Let M represent a certain mass of coal which we assume will...

Let $M$ represent a certain mass of coal which we assume will deliver 100 joules of heat when burned - whether in a house, delivered to the radiators or in a power plant, delivered at $1000^{\circ} \mathrm{C}$. Assume the plant is ideal (no waste in turbines or generators) discharging its heat at $30^{\circ} \mathrm{C}$ to a river. How much heat will $M$, burned at the plant to generate electricity, provide for the house when the electricity is: (a) delivered to residential resistance-heating radiators? (b) delivered to a residential heat pump (again assumed ideal) boosting heat from a reservoir at $0^{\circ} \mathrm{C}$ into a hot-air system at $30^{\circ} \mathrm{C}$ ?

Let $M$ represent a certain mass of coal which we assume will deliver 100 joules of heat when burned - whether in a house, delivered to the radiators or in a power plant, delivered at $1000^{\circ} \mathrm{C}$. Assume the plant is ideal (no waste in turbines or generators) discharging its heat at $30^{\circ} \mathrm{C}$ to a river. How much heat will $M$, burned at the plant to generate electricity, provide for the house when the electricity is:
(a) delivered to residential resistance-heating radiators?
(b) delivered to a residential heat pump (again assumed ideal) boosting heat from a reservoir at $0^{\circ} \mathrm{C}$ into a hot-air system at $30^{\circ} \mathrm{C}$ ?

Key Concepts

Carnot Efficiency

Carnot efficiency is the maximum possible efficiency with which a heat engine can convert heat from a high?temperature reservoir into work while expelling some waste heat to a low?temperature reservoir. It is determined solely by the absolute temperatures of the two reservoirs and represents an ideal limit that real engines can approach but never reach.

Heat Engine

A heat engine is a system that converts heat energy into mechanical work by exploiting the temperature differential between a hot and a cold reservoir. The efficiency of a heat engine is limited by the Carnot efficiency, meaning not all the heat energy can be converted into useful work; the rest is rejected to the colder reservoir.

Electrical Resistance Heating

Electrical resistance heating is a method where electrical energy is converted directly into heat via resistors. This process is nearly 100% efficient in terms of energy conversion, as virtually all the electrical energy is transformed into thermal energy, making it a useful benchmark for comparing other heating methods.

Heat Pump and Coefficient of Performance (COP)

A heat pump transfers heat from a lower temperature source to a higher temperature sink by using external work. Its effectiveness is measured by the Coefficient of Performance (COP), which is the ratio of the heat delivered to the work input. An ideal heat pump can deliver several times more heat energy than the electrical energy it consumes, making it more efficient than simple resistive heating in many scenarios.

Transcript

00:02 So to work out the maximum possible efficiency of this power plant, we need the equation for a kano engine, for the efficiency of kano engine, which is 1 minus the cold temperature divided by the hot temperature.

00:21 Now, these both have to be in kelvin, so we need to convert the temperatures given to us in degrees into kelvin.

00:29 We do this by adding 273 to the value.

00:33 So our cold temperature is 30 degrees.

00:36 Add 273 we get 303 kelvin.

00:42 And the hot temperature is 280 degrees.

00:45 So again, adding 273 we get 533 kelvin.

00:53 And this comes out to 0 .45, which is 45%.

01:02 This is the maximum possible efficiency for part a.

01:10 This is, we're giving in the question, the actual efficiency is 29%, so this is quite significantly lower than the real efficiency.

01:21 So part b to work out the waste heat discharge, we're told that 880 megawatts is supplied by the plant of useful energy.

01:35 And we know that it's at 29 % efficiency.

01:39 So 880 megawatts is 29 % of the total energy that's been extracted from the hot reservoir, not the steam.

01:55 So if we divide 880 by 29, we will get to work out the total energy extracted.

02:08 This is to work out 1 % of the energy that's been extracted, if you like...

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