The total evaporation loss of water is 87.873 ×
lbm /day.
<u>Explanation:</u>
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:According to NEC 220.12,
“the floor area shall be calculated from outside dimensions of the buildings, dwelling unit, or other area involved.
The area should not include open porches, garages or unused or unfinished spaces not adaptable for future use”.
pa brainliest plss..thankss
Explanati
Answer:
a)Patm=135.95Kpa
b)Pabs=175.91Kpa
Explanation:
the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation
Pabs=Pw+Patm(1)
Where
Pabs=absolute pressure
Pw=Water pressure
Patm=
atmospheric pressure
Water pressure is calculated with the following equation
Pw=γ.h(2)
where
γ=especific weight of water=9.81KN/M^3
H=depht
A)
Solving using ecuations 1 y 2
Patm=Pabs-Pw
Patm=185-9.81*5=135.95Kpa
B)
Solving using ecuations 1 y 2, and atmospheric pressure
Pabs=0.8x5x9.81+135.95=175.91Kpa
