The engineering log from a deep-sea survey vessel includes the energy route shown in Fig. 2. Its compact reactor allows the vessel to operate far from fuel ...
Question 1 Report
The engineering log from a deep-sea survey vessel includes the energy route shown in Fig. 2. Its compact reactor allows the vessel to operate far from fuel ports. Inside the reactor, fission releases energy from nuclear fuel. A separate loop carries this energy to a steam unit, which drives a turbine and generator. The generator supplies a motor through a cable at a potential difference of 4.0 kV. The motor turns the propeller and gives the vessel velocity through the water.
During one hour, the reactor releases 8.0 × 108 J of energy. The useful energy delivered to the propeller is 2.0 × 108 J.
(a) What store provides the original energy in the reactor fuel? [1]
(b) Complete the energy pathway for Fig. 2 using suitable energy stores or transfers: nuclear store → ............ → kinetic store of turbine → ............ → kinetic store of propeller. [3]
(c) Explain why the vessel uses a separate coolant loop rather than sending water from the reactor core directly to the turbine. [3]
(d) Calculate the efficiency of the energy transfer from reactor fuel to the propeller. [2]
(e) Give two reasons why a nuclear-powered vessel can be useful on a long survey voyage. [2]
Answer Details
(a) The original energy is in the nuclear energy store of the fuel nuclei. [1]
(b) The complete route is nuclear store → thermal/internal energy store of the coolant or steam → kinetic store of turbine → mechanical/kinetic energy transfer from the turbine → electrical energy transfer/current in the cable → kinetic store of propeller. The hot coolant/steam transfers energy to the turbine; the turning turbine drives the generator, which transfers energy electrically to the motor. [3]
(c) Water in the reactor core may become radioactive. A separate coolant loop prevents this water from reaching the turbine and other machinery, reducing contamination and radiation exposure during maintenance. [3]
(d) Efficiency compares useful output with total input:
\[\text{efficiency}=\frac{\text{useful energy output}}{\text{total energy input}}=\frac{2.0\times10^8\text{ J}}{8.0\times10^8\text{ J}}=0.25\]
Therefore the efficiency is 0.25, or 25%. [2]
(e) Two valid reasons are that a small mass of nuclear fuel releases a very large amount of energy, so refuelling is needed less often, and the vessel can therefore operate far from fuel ports. It also produces no carbon dioxide while operating. [2]
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