Problem 5.3.6

Water is flowing through an annulus that is formed by two tubes. These tubes have diameters of 1.5 inch and 2 inches. Both of these tubes have thickness rating of 14 BWG (Birmingham wire gauge). Inner diameter of 2-inch tube is 1.834 inches. Water is to be cooled from 160 degrees Fahrenheit to 120 degrees Fahrenheit.

Find out the values of Nusselt number for the following cases:

1. Water flowing at 50 pounds per hour with a pipe length of 10 feet.
2. Water flowing at 800 pounds per hour with a pipe length of 10 feet.

Viscosity of water wall temperature is reported as 1.406 pound per foot per hour

Solution:

(See Problem 5.3.2 also). Bulk temperature of water is 120 plus 160 over 2 equals 140 degrees Fahrenheit. So properties of water at this temperature can be seen as specific heat 0.9989 BTU per pound per degree Fahrenheit, density 61.55 pounds per cubic foot, viscosity 1.24 pounds per foot per hour, thermal conductivity 0.375 BTU per hour per foot per degree Fahrenheit. Further, viscosity correction factor, phi, is 0.9826.

Given the mass flow rate, we can find velocity through the annulus, once annular area is determined. For a circular tubes, this value is given by pi over 4 times (D_i squared minus d_o squared), where D_i is the inside diameter of the outer tube, and d_o is the outer diameter of the inside tube. Annular area is equal to 0.00607 square foot.

Reynolds number is V time d time rho over mu. For flow through annulus, we need to obtain the equivalent diameter. It is equal to 4 times the hydraulic radius, r_H, which in turn is a ratio of cross-sectional area and the heat-transfer perimeter. This reduces to (D_i squared minus d_o squared) over d_o. Equivalent diameter is different for frictional flow calculations. In that case, this expression reduces to (D_i minus d_o). Plugging in their values, we get equivalent diameter to be 0.06186 foot.

Velocity of water can be calculated by using the value of mass flow rate, and dividing it by density, giving us volumetric flow rate, and then dividing it by area. Now we can find Reynolds number by multiplying V by d time rho by mu. Prandtl number would be calculated by multiplying c by mu and dividing it by k. Its value is equal to 3.3.

For Case 1, Velocity could be found to be 133.75 feet per hour. Reynolds number is 410.7, which is less than 2,100 (laminar condition). The appropriate correlation is Nusselt number is 10.02 times Reynolds number raised to the power 0.45 times Prandtl number raised to the power 0.5 and other factors. Plugging in their values, Nusselt number is 18.07.

For Case 2, Velocity is again 2,140 feet per hour. Reynolds number is 6,572 (turbulent conditions). The appropriate correlation is Nusselt number is 0.020 times Reynolds number raised to the power 0.8 times Prandtl number 0.333 times other factors. Nusselt number is found to be 37.518.