Finding the weight of water lost from cloud [Fluid Mechanics]

[leafjerky]

Homework Statement

The rain cloud has an approximate volume of 6.50 mile3and an average height, top to bottom, of 350 ft.1 mile = 5280 ft.
If a cylindrical container 6 ft in diameter collects 2 in. of water after the rain falls out of the cloud, estimate the total weight of rain that fell from the cloud.

Homework Equations

1 mile = 5280 ft
1 cubic foot of water = 62.4 lb
6.5 mile³ = 9.57 * 10¹¹ ft³

The Attempt at a Solution

V_cloud = (pi)(r²)(h)
9.57 * 10¹¹ ft³ = (pi)(r²)(350 ft)
r_cloud = 29498.4 ft

2 in rain fell = 1.67 * 10^-1 ft

V_rain = (pi)(29498.4 ft)²(1.67 * 10^-1 ft)
V_rain = 4.57 * 10⁸ ft³ water

4.57 * 10⁸ ft³ water * 62.4 lb/ ft³ water = 2.85 * 10¹⁰ lb of water.

Does this seem correct? I have a habit of missing mundane details or sometimes the whole idea in general. Thanks for your help. -David

 

[Bystander]

Seems correct. Might be an order of magnitude here or there, but operations are all present.

 

[SteamKing]

Homework Statement

The rain cloud has an approximate volume of 6.50 mile3and an average height, top to bottom, of 350 ft.1 mile = 5280 ft.
If a cylindrical container 6 ft in diameter collects 2 in. of water after the rain falls out of the cloud, estimate the total weight of rain that fell from the cloud.

Homework Equations

1 mile = 5280 ft
1 cubic foot of water = 62.4 lb
6.5 mile³ = 9.57 * 10¹¹ ft³

The Attempt at a Solution

V_cloud = (pi)(r²)(h)
9.57 * 10¹¹ ft³ = (pi)(r²)(350 ft)
r_cloud = 29498.4 ft

2 in rain fell = 1.67 * 10^-1 ft

V_rain = (pi)(29498.4 ft)²(1.67 * 10^-1 ft)
V_rain = 4.57 * 10⁸ ft³ water

4.57 * 10⁸ ft³ water * 62.4 lb/ ft³ water = 2.85 * 10¹⁰ lb of water.

Does this seem correct? I have a habit of missing mundane details or sometimes the whole idea in general. Thanks for your help. -David

You've made the assumption that the cloud is a cylinder as well, which is not supported by the problem statement.

Since you know the total volume of the cloud and the average height, you can find the area of the cloud which gives an equivalent volume of 6.5 mi³

Once you do that, you can find out how many barrels would fit within this area, and knowing the amount of rain water collected in each barrel, calculate the weight of the rain from the cloud.

 

[CWatters]

I think it's even easier than that.

Once you have the area of the cloud just multiply by 2 inches to get volume of rain.
(In the right units obviously).

 

[leafjerky]

You've made the assumption that the cloud is a cylinder as well, which is not supported by the problem statement.

Since you know the total volume of the cloud and the average height, you can find the area of the cloud which gives an equivalent volume of 6.5 mi³

Once you do that, you can find out how many barrels would fit within this area, and knowing the amount of rain water collected in each barrel, calculate the weight of the rain from the cloud.

Okay so this is what I've come up with. The area of the cloud is equal to volume/height = 9.57 * 10^11 ft^3 / 350 ft = 2.73 * 10^9 ft^2

Area of barrel = 28.3 ft^2
Volume water in barrel = 4.72 ft^3
Weight water in barrel = 294 lb
Barrels that fit in area of cloud = 9.65 * 10^7
Total weight = 2.84 * 10^10 lb

Sorry for bad formatting I'm on my phone

 

[SteamKing]

Okay so this is what I've come up with. The area of the cloud is equal to volume/height = 9.57 * 10^11 ft^3 / 350 ft = 2.73 * 10^9 ft^2

Area of barrel = 28.3 ft^2
Volume water in barrel = 4.72 ft^3
Weight water in barrel = 294 lb
Barrels that fit in area of cloud = 9.65 * 10^7
Total weight = 2.84 * 10^10 lb

Sorry for bad formatting I'm on my phone

CWatters idea is even simpler than this.

 

[leafjerky]

CWatters idea is even simpler than this.

So that sounds fine? That was my general idea in my first one I was just assuming the cloud to be a cylinder for some reason. Thanks for the help everyone. I'm still coming up with 2.84 * 10^10 lb so I guess I'll give that a shot.