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 #5
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Jun 6, 2014
 #1
avatar+2353 
+5

This image might help;

Now the surface area basically consists of three parts, the top, the bottom and what i'll call the wrapper (I dont know the correct English word). Remember that you could calculate the surface of a circle with the formula  $$\pi r^2$$ where r is the radius.

Therefore if we divide the volume by the height, we'll know the surface of the top of the cylinder.

The height (What is described in your case as length, but this is just a matter of how you hold the cylinder) is 25m

Therefore the surface of the top circle is $$\frac{400m^3}{25m} = 16m^2$$

To find the 'wrapper' we first need to find the radius of the circle.

We know the surface of a circle is $$\pi r^2$$ therefore 

$$\pi r^2 = 16m^2 \Rightarrow r = \sqrt{\frac{16m^2}{\pi}} \approx 5.09m$$

 

Now to calculate the 'wrapper' think of this area as if I wanted to wrap a paper around this part, how much paper would I need?

If I'd wrap a perfectly matching paper around the side of the cylinder it would have the length of the circumference of the cylinder and the height would be the same as that of the cylinder.

 

Therefore, we want to find the circumference (let's call it C) of the circle.

The formula for this is

$$C = 2 \pi r$$

And if we want to know the entire surface of the wrapper (let's call this W) we need to multiply this by the height

$$W = 2 \pi r h$$

Therefore

$$W = 2 \pi \sqrt{\frac{16m^2}{\pi}}25m$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\sqrt{{\frac{{\mathtt{16}}}{{\mathtt{\pi}}}}}}{\mathtt{\,\times\,}}{\mathtt{25}} = {\mathtt{354.490\: \!770\: \!181\: \!103\: \!182\: \!2}}$$

Therefore the surface of the wrapper is approx. 354.5m^2

the surface of the top is 16m^2

the surface of the bottom is the same as the top and therefore also 16m^2

And hence

The surface area of the cylinder is

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{16}}{\mathtt{\,\small\textbf+\,}}{\mathtt{354.49}} = {\mathtt{386.49}}$$

386.49m^2

Jun 6, 2014
 #2
avatar+2353 
+5

What applies to this guy,

http://web2.0calc.com/questions/what-formula-do-i-use-to-find-the-volume-of-a-bead-when-you-onlyhave-the-hight-like-in-the-following-problem

is pretty much what applies to you.

If you're going to copy-paste your entire homework here we're not inclined to help you.

If you genuinly have no idea how to begin, then ask us a more specific question.

For example, you might not have a clue what a PV diagram is. 

I just filled it in in google and the first thing I found was a pressure volume diagram...

Okay, so what's next? 

I want to know how pressure relates to volume right?

let's google 'formula pressure volume'.

First hit; wikipedia- Boyle's law

What do I see?

I see Pressure*Volume = k

where k is a constant.

Hey! if k is a constant, than I can make a graph!.

But how do I calculate k?

Now you have a specific question;

You know this ; 'An engine operating on the diesel cycle with air as the working fluid has a maximum pressure of P = 1.50 MPa and maximum temperature T = 1200 K; the volume at this state is 1.3 L. The compression ratio is 8 and the heat addition is 1.0 kJ. The compression process is polytropic with n = 1.35, while all other processes are ideal.' and want to know how pressure and volume are related.

 

Sorry for beeing so harsh on you, but every now and then we simply are. 

 

reinout 

Jun 6, 2014