Data:
fresh water supply needed = 2x10^6 m3 per day
distance from antarctica = 10^4 Km
tug towing force = 5x10^6 N
Towing speed = 0.8 m s-1 for iceberg of volume 10^8 m3
total cost to tow iceberg of 10^8 m3 = £18 million

total cost to tow iceberg of 10^9 m3 = £35 million
total running cost of desalination = £1.5 per m3 of fresh water produced
Fuel cost of desalination = 70% of total running costs
cost of electricity = 8p per kW-hour
energy required to melt ice at 0 degrees Celsius - 3.4 x 10^5 J kg-1
Sun's radiation at earth's surface = 600 W m-2
Thickness of icebergs = 250 m
Number of seconds in a year = 3x10^7 s

1. estimate the number of icebergs of each size (10^8 or m3 or 10^9 m3) that need to be delivered to the desert country each year.

2. comment on which iceberg size you would advise for towing

3. compare quauantitively the energy required to melt the 10^8m3 iceberg with the energy required to tow it 10^4km.

4. discuss, with suitable calculations, the choice between allowing the ice to melt in the sun and melting it using electrical energy. Assume that a 250 MW generating station is available nearby.

5. compare the daily costs of desalination with those of electrical melting (with no energy input from the sun).

6. discuss how your comparison to question 5 will be changed if both electrical melting and solar energy melt the ice. What other measures could be employed to reduce the solar/electrical costs?

1. No. of icebergs needed =

First, use

Then, use

2. Use proportions;

If 10^8 m^3 is 18 million pounds, by proportion, you'll have to pay (18/10^8) 0.18 pounds, and for 10^9 m^3, then would be (0.18 * 10^9) giving 180 million pounds.

So, it's cheaper to bring the 10^9 iceberg.

3. Energy to melt = ( 3.4 x 10^5 x 10^8 ) J {3.4 x 10^13}

Energy to tow = work done = force x distance = (5x10^6 x 10^4 x 1000) J {5 x 10^13}

More energy is required for towing than for melting.

Next three coming...

4. Time to melt 10^8 iceberg with sun = Energy required to make iceberg melt/power supplied = (3.4 x 10^13)/600 = 5.67 x 10^10 s = 1776 years (assuming sun shines all 24 hours)

Time to melt 10^8 iceberg with station = (3.4 x 10^13)/(250 x 1000000) = 136000 s = 37.8 hours.

Now you discuss, that's truly easy.

5. Cost of desalination = 1.5 * 10^8 = 1.5 x 10^8 pounds

Cost of electricity = 250 x 1000 x 8p = 2000000p

Now, you can compare.

6. It's obvious that the cost would be reduced if the sun also helps in melting the iceberg. You have to guess about how to reduce costs, think about how to cheaply heat the iceberg. What causes the iceberg to melt slower or faster, etc.

Quote:

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Originally Posted by Unknown008

4. Time to melt 10^8 iceberg with sun = Energy required to make iceberg melt/power supplied = (3.4 x 10^13)/600 = 5.67 x 10^10 s = 1776 years (assuming sun shines all 24 hours)

Time to melt 10^8 iceberg with station = (3.4 x 10^13)/(250 x 1000000) = 136000 s = 37.8 hours.

Now you discuss, that's truly easy.

5. Cost of desalination = 1.5 * 10^8 = 1.5 x 10^8 pounds

Cost of electricity = 250 x 1000 x 8p = 2000000p

Now, you can compare.

6. It's obvious that the cost would be reduced if the sun also helps in melting the iceberg. You have to guess about how to reduce costs, think about how to cheaply heat the iceberg. What causes the iceberg to melt slower or faster, etc.

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thanks I now understand... can you also please help me on the u-values that posted as this is the one I find most difficult

Sorry, the U values, I've not done so yet at school. I'm still on SHM (simple harmonic motion) I'm going to do thermodynamics in the upcoming month.

Quote:

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Originally Posted by Unknown008

Sorry, the U values, I've not done so yet at school. I'm still on SHM (simple harmonic motion) I'm going to do thermodynamics in the upcoming month.

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OK that is fine I'll try working that out some other way but I was wondering can you help me on these:

3. many different energy units are used in industry. Two are the kilowatt hour (k W h) and the tonne-of-coal-equivalent (tce), where 1 tce = 3 x 10^8 J.
explain how many joules are equivalent to 1 k W h.
give the name of 2 other energy units.
why is it important to standardise on a common unit?

4. in the 19th century, coal was the principle fuel used to provide domestic hot water. Sugest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. Answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumtions made in arriving at your estimates.

5. 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/kWh efficiency cost of heat used/kWh cost/h

gas 1.5p 77%

Night storage 3p 100%
electricity

Normal price 7p 100%
electricity

oil 2.4p 68%

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


6. 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%

7. comment on the typical efficiencies of heat engines

:eek: I'll have for the night!!

1. 1 kWh = 1000 Wh

1 W = 1 J/s
1Ws = 1J
1Wh = 1*3600 J = 3600 J
1000 Wh = 3600 * 1000 = 3600000 J

We standardise so that there is no confusion when for example, a country sells energy to another and things like that.

Energy units... hmm... I know of Calories... can't think of another...

I'll do the rest tomorrow... night here and quite late...

Quote:

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Originally Posted by Unknown008

:EEK: I'll have for the night!!!

1. 1 kWh = 1000 Wh

1 W = 1 J/s
1Ws = 1J
1Wh = 1*3600 J = 3600 J
1000 Wh = 3600 * 1000 = 3600000 J

We standardise so that there is no confusion when for example, a country sells energy to another and things like that.

Energy units... hmm...I know of Calories... can't think of another...

---

Thanks lastly can you just help me on:

6. 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%

7. comment on the typical efficiencies of heat engines

Quote:

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Originally Posted by Nargis786

thanks lastly can you just help me on:

6. 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%

7. comment on the typical efficiencies of heat engines

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and I just need help on this aswell:

4. in the 19th century, coal was the principle fuel used to provide domestic hot water. Sugest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. Answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumtions made in arriving at your estimates.

Quote:

---

Originally Posted by Nargis786

and i just need help on this aswell:

4. in the 19th century, coal was the principle fuel used to provide domestic hot water. sugest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumtions made in arriving at your estimates.

---

5. 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/kWh efficiency cost of heat used/kWh cost/h

Gas 1.5p 77%

Night storage 3p 100%
Electricity

Normal price 7p 100%
Electricity

Oil 2.4p 68%

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


By the way I no I am asking quite a lot but I would really appreciate it as my assignment was supposed to be in for last week but I have been struggling. Also I need to hand it in in order to carry on with my 2nd year of college, thank you.

Quote:

---

Originally Posted by Nargis786

and i just need help on this aswell:

4. in the 19th century, coal was the principle fuel used to provide domestic hot water. suggest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumptions made in arriving at your estimates.

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First assuming that the temperature of the water should be 40 degress Celsius
20 people require (10 x 20) litres of hot water, that makes 200 L.
Temperature rise = 40 - 25 = 15 K (assuming ambiant temperature being 25 celcius)
Energy required to heat 200 L of water from 25 to 40 Celsius = 4 x 200 x 15 = 12000 kJ. (since 200 L = 200 kg)

Energy supplied by sun, assuming day is 12 hours, and the mean power is the 0.5 W every hour = 500 x 12 x 3600 = 21600 kJ for each m^2.

Now surface area needed = 12000/21600 = 0.55 m^2

More assupmtions include that no heat is lost, due to evaporation, nor to conduction and convection to surrounding, that there's no rain, no cloud, etc.