25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter and it is 1.0 ft long to the outside atmosphere at 25°C. There are 10 vents in the 1000-ft ditch. The outside air can be assumed to be dry. Calculate the total evaporation loss of water in lbm/d. Assume that the partial

The total evaporation loss of water is 87.873 ×  lbm /day.

Explanation:

Assume A is the water and B is air.

A is diffusing to non diffusing B.

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is .

Total pressure P = 1 atm = 101.325 KPa

= 23.76 mm Hg

=

= 0.03126 atm

= 3.167 K Pa

When air surrounded is dry air, then = 0 mm Hg

R = 8.314

=

= 101.325 - 3.167

= 98.158 K Pa

= 101.325 - 0

= 101.325 K Pa

= 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

=  

= 0.11077 × mol/ .s

= 0.11077 × × 18 × (60×60×24)

= 0.1723 / .day

Area of individual pipe is

A = 0.00051

= 0.1723 × 0.00051

= 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 ×  lbm /day.

answer:

not me

explanation:

answer:

the average distance would be 20ly

Explanation:

100,000ly times 5,000ly=20ly