| Date | November 2016 | Marks available | 5 | Reference code | 16N.2.HL.TZ0.10 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | Calculate and Explain | Question number | 10 | Adapted from | N/A |
Question
The following data are available for a natural gas power station that has a high efficiency.
| Rate of consumption of natural gas | = 14.6 kg s–1 |
| Specific energy of natural gas | = 55.5 MJ kg–1 |
| Efficiency of electrical power generation | = 59.0 % |
| Mass of CO2 generated per kg of natural gas | = 2.75 kg |
| One year | = 3.16 × 107 s |
Electrical power output is produced by several alternating current (ac) generators which use transformers to deliver energy to the national electricity grid.
The following data are available. Root mean square (rms) values are given.
| ac generator output voltage to a transformer | = 25 kV |
| ac generator output current to a transformer | = 3.9 kA |
| Transformer output voltage to the grid | = 330 kV |
| Transformer efficiency | = 96% |
(i) Calculate the current output by the transformer to the grid. Give your answer to an appropriate number of significant figures.
(ii) Electrical energy is often delivered across large distances at 330 kV. Identify the main advantage of using this very high potential difference.
In an alternating current (ac) generator, a square coil ABCD rotates in a magnetic field.
The ends of the coil are connected to slip rings and brushes. The plane of the coil is shown at the instant when it is parallel to the magnetic field. Only one coil is shown for clarity.
The following data are available.
| Dimensions of the coil | = 8.5 cm×8.5 cm |
| Number of turns on the coil | = 80 |
| Speed of edge AB | = 2.0 ms–1 |
| Uniform magnetic field strength | = 0.34 T |
(i) Explain, with reference to the diagram, how the rotation of the generator produces an electromotive force (emf ) between the brushes.
(ii) Calculate, for the position in the diagram, the magnitude of the instantaneous emf generated by a single wire between A and B of the coil.
(iii) Hence, calculate the total instantaneous peak emf between the brushes.
Markscheme
i
\(I = 0.96 \times \left( {\frac{{25 \times {{10}^3} \times 3.9 \times {{10}^3}}}{{330 \times {{10}^3}}}} \right)\)
Award [2] for a bald correct answer to 2 sf.
Award [1 max] for correct sf if efficiency used in denominator leading to 310 A or if efficiency ignored (e=1) leading to 300 A (from 295 A but 295 would lose both marks).
=280 «A»
Must show two significant figures to gain MP2.
ii
higher V means lower I «for same power»
thermal energy loss depends on I or is ∝I2 or is I2R so thermal energy loss will be less
Accept “heat” or “heat energy” or “Joule heating” for “thermal energy”.
Reference to energy/power dissipation is not enough.
i
«long» sides of coil AB/CD cut lines of flux
OR
flux «linkage» in coil is changed
«Faradays law:» induced emf depends on rate of change of flux linked
OR
rate at which lines are cut
“Induced” is required
Allow OWTTE or defined symbols if “induced emf” is given.
Accept “induced” if mentioned at any stage in the context of emf or accept the term “motional emf”.
Award [2 max] if there is no mention of “induced emf”.
emfs acting in sides AB/CD add / act in same direction around coil
process produces an alternating/sinusoidal emf
ii
Blv = 0.34×8.5×10–2×2 = 0.058 «V»
Accept 0.06V.
iii
160×(c)(ii) = 9.2 or 9.3 or 9.6 «V»
Allow ECF from (c)(ii)
If 80 turns used in cii, give full credit for cii x 2 here.
