Questions on heat engines and their efficiency

Question 1

State the energy transfers that take place in one cycle of a heat engine, and write an equation for the work done per cycle.

Accepted Answer

Each cycle the working substance absorbs heat Q_H from a hot source, converts part of it to useful work W, and rejects the rest, Q_C, to a cold sink (1). Energy is conserved, so W = Q_H − Q_C (1).

Question 2

Explain why no heat engine can be 100% efficient, even if all friction could be removed.

Accepted Answer

The working substance must be returned to its starting state to repeat the cycle, which means compressing it, and that requires heat Q_C to be rejected to a colder sink (1). Since Q_C can never be zero, W = Q_H − Q_C is always less than Q_H, so the efficiency is always below 1 (1).

Question 3

Write the equation for the maximum theoretical efficiency of a heat engine, and state the temperature scale that must be used in it.

Accepted Answer

Maximum theoretical efficiency = (T_H − T_C)/T_H (1), where T_H and T_C are the source and sink temperatures in kelvin (1).

Question 4

Define the coefficient of performance of a refrigerator and of a heat pump.

Accepted Answer

Refrigerator: COP = Q_C/W, the heat removed from the cold space per unit of work input (1). Heat pump: COP = Q_H/W, the heat delivered to the hot space per unit of work input (1).

Question 5

State how the net work done per cycle by an engine is found from its loop on a p-V diagram, and state the direction in which an engine traverses the loop.

Accepted Answer

The net work per cycle is the area enclosed by the loop (1), and an engine goes round it clockwise, expanding at high pressure and compressing at low pressure so that more work comes out than goes back in (1).

Question 6

State how the input power of an engine is calculated from its fuel.

Accepted Answer

Input power = calorific value of the fuel × the rate at which fuel is burned (1), in J kg⁻¹ × kg s⁻¹ to give watts.

Question 7

A refrigerator is a heat engine cycle driven in reverse. State the direction in which it is traversed on a p-V diagram, and name the two energy transfers that together make up the heat delivered to the hot side each cycle.

Accepted Answer

It is traversed anticlockwise, work being supplied to the working substance rather than taken from it (1). The hot side receives the heat Q_C extracted from the cold space together with the work W done on the working substance, so Q_H = Q_C + W (1).

Question 8

In each cycle an engine absorbs 1500 J of heat from its source and rejects 1050 J to its sink. Calculate the work done per cycle and the efficiency of the engine.

Accepted Answer

W = Q_H − Q_C = 1500 − 1050 = 450 J (1). Efficiency = W/Q_H (1) = 450/1500 = 0.30, that is 30% (1).

Question 9

An engine takes in heat from a source at 520 °C and rejects heat to a sink at 30 °C. Calculate its maximum theoretical efficiency.

Accepted Answer

Convert to kelvin first: T_H = 520 + 273 = 793 K and T_C = 30 + 273 = 303 K (1). Maximum efficiency = (T_H − T_C)/T_H (1) = (793 − 303)/793 = 0.62, about 62% (1).

Question 10

A four-cylinder engine runs at 30 cycles per second and the p-V loop for each cylinder encloses an area of 260 J. Calculate the indicated power of the engine.

Accepted Answer

Indicated power = area of loop × cycles per second × number of cylinders (1) = 260 × 30 × 4 (1) = 31200 W, that is 31.2 kW (1).

Question 11

The crankshaft of an engine delivers a torque of 145 N m at 3600 revolutions per minute. Calculate the brake power.

Accepted Answer

ω = 3600 × 2π/60 = 377 rad s⁻¹ (1). Brake power = Tω (1) = 145 × 377 = 5.47 × 10⁴ W, about 55 kW (1).

Question 12

An engine burns fuel of calorific value 43 MJ kg⁻¹ at a rate of 3.5 × 10⁻³ kg s⁻¹ and delivers a brake power of 52 kW. Calculate the input power and the overall efficiency.

Accepted Answer

Input power = calorific value × fuel flow rate = 43 × 10⁶ × 3.5 × 10⁻³ (1) = 1.5 × 10⁵ W, about 150 kW (1). Overall efficiency = brake power/input power = 52/150 = 0.35 (1).

Question 13

In each cycle a refrigerator extracts 250 J of heat from its cold compartment using 100 J of work. Calculate its coefficient of performance and the heat rejected into the kitchen per cycle.

Accepted Answer

COP_ref = Q_C/W (1) = 250/100 = 2.5 (1). Q_H = Q_C + W = 250 + 100 = 350 J (1).

Question 14

A heat pump with a coefficient of performance of 3.6 delivers heat to a house at 7.2 kW. Calculate the electrical power it consumes and the rate at which it extracts heat from the outside air.

Accepted Answer

COP_hp = Q_H/W, so W = Q_H/COP (1) = 7.2/3.6 = 2.0 kW (1). Q_C = Q_H − W = 7.2 − 2.0 = 5.2 kW drawn from the cold outdoor air (1).

Question 15

In each cycle a machine extracts 480 J of heat from a cold space and rejects 640 J into the room. Calculate its coefficient of performance judged as a refrigerator and judged as a heat pump, and show that the two differ by exactly 1.

Accepted Answer

W = Q_H − Q_C = 640 − 480 = 160 J (1). COP_ref = Q_C/W = 480/160 = 3.0 (1). COP_hp = Q_H/W = 640/160 = 4.0, which is greater by 1.0, because Q_H exceeds Q_C by exactly W and W/W = 1 (1).

Question 16

A combined heat and power station is supplied with 120 MW of energy by the fuel it burns. It generates 42 MW of electricity and delivers 54 MW of heat as hot water to the buildings around it. Calculate the electrical efficiency of the station, the fraction of the fuel's energy that is put to use, and the rate at which energy is still wasted.

Accepted Answer

Electrical efficiency = 42/120 = 0.35 (1). Energy put to use = 42 + 54 = 96 MW (1). Useful fraction = 96/120 = 0.80 (1). Wasted = 120 − 96 = 24 MW (1), rejected at too low a temperature to be worth piping. The second figure is an energy utilisation rather than an efficiency, since it adds work to low-grade heat.

Question 17

A six-cylinder engine burns fuel of calorific value 45 MJ kg⁻¹ at 4.0 × 10⁻³ kg s⁻¹. Each cylinder's p-V loop encloses 480 J and the engine completes 25 cycles per second. The output shaft turns at 2900 revolutions per minute and delivers a torque of 210 N m. Calculate the input power, the indicated power, the brake power, the friction power and the overall efficiency.

Accepted Answer

Input power = 45 × 10⁶ × 4.0 × 10⁻³ = 180 kW (1). Indicated power = 480 × 25 × 6 = 72 kW (1). ω = 2900 × 2π/60 = 304 rad s⁻¹ (1), so brake power = Tω = 210 × 304 = 63.8 kW (1). Friction power = indicated − brake = 72 − 63.8 = 8.2 kW (1). Overall efficiency = brake/input = 63.8/180 = 0.35 (1).

Question 18

In each cycle an engine absorbs 2400 J of heat from a source at 900 K and rejects 1560 J to a sink at 320 K, completing 20 cycles per second. Calculate the work per cycle, the output power, the actual efficiency and the maximum theoretical efficiency, and comment on the comparison.

Accepted Answer

W = Q_H − Q_C = 2400 − 1560 = 840 J per cycle (1). Power = 840 × 20 = 1.68 × 10⁴ W, that is 16.8 kW (1). Efficiency = W/Q_H = 840/2400 = 0.35 (1). Maximum = (900 − 320)/900 = 0.64 (1), the temperatures already being in kelvin. The engine reaches 0.35 against a ceiling of 0.64, a little over half of what is theoretically available; the shortfall is friction, turbulence and heat leaking out of the cylinder, none of which the ideal ceiling allows for (1).

Question 19

A householder can heat a room either with a 3.0 kW electric heater or with a heat pump of coefficient of performance 4.0 drawing the same 3.0 kW of electrical power. Calculate the heating power delivered by each, and explain why the heat pump does not break conservation of energy.

Accepted Answer

Electric heater: essentially all the electrical power becomes heat, so it delivers 3.0 kW (1). Heat pump: Q_H = COP × W = 4.0 × 3.0 = 12 kW (1), four times as much. Conservation holds because 9 kW of that is heat extracted from the cold outdoor air and merely moved indoors; only 3.0 kW is created from electrical work (1). The pump transfers heat rather than generating it, which is why a COP above 1 is normal (1). Its COP falls on a colder day, because the temperature gap it has to pump across is then wider.

Question 20

A refrigerator has a coefficient of performance of 2.4 and its compressor draws 90 W of electrical power. Calculate the rate at which heat is removed from the cold compartment and the rate at which heat is delivered to the kitchen, and explain why running the refrigerator with its door left open warms the kitchen rather than cooling it.

Accepted Answer

COP_ref = Q_C/W, so Q_C = 2.4 × 90 = 216 W (1). Q_H = Q_C + W = 216 + 90 = 306 W delivered to the kitchen (1). With the door open the heat removed from the compartment simply returns to the kitchen, so the room receives 306 W while losing 216 W (1), a net gain of 90 W, exactly the electrical work the compressor is doing, which ends up as heat (1).

Question 21

A power station is converted to combined heat and power: instead of rejecting heat into a river at 300 K it rejects into water at 360 K, which is piped to heat nearby buildings. Explain why the electrical output falls, why the share of the fuel's energy doing something wanted nevertheless rises, and why the plant's quoted 80% figure does not break the second law of thermodynamics.

Accepted Answer

Raising the sink temperature T_C lowers the ceiling on efficiency, (T_H − T_C)/T_H, so the plant generates less electricity from the same fuel (1). What that sacrifice buys is rejected heat leaving at a temperature high enough to be useful: it warms buildings instead of a river (1), so the share of the fuel's energy that ends up wanted climbs from around forty per cent towards eighty (1). The second law caps only the fraction of Q_H that can become work; using Q_C as heat is not turning it into work, so no cap applies: the 80% is an energy utilisation rather than an efficiency (1).

Question 22

A station's boiler supplies steam at 810 K. Working as a condensing station it rejects heat to a river at 290 K; converted to combined heat and power it rejects instead into water at 363 K, which is piped to heat nearby buildings. The fuel supplies 150 MW to the boiler. Calculate the maximum theoretical efficiency in each case, determine the electrical output given up in the conversion if the station works at its theoretical ceiling, and state what is gained in exchange.

Accepted Answer

Condensing: maximum efficiency = (810 − 290)/810 = 0.64 (1). Combined heat and power: (810 − 363)/810 = 0.55 (1), both temperatures already being in kelvin. Electrical output given up = (0.64 − 0.55) × 150 = 13.5 MW (1). In exchange the rejected heat leaves hot enough to be piped and used to warm buildings instead of a river, so a far larger share of the fuel's energy does something wanted (1).

Question 23

A four-stroke petrol engine and a four-stroke diesel engine each complete one cycle in four strokes of the piston. Name the four strokes in order and state what happens to the gas in the cylinder during each, and state where each stroke appears on the theoretical p-V cycle of the engine. Your answer should make clear how the petrol and the diesel cycle differ in the way the heat Q_H enters.

Accepted Answer

Induction: the inlet valve opens and the descending piston draws the charge in at about atmospheric pressure, which on the theoretical cycle is the outward line along the bottom, the volume rising at constant low pressure (1). Compression: both valves shut and the rising piston squeezes the charge into a small volume, raising its pressure and temperature with too little time for heat to escape, which is the adiabatic compression curve (1). Power: the charge burns and the pressure jumps, and the hot gas drives the piston down, expanding and cooling as the crankshaft takes the work away; on the cycle this is the heat input followed by the adiabatic expansion back out to the largest volume (1). Exhaust: the exhaust valve opens and the pressure drops to atmospheric while the volume has no time to change, drawn as the fall in pressure at constant maximum volume where Q_C leaves, and the rising piston then pushes the burnt gas out along the bottom line (1). Petrol: a mixture of fuel and air is drawn in and compressed about tenfold, and a spark ignites it at the top of the compression stroke; the burn is over before the piston has moved appreciably, so Q_H enters at constant volume, a vertical rise in pressure at the smallest volume (1). Diesel: air alone is drawn in and compressed fifteen to twenty times, hot enough to ignite the fuel sprayed in at the top of the stroke, and that fuel burns as fast as it arrives while the piston is already descending, so Q_H enters at constant pressure, a horizontal line across the top of the cycle (1). Four strokes are two revolutions of the crankshaft, and only one of them is a power stroke.