Throttling, mixing and friction

Some of the most persistent losses in a plant leave no trace in an energy balance. A pressure regulator takes compressed air in at 8 bar and lets it out at 5 bar, with no heat lost and nothing thrown away: the energy entering equals the energy leaving. A tempering valve mixes hot and cold water to a comfortable temperature, again with no energy lost. A filter or a partly closed valve turns some of a flow’s pressure into a little warmth, too little to measure. The first law calls all three perfect. The second law shows that each destroys work potential, often continuously, every hour the plant runs.

This chapter calculates the entropy generated by throttling, mixing and friction, applies it to the running case’s air regulators, tempering valve and filters, and shows why the cure is nearly always to avoid creating the pressure or temperature in the first place. It ends with a compressor whose pressure had been raised years earlier for a reason nobody remembered.

Throttling

A throttle is any restriction that drops a fluid’s pressure without extracting work: a pressure regulator, a partly closed valve, an orifice plate, a refrigerant expansion valve. For steady flow with no heat transfer and no work, the first law says the enthalpy leaving equals the enthalpy entering. A throttling process is isenthalpic.

For an ideal gas, enthalpy depends only on temperature, so the temperature after the throttle equals the temperature before it. Nothing seems to happen except a fall in pressure. But the entropy rises, by the amount Chapter 4 calculated for the equivalent slow isothermal expansion:

Δs = R ln(p₁ ÷ p₂)

and because no heat crosses the boundary, all of that rise is entropy generated.

The running case: the packaging line regulators

This is an illustration. The compressor delivers 0.06 kg/s of air at 8 bar absolute, and regulators reduce it to 5 bar absolute for the packaging line. Each kilogram of air generates 0.287 × ln(8 ÷ 5) = 0.1349 kJ/kg·K of entropy in the regulators, 0.00809 kW/K for the whole flow, and the lost work is 293.15 × 0.00809 = 2.37 kW.

Figure 1Lost work in throttling an ideal gas

W_lost = ṁ × R × T₀ × ln(p₁ ÷ p₂)

ṁ

mass flow, kg/s

R

specific gas constant, 0.287 kJ/kg·K for air

p₁, p₂

absolute pressures before and after the throttle

1. 8 to 5 bar: 0.06 × 0.287 × 293.15 × ln(1.6) = 2.37 kW
2. 7 to 5 bar: 0.06 × 0.287 × 293.15 × ln(1.4) = 1.70 kW
3. 6 to 5 bar: 0.06 × 0.287 × 293.15 × ln(1.2) = 0.92 kW

The 2.37 kW is the difference between the air’s exergy at 8 bar and at 5 bar, 10.43 and 8.06 kW, found in Chapter 7. It equals 10.8% of the compressor’s 22 kW electrical input. The compressor did the work to raise the air to 8 bar; the regulators throw away the part that the packaging line does not use.

Real gases and the Joule–Thomson effect

A real gas does change temperature slightly in a throttle, because its enthalpy depends a little on pressure. The rate is the Joule–Thomson coefficient. For nitrogen at 0.8 MPa and 20 °C, the NIST WebBook gives 2.18 K per MPa, so a drop of 0.3 MPa cools the gas by about 0.65 K. Air behaves similarly. The cooling is real, and at large pressure drops it becomes useful: it is the basis of the Linde process for liquefying gases. For a compressed-air regulator it is a curiosity, too small to affect the lost work calculated above.

Throttling in refrigeration

Every vapour-compression refrigerator throttles its liquid refrigerant through an expansion valve, from condensing to evaporating pressure, and the same isenthalpic process generates entropy there. It is one of the losses between Chapter 3’s internal Carnot COP of 5.96 for the cool room and its measured 2.5. Large systems sometimes recover part of it with an expander or an economiser stage; small cool-room units accept it, because the hardware to recover it costs more than it would save. Recognising the throttle as a loss is still the first step to judging whether it is worth tackling.

Mixing

When two streams at different temperatures mix, heat flows from the hotter to the colder within the mixture, and the result is the same entropy generation as heat transfer across a temperature difference, with no heat exchanger in between.

This is an illustration. The plant’s tempering valve mixes 0.05 kg/s of 60 °C hot water with 0.05 kg/s of 20 °C cold water to give 0.10 kg/s of hand-wash water at 40 °C. The energy balance closes: 0.05 × 251.42 + 0.05 × 84.194 = 16.781 kW against 0.10 × 167.79 = 16.779 kW from the NIST enthalpies. The entropy balance does not: the hot water loses 0.05 × (0.83114 − 0.57229) = 0.01294 kW/K, the cold water gains 0.05 × (0.57229 − 0.29642) = 0.01379 kW/K, and the difference, 0.000851 kW/K, is generated. The lost work is 293.15 × 0.000851 = 0.25 kW.

Figure 2Three losses with no heat lost, running case

Item	What happens	Entropy generated	Lost work	First-law loss
Air regulators, 8 to 5 bar	air expands with no work	0.00809 kW/K	2.37 kW	none
Air filter, 0.5 bar drop at 8 bar	air expands with no work	0.00111 kW/K	0.33 kW	none
Tempering valve, 60 °C with 20 °C	heat flows within the mixture	0.000851 kW/K	0.25 kW	none
Throttled pump valve, 150 kPa at 0.5 kg/s	friction in the valve	0.00023 kW/K	0.07 kW	none

The tempering valve’s loss is small, but it points to a larger question. The plant heats water to 60 °C, at an exergy cost, and then mixes some of it down to 40 °C. Hand-wash water could be heated directly to 40 °C by a heat pump at a higher COP, because the lift is smaller, or taken from the effluent recovery exchanger of Chapter 6, which already delivers water at 38 °C to 42 °C. Water heated to 60 °C for food safety in some duties may need to stay at 60 °C; the second law asks only whether each duty truly needs the temperature it is given.

Friction in pipes, valves and filters

Friction in a flowing fluid converts pressure, a form of exergy, into a small temperature rise, which is nearly worthless. For a liquid, whose density barely changes, the lost work is close to the volume flow times the pressure drop. For a gas, it follows the throttling formula.

• The wash-water pump delivers 0.50 kg/s through a valve partly closed to set the flow, dropping 150 kPa. The volume flow at 60 °C is 0.50 ÷ 983 = 0.000509 m³/s, and friction in the valve dissipates 0.000509 × 150,000 = 76 W of pumping work as heat in the water. A variable-speed drive that sets the flow by slowing the pump would avoid most of it.
• The compressed-air filter, if it drops 0.5 bar at 8 bar when partly clogged, loses 0.06 × 0.287 × 293.15 × ln(8 ÷ 7.5) = 0.33 kW. A clean filter dropping 0.1 bar loses a fifth of that. Filter elements are cheap; the pressure they waste is paid for at the compressor every hour.

For a liquid, the entropy generated by friction is the dissipated work divided by the fluid’s absolute temperature. For the pump valve that is 76 W at 333.15 K, 0.00023 kW/K, and the lost work is T₀ times that, 293.15 × 0.00023 = 67 W. The other 9 W survives as a trace of extra heat in water that is hotter than the surroundings. For liquids near ambient temperature the distinction vanishes, and the lost work is simply the volume flow times the pressure drop.

Friction losses in liquid systems are usually small in exergy terms, because pumping power is small compared with heating power. In compressed air they are large, because every pressure drop between the compressor and the point of use is pressure that was expensive to make.

Leaks: throttling all the way down

A compressed-air leak is a throttle that drops the air from line pressure all the way to atmospheric, outside any machine that could use it. Every kilogram that leaks at 8 bar destroys its whole exergy, 173.8 kJ/kg, the most a regulator could ever lose.

This is an illustration. Suppose the plant’s distribution system leaks 0.005 kg/s, 8.3% of the compressor’s output, a modest figure for an older system. The exergy destroyed in the leaks is 0.005 × 173.8 = 0.87 kW. But the compressor took about 2.1 kW of electricity for every kilowatt of air exergy it produced, from Chapter 7’s balance, so the leaks cost about 1.83 kW of electrical input, continuously, whether or not the plant is producing anything. Seen through the second law, a leak is not a small loss of air but the complete loss of everything the compressor did to that air.

Leak losses also scale with pressure, twice over. A higher system pressure pushes more air through the same leak, and each kilogram that escapes carries more exergy. Lowering the setpoint, as in the steps below, cuts leak losses as well as regulator losses.

When throttling is the right answer

Not every throttle is a mistake. A control valve must drop some pressure to have any authority over the flow, and a valve that is never partly closed cannot control anything. Pressure regulators protect equipment rated below the system pressure, and a single central compressor with regulators at each user is often simpler and more reliable than several machines at different pressures. Tempering valves prevent scalds.

The second law does not say these devices should be removed. It says their losses should be known, priced and kept to what their purpose requires: a regulator that trims a few tenths of a bar to hold a steady pressure is doing its job; one that drops three bar because the compressor is set too high is a symptom.

The cure: do not create what you will throw away

Throttling, mixing and friction share one cure. Each destroys exergy that was created upstream, by a compressor raising pressure or a heater raising temperature, and then not used. Reducing the loss at the throttle or mixer helps a little; avoiding the excess at its source helps much more, because the source machine also stops spending work to create it.

Figure 3Reducing the packaging line's regulator loss, in order of value

1. 1Find the true minimum pressurethe lowest pressure at which every user works reliably, with a stated margin
2. 2Fix the pressure drops in betweenfilters, undersized pipework and leaking fittings that force a higher setpoint
3. 3Lower the compressor setpointto the minimum plus the remaining distribution drop
4. 4Separate any high-pressure usersupply a small, occasional high-pressure duty locally instead of raising the whole system
5. 5Keep regulators only for controllet them trim, not drop several bar

The error: a setpoint nobody remembered

This is an illustration. During the audit, the engineer asked why the compressor ran at 8 bar absolute when the packaging line needed 5 bar and no other user needed more than 6. The maintenance log gave the answer: some years earlier the setpoint had been raised from 7 to 8 bar to cure low pressure at the packaging line, which turned out later to be a clogged filter. The filter had long since been replaced. The setpoint had not.

The second law put a price on the leftover bar. At 7 bar, the regulators would lose 1.70 kW instead of 2.37 kW, and the compressor’s minimum work would fall from 10.43 kW to 9.76 kW. Scaled in proportion, the real compressor’s 22 kW would fall to about 20.6 kW, a saving of about 1.4 kW for as long as it runs. A step further, to 6 bar, would save more, provided the line’s distribution losses were checked first.

The fix cost nothing but a setpoint change and a week of monitoring the packaging line’s pressure. The plant added the compressor setpoint to its list of items reviewed at every change to the air system, with the reason for the current value written beside it. A setpoint with no recorded reason is one that nobody will dare to lower.

How this applies to a small Australian business

Compressed air is one of the most expensive forms of energy in a small plant, and much of its cost is spent making pressure that is then throttled away, leaked or lost in filters and long pipe runs. Count the air and what it costs shows how to price the air itself. The second law adds a sharper question for every regulator and pressure drop in the system: how many kilowatts does this restriction destroy, and is anyone using the pressure it destroys?

The same thinking applies to hot water. Mixing valves are often necessary for safety, to deliver water at a temperature that will not scald, and they should stay. But a plant that heats everything to the highest temperature any duty needs, then mixes down for the rest, pays the second law’s price on every litre of mixed water.

Exercise: a different tempering mix

A second tempering valve mixes 0.04 kg/s of 60 °C hot water with 0.06 kg/s of 20 °C cold water. Using a constant specific heat of 4.18 kJ/kg·K, find the mixed temperature, the entropy generated and the lost work.

Worked answer. With equal specific heats, the mixed temperature is the flow-weighted mean: (0.04 × 60 + 0.06 × 20) ÷ 0.10 = 36 °C, or 309.15 K.

The entropy generated is the sum of the two streams’ entropy changes:

4.18 × [0.04 × ln(309.15 ÷ 333.15) + 0.06 × ln(309.15 ÷ 293.15)] = 4.18 × [−0.002991 + 0.003189] = 0.000827 kW/K

The lost work is 293.15 × 0.000827 = 0.24 kW. A smaller share of hot water gives a lower mixed temperature, but the loss is similar, because the temperature gap between the two inlet streams, which drives the irreversibility, is unchanged.