Check out some similar questions!

Nargis786 · Jun 14, 2009, 05:44 AM

1. One possible design for a storage power station is to use off-peak energy to compress air into a sealed underground canve and then, during periods of high demand, to reverse the system so that the air can drive a generator.
one system of this type compresses air from the atmospheric pressure (10^5 Pa) to 7 MPa in an underground cave of volume 3 x 10^5 m3.

a) calculate the volume of this air before compression

b) sketch a graph of pressure against volume for the air, assuming that the temperature is constant... (I ASSUME THAT THIS WOULD BE A CURVED GRAPH, I WOULD JUST LIKE TO DOUBLE CHECK).

c) the area under the curve is given by the formula:
area = P1 V1 In (P2/P1), where P1, V1 are the pressure and volume respectively before the compression and P2 is the pressure after the compression.
use the formula to estimate the stored energy in the air.

2. batteries of storage cells could also be used. Lead-acid batteries can store about 50 W-h kg-1 of their mass and generate a pek power of 70 W kg-1.

a) claculate the energy stored per kilogram mass of the battery.

3. a third type of storage system pumps water at periods of low demand from an underground reservoir to a lake high in the mountains. At peak demand times the water is allowed to flow back down to the reservoir through a turbine and generating system. The stored water (density kg m-3) drops through a vertical distance of 370m as it does so.

a) estimate the volume of water that is required to flow in order that this station will equal the energy storage capacity of the air system described in question 1.

b) the pipe down which the water flows is 1.6m in diameter. Estimate the speed of the water flow.

Unknown008 · Jun 14, 2009, 06:09 AM

1.a) Use

$P_1V_1=P_2V_2$

b) Yup, got it right. Pressure is inversely proportional to volume. The actual curve has the form of a $y=\frac{a}{x}$ equation.

c) Just substitute the figures you've got. You just got V_1 from a).

Nargis786 · Jun 14, 2009, 06:11 AM

Originally Posted by Unknown008

1.a) Use

$P_1V_1=P_2V_2$

b) Yup, got it right. Pressure is inversely proportional to volume. The actual curve has the form of a $y=\frac{a}{x}$ equation.

c) Just substitute the figures you've got. You just got V_1 from a).

can you please explain the questions furthers as i do not understand with some working out. Thank you

Unknown008 · Jun 14, 2009, 06:15 AM

2. Simple proportions;

50 W in an hour
Time to reach 70 W = $\frac{1}{50} \times 70= 1.4h=5040 s$
If 70 W are stored in 5040 seconds, then the energy is 5040 * 70 = 352800 J = 352.8kJ

Unknown008 · Jun 14, 2009, 06:18 AM

Originally Posted by Nargis786

can you please explain the questions furthers as i do not understand with some working out. thank you

Oops, ok;

$P_1V_1 = P_2V_2$ , P_1 is the initial pressure, V_1 the initial volume, P_2 the final pressure and V_2 the final volume.

Substituting the values you've got, you have;

$(10^5)V_1 = (5\times10^6)(3\times10^5)$

$V_1 = \frac{(5\times10^6)(3\times10^5)}{(10^5)}$

Nargis786 · Jun 14, 2009, 06:28 AM

Originally Posted by Unknown008

Oops, ok;

$P_1V_1 = P_2V_2$ , P_1 is the initial pressure, V_1 the initial volume, P_2 the final pressure and V_2 the final volume.

Substituting the values you've got, you have;

$(10^5)V_1 = (5\times10^6)(3\times10^5)$

$V_1 = \frac{(5\times10^6)(3\times10^5)}{(10^5)}$

sorry I'm kind of confused :confused: the answer you just gave it it for 1.a. can you laso help me with 1.c and 2 and 3. i would appreciate it. Thank you. :)

Unknown008 · Jun 14, 2009, 06:36 AM

I'm typing for 3... 1 c) is easy, just plug in the numbers!

area (energy) = P1 V1 In (P2/P1)

area (energy) = (10^5)(1.5 x 10^ 7) In ((7000000)/(10^5))

2. I have already posted the answer and the how to do it.

Since this includes only a kilogram everywhere, you can ignore the kilogram. It stores 50 W in an hour (3600 seconds). It'll store 70 W in 5040 seconds (its max capacity). Since power = energy / time, you get energy by multiplying power with time.

Unknown008 · Jun 14, 2009, 06:41 AM

3. a)You need the density of water... 1000 kg/m^3 (roughly)

With that, use the formula for gravitational potential energy to find the mass of water required.

$E_p = mgh$

$m=\frac{E_p}{gh}$

E_p is the energy you got in question 1, g is the acceleration due to free fall, 9.81 m/s^2, and h is the height, 370m.

When you get the mass, find the volume, by dividing the mass you obtained by the density of water.

I'll post for b later. Have to go... :( Will be here in half an hour or an hour.

Nargis786 · Jun 14, 2009, 06:47 AM

Originally Posted by Unknown008

I'm typing for 3... 1 c) is easy, just plug in the numbers!

area (energy) = P1 V1 In (P2/P1)

area (energy) = (10^5)(1.5 x 10^ 7) In ((7000000)/(10^5))

2. I have already posted the answer and the how to do it.

Since this includes only a kilogram everywhere, you can ignore the kilogram. It stores 50 W in an hour (3600 seconds). It'll store 70 W in 5040 seconds (its max capacity). Since power = energy / time, you get energy by multiplying power with time.

Sorry its very confusing... can you explain 1 and 3 again but seperatly so I can understand it. Thanks

Nargis786 · Jun 14, 2009, 06:49 AM

Originally Posted by Nargis786

sorry its very confusing.... can you explain 1 and 3 again but seperatly so i can understand it. thanks

Don't worry I got the answers abit late... you helped me a lot... I will just need the answer to
3.b when you come back :-)

Unknown008 · Jun 14, 2009, 07:10 AM

Ok, I'm refreshed, and can think clearer. Typing is exhausting! :p

Ok, you have the volume of water, yes? You have the area of the cross section of the pipe and so, you can find the 'height' of water in the pipe.

Now, you can use the formula for kinetic energy,

$E_k = \frac{1}{2} mv^2$

You have the energy, since all the potential energy is (considered to be) converted to kinetic energy . The mass also, you have. You can find v, the speed of water falling.

Now, this speed is for a 'height' of water. So, to find the volume of water entering per second, multiply by the area of the cross section of the pipe.

Voilà!

Nargis786 · Jun 14, 2009, 07:36 AM

Originally Posted by Unknown008

Ok, I'm refreshed, and can think clearer. Typing is exhausting! :p

Ok, you have the volume of water, yes? You have the area of the cross section of the pipe and so, you can find the 'height' of water in the pipe.

Now, you can use the formula for kinetic energy,

$E_k = \frac{1}{2} mv^2$

You have the energy, since all the potential energy is (considered to be) converted to kinetic energy . The mass also, you have. You can find v, the speed of water falling.

Now, this speed is for a 'height' of water. So, to find the volume of water entering per second, multiply by the area of the cross section of the pipe.

Voila!

sorry can you show me how to do this I'm kind of confused. Sorry about all this

Unknown008 · Jun 14, 2009, 08:10 AM

Ok, no prob

$E_k = \frac{1}{2} mv^2$

Energy is 6.37 x 10 ^12 J
Mass is 1.76 x 10^ 9 kg
v^2 = ((6.37 x 10^12) * 2)/1.76 x 10^9 = 7259.4
v = 85.2 m/s

Now, rate of flow of water = 85.2 * 1.6 = 136.3 m^3 /s

Voilà!

Nargis786 · Jun 14, 2009, 08:18 AM

Originally Posted by Unknown008

Ok, no prob

$E_k = \frac{1}{2} mv^2$

Energy is 6.37 x 10 ^12 J
Mass is 1.76 x 10^ 9 kg
v^2 = ((6.37 x 10^12) * 2)/1.76 x 10^9 = 7259.4
v = 85.2 m/s

Now, rate of flow of water = 85.2 * 1.6 = 136.3 m^3 /s

Voila!

thank you for this, i can understand it now. I have also posted some stuff on u-values and more stuff on energy and latent heat. I was wondering if you could take a look at them as well and help me out with them. Thank you

Unknown008 · Jun 14, 2009, 08:19 AM

Yup, sure. Could I ask you which level of Physics you're doing? O level, or A level? Your questions seem to be increasing in difficulty...

Nargis786 · Jun 14, 2009, 08:27 AM

Originally Posted by Unknown008

Yup, sure. Could I ask you which level of Physics you're doing? O level, or A level? Your questions seem to be increasing in difficulty...

Yep the questions are very hard as my teacher also pointed out... I am doing a levels in college where I am studying my 1st year of applied science, biology, chemistry and physics. I have no problems with the other subjects but I just struggle with physics as it has a lot of maths in it.

Unknown008 · Jun 14, 2009, 08:33 AM

Ah?? I'm doing A level too! :p I do French instead of Biology though... I'll answer our questions in a few. I'm playing chess against a good friend of mine.

Nargis786 · Jun 14, 2009, 08:40 AM

Originally Posted by Unknown008

Ah??? I'm doing A level too! :p I do French instead of Biology though... I'll answer our questions in a few. I'm playing chess against a good friend of mine.

OK thank you

Nargis786 · Jun 14, 2009, 09:16 AM

Originally Posted by Unknown008

Oops, ok;

$P_1V_1 = P_2V_2$ , P_1 is the initial pressure, V_1 the initial volume, P_2 the final pressure and V_2 the final volume.

Substituting the values you've got, you have;

$(10^5)V_1 = (5\times10^6)(3\times10^5)$

$V_1 = \frac{(5\times10^6)(3\times10^5)}{(10^5)}$

i would just like to double check on this one again...
you said V1 = (5x10^6) (3x10^5) divide by 10^5
so does that mean the answer will be 15 x 10^11 divide by 10^5 which equals to 15 10^6. can you let me no if this is right please.

Nargis786 · Jun 14, 2009, 09:17 AM

Originally Posted by Nargis786

i would just like to double check on this one again....
you said V1 = (5x10^6) (3x10^5) divide by 10^5
so does that mean the answer will be 15 x 10^11 divide by 10^5 which equals to 15 10^6. can you let me no if this is right please.

I would just like to double check on this one again...
you said V1 = (5x10^6) (3x10^5) divide by 10^5
so does that mean the answer will be 15 x 10^11 divide by 10^5 which equals to 15x10^6. can you let me no if this is right please.