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.

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).

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

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

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)}

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. :)

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.

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.

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

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 :-)

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!

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

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!

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

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...

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.

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.

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

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.

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.

Yup.:)

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.

you wrote:
area (energy) = P1 V1 In (P2/P1)

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

How do I work out the final answer from this. I put all the numbers in the formula but I am struggling to get the answer

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.

i need help on this one... im getting mixed up with the formula and the numbers... i need help in using the formula and finding out the answer

The stored energy is in that formula.

first, 7000000 / 10^5 = 70

(10^5) * (1.5 x 10^7) = 1.5 x 10^12

1.5 x 10^12 * ln 70 = 1.5 x 10^12 * 4.248... = 6.37 x 10^12

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.

can you please help me on this one... im confused with the formula and don't no how to use it to get the answer.
also can you help me with these following questions:

1. 100% = 1 = 1 - low temp/400k
what would the low temp be to get 100% efficiency.

2. a power station has an electrical output of 1200 MW. Its actual efficiency is 36%.
a) how much heat is provided by the burning coal each second?
b) how muc heat is wasted by the power station each second?

Aww... you're still stuck with this one? You really need spoon feeding! I wonder how you did pass your O level??

I'll take the question, again,


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.

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

I already posted the answer to the mass and energy here:
https://www.askmehelpdesk.com/math-sciences/energy-power-latent-heat-data-analysis-power-storage-systems-364779.html#post1795935

Height = 370 m

Gravity = 9.81 m/s^2

Mass = \frac{E_p}{gh} = \frac{6.37 \times 10 ^{12}}{(370)(9.81)}

and you calculate that on your calculator.

~~~~~

1. For the first question, I don't quite understand what you mean..

100% = 1 = 1 - low temp/400k

2. Use the formula

Efficiency = \frac{Output}{Input}

Since the efficiency is 36%, and the output is 1200 MW, the input power is 3373.3 MW.

\frac{Output}{Efficiency}= Input

Now, energy = power x time. So heat energy supplied = 3373.3 J/s

If the efficiency is 36%, therefore, only 36% is used. The rest is lost, that is 64%. What is 64% of the total energy, that's (64% x 3373.3) 2173.3 J/s

Aww... you're still stuck with this one? You really need spoon feeding! I wonder how you did pass your O level???

I'll take the question, again,



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

I already posted the answer to the mass and energy here:
https://www.askmehelpdesk.com/math-sciences/energy-power-latent-heat-data-analysis-power-storage-systems-364779.html#post1795935

Height = 370 m

Gravity = 9.81 m/s^2

Mass = \frac{E_p}{gh} = \frac{6.37 \times 10 ^{12}}{(370)(9.81)}

and you calculate that on your calculator.

~~~~~

1. For the first question, I dont' quite understand what you mean..


2. Use the formula

Efficiency = \frac{Output}{Input}

Since the efficiency is 36%, and the output is 1200 MW, the input power is 3373.3 MW.

\frac{Output}{Efficiency}= Input

Now, energy = power x time. So heat energy supplied = 3373.3 J/s

If the efficiency is 36%, therefore, only 36% is used. The rest is lost, that is 64%. What is 64% of the total energy, that's (64% x 3373.3) 2173.3 J/s


thanks that helped a lot. But for question 1 i am asking what would the low temp have to be in order to get 100% efficiency. The high temp is already provided which is 400

I still don't get what you're looking for...

Efficiency=\frac{output}{Input}

If output = input, then efficiency = 100%, since 100% = 1

I still don't get what you're looking for...

Efficiency=\frac{output}{Input}

If output = input, then efficiency = 100%, since 100% = 1

i already no what the efficiency is as it is 100%. To get 100% i will divide low temp with high temp as shown below.

100% = low temp/400

(400 is the high temp, as i already no what the high temp is, i now need to no what the low temp would be, so if i divide the low temp by 400 i would get the 100% efficiency).

what would the low temp be?

But then, use the equation I gave you. I gave it to you , for that's all you need to solve that problem. That means that the low temperature has to be 400, so that 400/400 = 1 = 100%

But then, use the equation I gave you. I gave it to you , for that's all you need to solve that problem. That means that the low temperature has to be 400, so that 400/400 = 1 = 100%

Oops sorry my mistake I misunderstood the question, I get it now. Lastly I need help on 2 more questions:

1. table 4.1 lists the eficiencies of a variety of energy conversion devices. Regroup the devices according to the energy conversions which they perform. Group together hose which convert fuel (chemical energy) to heat, those which convert between mechanical and electrical energy, and those which convert heat to mechanical energy (heat engines).

Table 4.1 - energy conversion efficiencies

Device

Power station boiler - 90%
Hydroelectric turbine - 90%
Large electric motor - 90%
Large electric generator - 90%
Domestic gas fired boiler - 75%
Washing machine motor - 70%
Domestic ocal fired boiler - 60%
Steam turbine (power station) - 45%
Diesel engine - 40%
Car (petrol) engine - 30%
Steam locomotive - 10%

a)comment on the typical efficiencies of heat engines


2. the following figures were gven by a central heating consultant to a customer who required a central heating system capable of supplying 25 kW.

Supplied cost/kW h Efficiency Cost of heat used/kW h Cost/h

gas 1.5p 77%

Night storage 3p 100%
electricity

Normal price 7p 100%
electricity

Oil 2.4p 68%


Find the cost of each kWh of heat and the cost of runnig each system for an hour.

I'll take them tomorrow. I'm really tired now... Sorry :(

1. table 4.1 lists the eficiencies of a variety of energy conversion devices. regroup the devices according to the energy conversions which they perform. group together hose which convert fuel (chemical energy) to heat, those which convert between mechanical and electrical energy, and those which convert heat to mechanical energy (heat engines).

table 4.1 - energy conversion efficiencies

device

power station boiler - 90%
hydroelectric turbine - 90%
large electric motor - 90%
large electric generator - 90%
domestic gas fired boiler - 75%
washing machine motor - 70%
domestic ocal fired boiler - 60%
steam turbine (power station) - 45%
diesel engine - 40%
car (petrol) engine - 30%
steam locomotive - 10%

Now now, have you ever seen these devices? You should know what energy changes are involved.

Power station boiler - 90% - from chemical (fuel) to heat (to boil)
Hydroelectric turbine - 90% - from mechanical (water moving) to electrical (hydroelectric power station)
Large electric motor - 90% - motors convert electrical to various other forms, but mainly mechanical (motor in a toy, etc)
Large electric generator - 90% - makes electricity, from mechanical, as most generators do.
Domestic gas fired boiler - 75% -
Washing machine motor - 70%
Domestic ocal fired boiler - 60%
Steam turbine (power station) - 45%
Diesel engine - 40%
Car (petrol) engine - 30%
Steam locomotive - 10%

Can you try the others out?
Engine/motor - convert electricity or chemical to mechanical mainly
Generators/turbines - generates electricity, mostly from mechanical.
Boilers - to boil means to heat... therefore from fuel (sometimes electricity) to heat.

You'll have to put your own comments, hint; view all from their groups.

2. the following figures were gven by a central heating consultant to a customer who required a central heating system capable of supplying 25 kW.

Supplied cost/kW h Efficiency Cost of heat used/kW h Cost/h

gas 1.5p 77%

Night storage 3p 100%
electricity

Normal price 7p 100%
electricity

oil 2.4p 68%


find the cost of each kWh of heat and the cost of runnig each system for an hour.

To find the cost of heat used, multiply the supplied cost by the efficiency.

Find the energy required for an hour, that makes 25kW x 3600 = 90000 kWh

Divide that by the supplied cost to have the cost.