputs as Example 2.11 that was solved using a dimensional correlation.
First, the Reynolds number is determined from Equation 12.14 as
Re = \frac{\rho v D}{\mu} = \frac{1000 kg/m^{3} \times 2.5 m/s \times 0.008 m}{0.00131 Pa \cdot s}= 15,267
The flow is clearly turbulent.
Next, the Prandtl number is computed from Equation 12.15 as follows:
P r = \frac{\mu c_{p}}{k} (12.15)
Pr = \frac{4190 J/kg \times 0.00131 Pa \cdot s}{0.573 W/(m \cdot K)} = 9.6We then use Equation 12.20 with n = 0.3 to find
Nu_{D} = 0.023 \times 15,267^{0.8} \times 9.6^{0.3} = 100.8from which the convective heat transfer coefficient is deduced from Equation 12.16 as
Nu = \frac{h_{con} D}{k} (12.16)
h_{con} = \frac{0.573 W/(m \cdot K) \times 100.8 }{0.008 m} = 7220 W/(m^{2} \cdot K)Comments
This example is identical to Example 2.11 that was solved using simplied dimensional equations.
In that case, h_{con} = 5171 W/(m6{2} K), which is about a 25% difference from the value found using the more accurate dimensionless equation.