Yup.:)

Quote:

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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.

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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

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Originally Posted by Unknown008

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 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.

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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

Quote:

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Originally Posted by Unknown008

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 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.

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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,

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Originally Posted by Nargis786

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.

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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-s...ml#post1795935

Height = 370 m

Gravity = 9.81 m/s^2



and you calculate that on your calculator.

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

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

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2. Use the formula



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



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

Quote:

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Originally Posted by Unknown008

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-s...ml#post1795935

Height = 370 m

Gravity = 9.81 m/s^2



and you calculate that on your calculator.

~~~~~

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


2. Use the formula



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



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

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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...



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

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Originally Posted by Unknown008

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



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

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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%

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Originally Posted by Unknown008

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%

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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 :(

Quote:

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Originally Posted by Nargis786

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%

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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.

Quote:

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Originally Posted by Nargis786

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.

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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.