Physics HL/SL Help

[hedwig]

A typical family home uses a total of about 15 kWh of electrical energy per day and needs to heart about 45 kg (litres) of water from 12^oC to 80^oC for showers etc. If a house is in an area where it can receive, on average, about 400 W m^-2 of solar radiation for about 8 hours per day calculate:

a) The collecting area of the solar water heater if it has an efficiency of 40%;

b) The collecting area of the photovoltaic cells if their efficiency is 18%. (1 kWh = 3.6 x 10⁶J)

(Note: these calculations do not include energy used to heat the house during the winter.)

I'd appreciate it if you could explain step by step how to solve this and mention the formulas you used as well. We didn't have time to cover topic 8 in the school year so we have to do it on our own in the summer. I'm rather lost and the textbook is no help... Thank you!

b) The energy used (and thus required) is 15kWh (15kW per hour) which equals 15(3.6 x 10⁶J) = 54000kJ per hour. Over 1 day (24 hours) that becomes 1296000kJ (1296MJ) of enegry needed per day.

Using the value for the energy we need and the efficiency of the photovoltaic cells (18%), we can find the actual energy needed .

1296 x 10⁶(1/0.18) = 7200 MJ

Using the energy gained per square metre per day from part a) (11520 kJ = 11.52MJ), we can find the area needed.

Area required = energy needed/enegry per unit area

A = 7200000000/11520 = 625000m²

(which, to be completely honest, is ridiculous and impossible and just what the IB would ask...)

So, hope that helped in your understanding!

GL, HF, DD

The photovoltaic cells of efficiency 18% can supply (0.4 x 0.18 x 8) = 0.576 kWh of electrical energy per square metre per day.

Hence the area of cells required to provide 15kWh per day is 15/0.576 = 26 m2.

Thank you so much, The Not-Zetta Slow one! I think I understood just about everything. But isn't 625000 m² a little too much? That's what worries me.

26 m² like Teacher said makes more sense, but I can't understand how you got there. Could you please explain a little more?

Edited August 26, 2012 by hedwig

[HiggsHunter]

The photovoltaic cells of efficiency 18% can supply (0.4 x 0.18 x 8) = 0.576 kWh of electrical energy per square metre per day.

Hence the area of cells required to provide 15kWh per day is 15/0.576 = 26 m2.

Thank you so much, The Not-Zetta Slow one! I think I understood just about everything. But isn't 625000 m² a little too much? That's what worries me.

26 m² like Teacher said makes more sense, but I can't understand how you got there. Could you please explain a little more?

Breaking it down into smaller steps:

During daylight the photovoltaic cells generate an average of (400 x 0.18)W = 72W of electricity per square metre.

Hence with 8 hours of solar radiation they generate (8 x 72)Wh = 576Wh = 0.576kWh of electricity per square metre per day.

Thus to generate 15kWh of electricity per day the total area of photovoltaic cells required is (15/0.576)m2 = 26 m2.

[hedwig]

Breaking it down into smaller steps:

During daylight the photovoltaic cells generate an average of (400 x 0.18)W = 72W of electricity per square metre.

Hence with 8 hours of solar radiation they generate (8 x 72)Wh = 576Wh = 0.576kWh of electricity per square metre per day.

Thus to generate 15kWh of electricity per day the total area of photovoltaic cells required is (15/0.576)m2 = 26 m2.

Thank you very much! That makes sense, I understand it now

[SerUmbras]

Breaking it down into smaller steps:

During daylight the photovoltaic cells generate an average of (400 x 0.18)W = 72W of electricity per square metre.

Hence with 8 hours of solar radiation they generate (8 x 72)Wh = 576Wh = 0.576kWh of electricity per square metre per day.

Thus to generate 15kWh of electricity per day the total area of photovoltaic cells required is (15/0.576)m2 = 26 m2.

Thank you very much! That makes sense, I understand it now

Yeah, sorry about this mis-calc. I guess I was tired and what-not. Anyway, glad to help!

[CkyBlue]

Can someone help with physics nuclear reaction problem?

1 H + 7 Li = 4 He + 4 He

1 3 2 2

Find the energy released. Atomic mass of lithium is 7.016u

I'll try to pick up the pieces here. It's never fun being left out

I'm going to assume it's a nuclear decay question, and that the atomic masses otherwise not specified are the values you put in the equation (I'm looking at my periodic table and I'm finding that the left side actually has a smaller mass than the right side). The change in mass is 0.016u, and this is equivalent to the energy of 0.016 x 931.5MeV =14.9MeV.

[xXRainbowDashXx]

Question:

Person won a bike race; took 14 minutes and 34 seconds

Length of the track: 4.4 km

Mean speed: 18.12 km/h

Is this correct? In which barriers could her mean speed have been?

Can someone explain how to calculate this? Thanks !

[eelnedross]

Can anyone please explain the Huygens principle to me? I can't understand it by any means!

[HiggsHunter]

Can anyone please explain the Huygens principle to me? I can't understand it by any means!

If you have Tsokos' book "Physics for the IB Diploma", you will find a concise explanation of Huygens' Principle on pages 234-236.

[eelnedross]

Can anyone please explain the Huygens principle to me? I can't understand it by any means!

If you have Tsokos' book "Physics for the IB Diploma", you will find a concise explanation of Huygens' Principle on pages 234-236.

I've already checked it, and nope. Still can't comprehend it. What i need is the application of the Huygens Principle for diffraction.

[eelnedross]

I can't even solve it with right hand or left hand rule! I'm so depressed, i'm hopeless in magnetism...

Edited September 10, 2012 by eelnedross

[HiggsHunter]

I can't even solve it with right hand or left hand rule! I'm so depressed, i'm hopeless in magnetism...

With the current directed into the paper the magnetic field lines are circles centred on the wire and the direction of the field is clockwise. So the correct answer is A. (W)

[eelnedross]

I can't even solve it with right hand or left hand rule! I'm so depressed, i'm hopeless in magnetism...

With the current directed into the paper the magnetic field lines are circles centred on the wire and the direction of the field is clockwise. So the correct answer is A. (W)

So the direction of the magnetic field at that point is a tangent to the magnetic field lines?

[HiggsHunter]

I can't even solve it with right hand or left hand rule! I'm so depressed, i'm hopeless in magnetism...

With the current directed into the paper the magnetic field lines are circles centred on the wire and the direction of the field is clockwise. So the correct answer is A. (W)

So the direction of the magnetic field at that point is a tangent to the magnetic field lines?

Yes, that's correct.

[eelnedross]

I can't even solve it with right hand or left hand rule! I'm so depressed, i'm hopeless in magnetism...

With the current directed into the paper the magnetic field lines are circles centred on the wire and the direction of the field is clockwise. So the correct answer is A. (W)

So the direction of the magnetic field at that point is a tangent to the magnetic field lines?

Yes, that's correct.

Thank you Mr. Higgs!

On the other hand, does anyone know how to resolve this question? I don't know how velocity and displacement are related in this question... I only recall the SMH displacement, velocity and acceleration graphs.

[HiggsHunter]

On the other hand, does anyone know how to resolve this question? I don't know how velocity and displacement are related in this question... I only recall the SMH displacement, velocity and acceleration graphs.

Since the wave pulse is travelling to the right and the crest has just passed P, that point is moving downwards (ie, answer B)

[eelnedross]

On the other hand, does anyone know how to resolve this question? I don't know how velocity and displacement are related in this question... I only recall the SMH displacement, velocity and acceleration graphs.

Since the wave pulse is travelling to the right and the crest has just passed P, that point is moving downwards (ie, answer B)

The point P remains stationary or does it move along with the wave pulse?

[HiggsHunter]

On the other hand, does anyone know how to resolve this question? I don't know how velocity and displacement are related in this question... I only recall the SMH displacement, velocity and acceleration graphs.

Since the wave pulse is travelling to the right and the crest has just passed P, that point is moving downwards (ie, answer B)

The point P remains stationary or does it move along with the wave pulse?

The point P is supposed to be a fixed point on the string. So it doesn't move along with the wave pulse, but moves upwards and then downwards as the pulse passes through it.

[eelnedross]

Hey another question:

I was wondering, why are the waves slower in shallow water? Shouldn't it be the opposite as it is a less dense medium? And would the wavefronts be refracted away from the normal?

Thank you!

[Drake Glau]

Hey another question:

I was wondering, why are the waves slower in shallow water? Shouldn't it be the opposite as it is a less dense medium? And would the wavefronts be refracted away from the normal?

Thank you!

I don't know about the shallow water thing but I can tell you that shallower water doesn't make it less or more dense, it's still water

The little lines you have drawn are correct if that's what you were asking and I believe those are towards from the normal. The straight arrow that runs through the waves in region A is the normal and the waves you have slightly drawn in region B redirect towards it.

Edited September 16, 2012 by Drake Glau

[eelnedross]

Hey another question:

I was wondering, why are the waves slower in shallow water? Shouldn't it be the opposite as it is a less dense medium? And would the wavefronts be refracted away from the normal?

Thank you!

I don't know about the shallow water thing but I can tell you that shallower water doesn't make it less or more dense, it's still water

The little lines you have drawn are correct if that's what you were asking and I believe those are away from the normal. The straight arrow that runs through the waves in region A is the normal and the waves you have slightly drawn in region B redirect away from it.

If the waves have less speed in medium B, shouldn't they be refracted towards the normal?

Thanks for answering!