In desert climates, rainfall is not a common occurrence since the rain droplets formed in the up

step 1 Hi, so we have two questions here. First one, we need to calculate the time it will take for the

In desert climates, rainfall is not a common occurrence...

Key Concepts

Convection Heat Transfer

This concept involves the process by which heat is transferred between a fluid (such as air) and a surface (such as the raindrop) due to a temperature difference. It is characterized by the convective heat transfer coefficient and is essential for determining the rate at which heat enters the raindrop from the surrounding air.

Phase Change and Latent Heat

This concept refers to the energy required for a substance to change phases, such as from liquid to gas, without a change in temperature. The latent heat of vaporization is a key parameter that dictates how much energy is needed for the raindrop to evaporate, directly influencing the rate of mass loss.

Energy Balance

An energy balance involves matching the heat absorbed by the raindrop from the ambient air with the energy required for the phase change (evaporation). This balance is used to quantitatively determine how long it takes for the raindrop to reduce in size due to loss of mass.

Spherical Geometry in Evaporation

Since raindrops are approximately spherical, the relationships between surface area and volume are crucial. This geometry influences the rate of heat transfer and evaporation, as the surface area directly affects how much heat is available for the phase change, ultimately impacting the time required for the raindrop to shrink.

Transcript

00:01 Hi, so we have two questions here.

00:05 First one, we need to calculate the time it will take for the liquid level in the tube to drop three inches and then make a plot of the liquid level in the tube as a function of time for this period.

00:19 So remember that in a system or component a undergoes unimolecular diffusion through a stationary film of component b.

00:27 The diffusion process is governed by fixed law, which states that the total molar flux can be expressed in terms of the total concentration, the average concentration gradient, and the diffusion coefficient.

00:41 So let me just write the formula first.

00:43 This is the total molar flux, n sub a, then we have negative c.

00:49 C is the total concentration multiplied by the diffusion coefficient b, sub a, over the log mean of, 1 minus y sub a, lm, and multiplied by the average concentration gradient delta z.

01:14 Now the log mean of 1 minus y sub a formula is y sub a minus y sub a 2 over ln of 1 minus y sub a 2 over 1 minus y sub a 2 over 1 minus y sub a 1.

01:40 Now n sub a can also be expressed as the rate of change of the moles per cross -sectional area.

01:48 So we have this equation n sub a is given to one over cross -sectional area a and then rate of change of the moles, b and then sub a over t -t.

02:05 Now the change in moles can be further expressed in terms of the liquid's mass, molar mass, then and volume as well as the area and the height of the liquid measured from the top of the tube.

02:19 So let's write that in here.

02:21 B and sub a, this is equivalent to d.

02:30 Then m, small letter m, this is the liquid's mass over the molar mass, capital m, then b density multiplied by the volume over the molar mass.

02:48 M should be equivalent to density over molar mass bv.

03:06 So volume v can be expressed in terms of the area and the height z.

03:15 So we can rewrite this area multiplied by the height.

03:22 We can simplify this.

03:25 We'll have density multiplied by the area over the molar mass m, b, z.

03:33 And then we'll plug in b and sub a in this equation.

03:40 So we now have n sub a as equivalent to 1 over a, density a over m b, z over b c.

03:56 Let's simplify this.

03:59 They can cancel out area.

04:05 So we have density over the molar mass, b, z over.

04:10 Over d t.

04:13 Now delta z, this would be equivalent to z minus z sub o.

04:20 And for reference, c z o would be equivalent to zero.

04:28 So it's equivalent to z.

04:32 This is, let's write the value.

04:38 This is, let's just write here zero.

04:44 Then we'll equate these two equations, this one, this.

04:55 So, and sub a is equivalent to negative c, d sub a over 1 minus y sub a, log mean, delta y sub a, the delta z, mdz over d, d, c.

05:16 Again, delta z is equivalent to z.

05:20 So rearranging this.

05:25 Let's write this in black, negative m, c, d, a, over density, and then log, of 1 minus y sub a b t and then on the right side we have z b z and we'll have to integrate and evaluate this we have this limit 0 t b1 z 2 negative mc b a over density log mean of 1 minus y sub a also y and then for d t this is integral of d t evaluated from zero to c and then we have integral of z d z evaluated from c sub 1 to c sub 2 so we'll have mc t a over density multiplied by the log mean of 1 minus y sub a delta y sub a and then integral of bt evaluated from from 0 to t is t and integral of z d z evaluated from z one to z two will have one half multiplied by z z one raised to the power of two minus z two raise to the power of two and let's solve for t let's write that in here let's isolate t t would be equivalent to density multiplied by one minus y sub a, log mean over 2.

07:47 M is the molar mass, c, total concentration, delta y sub a, and then the coefficient, fusion coefficient, c sub 1 squared, c sub 2 squared.

08:04 So again, in this problem, we are to determine the duration required for the level of water to decrease to a specified height.

08:13 Due to evaporation into the atmosphere from an open tube under the assumption of unimolecular diffusion.

08:21 Now, given that the tube is open, the total pressure is the atmospheric pressure, which is 1 atm.

08:30 So now let's calculate the total concentration of water vapor, assuming ideal gas behavior.

08:37 And since we have the total pressure, which is the atmospheric, pressure so that's 1 atm let's write that information here one atm and the temperature we have 25 degrees celsius so in kelvin that's 298 .15 then using this formula for the total concentration of water vapor we have pressure over the gas constant are multiplied by the temperature so plugging in the known volume we have 1 atm over 0 .0821.

09:27 This is the value of r and then the temperature in kelvin is 298 .15.

09:35 So calculating this we'll get 0 .0409.

09:40 This is moles per liter or converting this to moles per cubic centimeter, which is 4 .09 times 10 to the negative 5.

09:53 We'll just remember that one liter is equivalent to 8 ,000 cubic centimeter.

10:03 So again, for 0 .0409 moles per liter is equivalent to 4 .09 times 10 to negative 5 moles per cubic centimeter.

10:13 And then we'll compute the mole fraction of water vapor directly above the surface in the vapor liquid interface using rawlslaw.

10:22 This is the formula that we will use.

10:29 Saturation vapor pressure over the total pressure multiplied by the mole fraction in the liquid phase.

10:40 And given that the liquid is composed entirely of water, the mole fraction of water in the liquid phase, or x sub a, would be equivalent to one.

10:50 Let's write that in here.

10:57 With the vapor pressure of water at 25 degrees celsius being 23 .8 tor.

11:13 We can look this up in standard reference materials.

11:21 We can now substitute the values here.

11:25 So we have 23 .8 over the total 1 atm.

11:31 But we have to convert this to tor since the saturation vapor pressures in terms of thor.

11:42 So 1 .80m is 760.

11:48 So we have 23 .8 divided by 760 and then multiply the mole fraction in the liquid phase.

11:56 It's one.

11:58 Then we calculate.

12:00 We'll get 0 .03313.

12:04 Now in the bulk air, the mole fraction of water vapor can be determined.