jadensababe9151

14.01.2021 •

Engineering

Use the concept of reverse engineering to explain how you might reverse engineer something in your home. Choose an object and describe, in detail, your process. Be sure to use correct grammar, spelling, and punctuation. HELP!!

Solved

Show answers

Ответ:

emmiles2005

12.06.2022

Engineering

1.25 cm/day

Explanation:

An air thickness , (l) = 0.15 cm

Air Temperature =

$(T_a)=20^0C = (20+273)K\\(T_a)=293K$

Mass Diffusion coefficient (D) = 0.25cm^2/sec

If the air pressure $(P_a) = 0.5 P_{sat}$

We are to determine how fast will the water (H_2O) level drop in a day.

From the property of air at T = 20° C

$P_{sat} = 2.34$ from saturated water properties.

The mass flow of (H_2O) can be calculated as:

$H_2O = \frac{D}{\phi} \delta C$

where:

$\delta C = \frac{P_{sat}*P_a}{RT }$

R(constant) = 8.314 kJ/mol.K

$\delta C = \frac{2.34*0.5}{8.314*293 }$

$\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3$

Since 1 mole = 18 cm ³ of water

$0.48*10^{-6}mol/cm^3$ will be: $(0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3$

$\delta C = 8.64 * 10^{-6}$

Again:

$H_2O = \frac{D}{\phi} \delta C$

$= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}$

$=1.4481*10^{-5} \frac{cm^2/sec}{cm}$

Converting the above value to cm/day: we have:

$1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}$

= 1.25 cm/day

∴ the rate at which the water level drop in a day = 1.25 cm/day

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Engineering

Ask an AI advisor a question

I want advice