Entropy generated by heat transfer

Heat flowing from a hot body to a cold one is the most familiar irreversible process in a plant, and the most common. It happens in every boiler, heat exchanger, cooking kettle, cool-room coil and uninsulated pipe. The heat arrives intact; the first law is satisfied. But the heat that left at a high temperature arrives at a lower one, and in doing so it loses part of its ability to do work, permanently.

This chapter calculates that loss. It starts with heat passing between two bodies at fixed temperatures, applies the result to the running case’s cooking kettle and effluent heat recovery, and shows how the temperature difference, the flow arrangement and the order in which heat sources are used decide how much entropy is generated. It ends with a heat exchanger plumbed the wrong way round.

Heat across a temperature difference

Suppose heat Q flows from a body at T_h to one at T_c, both large enough that their temperatures do not change. The hot body’s entropy falls by Q ÷ T_h and the cold body’s rises by Q ÷ T_c. Because T_c is lower, the rise is larger than the fall, and the difference is entropy generated by the transfer.

Figure 1Entropy generated when heat crosses a temperature difference

S_gen = Q × (1 ÷ T_c − 1 ÷ T_h) ≈ Q × ΔT ÷ T²

Q

heat transferred, kW

T_h, T_c

absolute temperatures of the hot and cold bodies, K

S_gen

entropy generated, kW/K

1. Kettle, 45.0 kW from steam at 375.68 K to food at 363.15 K: 45.0 × (1 ÷ 363.15 − 1 ÷ 375.68) = 0.00413 kW/K
2. Lost work, T₀ × S_gen = 293.15 × 0.00413 = 1.21 kW

Two features of the formula guide design. The entropy generated is proportional to the heat and to the temperature difference, so halving the difference roughly halves the loss for the same heat. And it is inversely proportional to the square of the temperature, so a given temperature difference costs far more at low temperatures than at high ones. A 10 K difference at 300 K generates about four times as much entropy per kilowatt as the same difference at 600 K. This is why refrigeration plant, which works near and below ambient, is so sensitive to coil temperature differences, as Chapter 3 found.

The running case: the cooking kettle

This is an illustration. The kettle’s jacket condenses 0.02 kg/s of steam at 102.53 °C, releasing 45.0 kW, which passes through the kettle wall into food held at about 90 °C. The steam’s temperature is constant while it condenses, and the food is stirred and held at a near-constant temperature, so the simple formula applies: 0.00413 kW/K of entropy is generated, and 1.21 kW of work potential is lost.

That loss is small compared with the 9.9 kW of work potential the steam carried, which Chapter 2 calculated, because the temperatures are close: 102.5 °C against 90 °C. The kettle uses its steam well. The large loss in the steam system happened earlier, in the boiler that raised the steam from fuel burning at well over 1,000 °C, where the temperature difference was hundreds of kelvin.

A heat exchanger between two streams

When both fluids change temperature, as in the plant’s proposed effluent heat recovery, the entropy generated is found from each stream’s entropy change, calculated with the formula of Chapter 4.

This is an illustration. A counterflow plate heat exchanger takes 0.50 kg/s of effluent at 45 °C and 0.50 kg/s of incoming wash water at 20 °C. With equal flows and specific heats, the two streams change temperature by the same amount. If the exchanger is sized for a 7 K approach, the effluent leaves at 27 °C and the wash water at 38 °C. Using c = 4.18 kJ/kg·K:

• heat recovered: 0.50 × 4.18 × 18 = 37.6 kW
• wash water’s entropy gain: 0.50 × 4.18 × ln(311.15 ÷ 293.15) = 0.12455 kW/K
• effluent’s entropy loss: 0.50 × 4.18 × ln(300.15 ÷ 318.15) = −0.12173 kW/K
• entropy generated: 0.00282 kW/K, and lost work 293.15 × 0.00282 = 0.83 kW

The approach temperature

A larger exchanger narrows the approach, recovering more heat and generating less entropy for each kilowatt. With a 3 K approach, the wash water leaves at 42 °C and the effluent at 23 °C.

Figure 2The effluent heat exchanger: approach and arrangement

Design	Wash water out	Effluent out	Heat recovered	Entropy generated	Lost work
Counterflow, 7 K approach	38 °C	27 °C	37.6 kW	0.00282 kW/K	0.83 kW
Counterflow, 3 K approach	42 °C	23 °C	46.0 kW	0.00148 kW/K	0.43 kW
Parallel flow, infinite area	32.5 °C	32.5 °C	26.1 kW	0.00350 kW/K	1.03 kW

The table shows the counterflow arrangement’s advantage plainly. In counterflow, the hot stream’s outlet meets the cold stream’s inlet, so the temperature difference can be kept small along the whole length. In parallel flow, the two inlets meet, the difference starts at 25 K and the streams can at best end at a common temperature halfway between. Parallel flow recovers less heat and generates more entropy for what it does recover. Heat exchangers covers the sizing of the area for each case; the second law explains why the counterflow design is worth its slightly more awkward pipework.

Matching temperatures

The kettle and the heat exchanger point to one principle: supply heat at a temperature only slightly above where it is needed. Every kelvin of excess temperature generates entropy that is gone for good. Applied to the whole plant, it leads to a simple test of any heating arrangement: compare the temperature of the source with the temperature of the duty.

Figure 3Lost work in supplying heat to the 20 °C to 60 °C wash water

Steam at 102.5 °C supplying all 83.6 kW13.1 kW

Effluent recovery, 7 K approach, first 37.6 kW0.83 kW

Effluent recovery, 3 K approach, first 46.0 kW0.43 kW

The steam figure comes from the entropy balance: the water gains 0.50 × 0.53472 = 0.26736 kW/K, the steam loses 83.6 ÷ 375.68 = 0.22257 kW/K, and the difference, 0.04480 kW/K, times 293.15 K is 13.1 kW. High-temperature heat used for a low-temperature duty is the largest source of avoidable loss in most heating systems.

Below ambient: the cool room’s evaporator

The same formula applies below ambient temperature, where the inverse-square effect is strongest. This is an illustration. The cool room’s evaporator absorbs 12 kW from room air at 2 °C into refrigerant evaporating at −5 °C. The entropy generated in the coil is 12 × (1 ÷ 268.15 − 1 ÷ 275.15) = 0.00114 kW/K, a lost work of 0.33 kW.

That looks small beside the kettle’s 1.21 kW, but it must be compared with what the cool room needs: Chapter 3 found a minimum work of only 0.785 kW for the whole duty. The evaporator’s temperature difference alone wastes work equal to more than 40% of that minimum. A coil with enough area and airflow to evaporate at −1 °C instead, a 3 K difference, would halve the loss to 0.19 kW and, more importantly, raise the refrigerant’s internal Carnot COP from 5.96 to 6.46 by narrowing the lift. At low temperatures, generous heat transfer surfaces pay twice: less entropy generated in the coil, and less work for the compressor.

When a large difference is unavoidable

Some temperature differences cannot be designed away. A flame must be far hotter than the water it heats, and combustion itself generates a great deal of entropy before any heat transfer happens. The second-law response is not to fight the difference but to use it: put the high-temperature heat to work first, in an engine, a turbine or a high-temperature process, and pass the heat down to lower-temperature duties afterwards. That is the logic of cogeneration, which uses a gas engine’s exhaust and cooling water for heating after the engine has produced electricity.

For a small plant without a high-temperature duty, the practical conclusion is usually the opposite one: avoid creating very hot heat for a low-temperature job at all. A heat pump lifts heat from the surroundings by only the temperature range the duty needs, so the large differences of combustion never arise.

Measuring it on site

Entropy generation in a heat exchanger needs only flows and temperatures, which most plants can measure with clamp-on or insertion sensors.

1. Measure the flow and the inlet and outlet temperatures of both streams, at steady conditions.
2. Check the energy balance: the heat lost by the hot stream should match the heat gained by the cold one within the measurement uncertainty, typically a few per cent. A large mismatch means a faulty sensor or an unmeasured flow.
3. Calculate each stream’s entropy change with m c ln(T_out ÷ T_in), in kelvin, and add them. The sum must be positive; a negative result means a measurement is wrong.
4. Multiply by T₀ to get the lost work, and record it with the date so that drift can be seen.

Staging the sources

When a duty spans a temperature range, as wash water heated from 20 °C to 60 °C does, the best arrangement uses the lowest-grade source first and the highest-grade source last, each over the part of the range it suits.

1. Effluent recovery first, from 20 °C to 38 °C, using heat that would otherwise go down the drain. It covers 37.6 kW of the 83.6 kW.
2. The heat pump second, from 38 °C to 60 °C, covering the remaining 46.0 kW. Its duty now needs only 4.12 kW of exergy, against 5.24 kW for the whole range.
3. The boiler, if kept, only as back-up, or for a duty above the heat pump’s reach.

At the same COP of 3.2, the heat pump’s work falls from 26.1 kW to 14.4 kW. In practice the saving is a little less certain than that figure suggests, because a heat pump heating water from 38 °C rather than 20 °C has a smaller temperature glide to exploit, and its COP depends on the design. The supplier must give performance for the actual inlet temperature. But the order is right whatever the exact COP: recovered heat goes in at the cold end, and the machine that needs work does only the part that recovered heat cannot.

Reversing the order would waste the recovery. If the heat pump heated water from 20 °C to 60 °C and the effluent were then used for something else, the heat pump would do 83.6 kW of work at its lowest lift, the recovered heat would have nowhere useful to go, and the plant would pay for heat that the drain was already throwing away.

The error: a heat exchanger plumbed in parallel

This is an illustration. The plant installed the effluent heat exchanger over a shutdown. At commissioning the wash water left at 31.0 °C and the effluent at 34.0 °C, recovering 0.50 × 4.18 × 11 = 23.0 kW, well short of the 37.6 kW design.

The engineer’s first suspicion was fouling, but the clue was in the outlet temperatures themselves. In a counterflow exchanger, the cold stream can leave hotter than the hot stream leaves, and at the design point it should: 38 °C against 27 °C. Here the cold outlet, 31.0 °C, was below the hot outlet, 34.0 °C, which is what parallel flow produces and counterflow would not, at this duty. The inspection found that the effluent connections had been swapped between top and bottom ports during installation, so both streams entered at the same end.

Re-piping the effluent connections restored counterflow. The exchanger then delivered wash water at 37.8 °C, within measurement uncertainty of the design. The plant’s commissioning sheet for heat exchangers now includes a single line: “Counterflow check: cold outlet above hot outlet at design flow.” It catches the most common installation error in seconds.

How this applies to a small Australian business

Small plants often heat everything from one central source, typically a boiler supplying steam or very hot water, because it is simple to install and maintain. The second law shows the cost: every duty at a lower temperature than the source loses work potential in proportion to the temperature gap. Low-temperature duties such as washing, cleaning, space heating and preheating are the best candidates for heat recovery and heat pumps, with high-temperature sources reserved for duties that need them.

Heat exchanger checks are worth building into commissioning and maintenance. An outlet temperature that has drifted from its design value is often the first sign of fouling, a stuck bypass valve or a flow problem, and each kelvin of lost approach is both less heat recovered and more entropy generated.

Exercise: an uninsulated hot water pipe

A length of uninsulated pipe carrying the plant’s 60 °C hot water loses 10 kW to the plant room at 20 °C. Find the entropy generated and the work potential lost. Compare the lost work with the heat lost, and explain why insulating the pipe matters more for the first law than the second.

Worked answer. Treating the water and the room as reservoirs at constant temperature, S_gen = 10 × (1 ÷ 293.15 − 1 ÷ 333.15) = 10 × (0.0034112 − 0.0030017) = 0.00410 kW/K. The lost work is 293.15 × 0.00410 = 1.20 kW.

The heat lost is 10 kW, but only 1.20 kW of it was work potential, because 60 °C water is low-grade heat. The pipe loss matters mainly as 10 kW of heat that the boiler or heat pump must replace; its second-law cost is the exergy of that heat, which the plant paid for in fuel or electricity. Insulation saves both, and Heat Transfer by the Numbers shows how to choose its thickness.