Compute the cloud liquid water content in kg/m3 for the following cloud: A.A mar

Question

Compute the cloud liquid water content in kg/m3 for the following cloud: A.A marine stratocumulus cloud containing 80 droplets/cm3, each with a radius of 12 pm. B.A continental stratocumulus cloud containing 250 droplets/cm3, each with a radius of 7 pm. C.Convert the liquid water content in the marine stratocumulus cloud to g(water)/kg(air). D.If the marine stratocumulus cloud is 200 m deep, how deep would the continental stratocumulus cloud be if it contained exactly the same total liquid water as the marine stratocumulus cloud?

Explanation / Answer

Solution:

Part (A)

A marine stratocumulus cloud contains 80 droplets/cm3

n = 80 droplets/cm3 = (80 droplets/cm3)*(100cm/1m)3 = 80*106 droplets/m3

It means that a marine stratocumulus cloud contain 80*106 droplets in 1 m3 volume of air.

Now radius of each droplet is r = 12 m

r = 12 m = 12*10-6 m

Volume of each droplet is

V = (4/3)**r3

V = (4/3)*(3.1416)*( 12*10-6 m)3

V = 7.2382*10-15 m3

Density of water is = 1000 kg/ m3

= m/V

m = *V

Thus mass of each droplet is,

m = (7.2382*10-15 m3)*( 1000 kg/ m3)

m = 7.2382*10-12 kg

Since each 1 m3 of cloud contains n = 80*106 droplets with mass of each droplet being m = 7.2382*10-12 kg

Thus the water content of marine stratocumulus cloud is m = n*m

m = (80*106 droplets/m3)*( 7.2382*10-12 kg)

m = 5.7906*10-4 kg/m3

Hence the answer is 5.7906*10-4 kg/m3

----------------------------------------------------------------------------------------------------------------------------------------------

Part (B)

A continental stratocumulus cloud contains 250 droplets/cm3

n = 250 droplets/cm3 = (250 droplets/cm3)*(100cm/1m)3 = 250*106 droplets/m3

It means that a continental cloud contain 250*106 droplets in 1 m3 volume of air.

Now radius of each droplet is r = 7 m

r = 7 m = 7*10-6 m

Volume of each droplet is

V = (4/3)**r3

V = (4/3)*(3.1416)*( 7*10-6 m)3

V = 1.4368*10-15 m3

Density of water is = 1000 kg/ m3

= m/V

m = *V

Thus mass of each droplet is,

m = (1.4368*10-15 m3)*( 1000 kg/ m3)

m = 1.4368*10-12 kg

Since each 1 m3 of cloud contains n = 250*106 droplets with mass of each droplet being m = 1.4368*10-12 kg

Thus the water content of continental stratocumulus cloud is n*m

c = (250*106 droplets/m3)*( 1.4368*10-12 kg)

c = 3.5919*10-4 kg/m3

Hence the answer is 3.5919*10-4 kg/m3

--------------------------------------------------------------------------------------------------------------------------------------------------

Part (C)

The density of air is air = 1.225 kg/m3

Thus the mass of 1 m3 of air is 1.225 kg.

Since in the case of marine stratocumulus cloud, 1 m3 of air contained 5.7906*10-4 kg of water droplets, it means that 1.225 kg of air contained 5.7906*10-4 kg of liquid water.

Water content = 5.7906*10-4 kg/m3

Water content = 5.7906*10-4 kg (liquid water) / 1m3 (air)

Water content = 5.7906*10-4 kg (liquid water) / 1.225 kg (air)

Water content = 4.7270*10-4 kg (liquid water) / 1 kg (air)

Water content = 4.7270*10-4 kg*(1000g/1kg) (liquid water) / 1kg (air)

Water content = 0.4727 g (water)/kg (air)

Hence the answer is 0.4727 g(water) / kg(air)

----------------------------------------------------------------------------------------------------------------------------------------------

Part (D)

The height of marine stratocumulus cloud is h = 200 m

We may consider a cylindrical shaped cloud who’s height is h = 200 m and area is A m2

Thus the volume of marine stratocumulus cloud is,

v = h*A

Now the total liquid water content in this marine stratocumulus cloud is given by,

m = m*v

m = (5.7906*10-4 kg/m3)*(v m3) = v*5.7906*10-4 kg = h*A*5.7906*10-4 kg

In the case of continental stratocumulus cloud, let it has same total liquid water as of above and same area A. Then the height H of continental stratocumulus cloud can be given by,

total liquid water in kg in continental stratocumulus cloud = total liquid water in kg in marine stratocumulus cloud

M = m

c*V = m*v

c*H*A = m*h*A

H = m*h/ c

H = (5.7906*10-4 kg/m3)*(200 m)/( 3.5919*10-4 kg/m3)

H = 322.43 m

Thus the height of continental stratocumulus cloud should be 322.43 m.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.