EXAMPLE ABOUT WASTE-HEAT RECOVERY
(I made this analysis for Henkel - Indonesia 1996)
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Conditions: The customer in the process industry operates an industrial plant, in which process tanks continuously are kept on an average temperature of 60 degr. C. For this purpose, presently oil burners are used to heat water to a temperature of around 90 degr.C, that is fed through the heating coils of the tanks. The waste heat of an existing diesel generator, with a shaft power of 200 kW (W2 in the sketch), should be recovered in a suitable manner, as to supply heat to the process tanks.

This was all the information I got, but it was enough for me to derive the following system-data:

1) Typical heat losses from the engine to the environment: W1= 60 kW
2) Typical cooled-off heat in the engine's radiator: W3= 500kW
3) Typical waste heat in the exhaust gases: W4= 440 kW
4) Calculated flow of cooling water through the radiator: 6 lit/sec.

Solution:
1) The heat-transfer between gases and liquid in a normal heat-exchanger is not very good, requiring a very large heat-exchanger in order to get a large enough heat-exchange area. Furthermore, the flow-resistance in the exhaust system should be as low as possible. The need for a low flow-resistance and for a large heat-exchange area, let me decide to use a flash tank, in which water and exhaust gases are brought in very intensive heat-contact (spraying in fine droplets) with each other. Thus heated water is collected in the bottom of the tank and through a pump P1 fed to the heating coils of the process tanks. As some water will evaporate in the flash tank, a level valve V4, for refill with fresh water is applied. The volume of the water in the bottom of the flash tank can have any desired value; the larger it is, the more stable the temperature of the water will be (heat reservoir). If large enough, it can even serve the process tanks during a certain time in which the diesel engine would be stopped for maintenance/repair.

Waste heat recovery schematic set-up for the recovery of the cooling heat from a diesel generator to heat the coils of process tanks.

2) Assumed that the exhaust gases leave the flash tank at 100 degr.C, the power absorbed by the water is 380 kW. The flashed gas and vapors then contain W6 = 60 KW, that is wasted to the environment. The applied, recovered heat W5 to the coils of the process tanks is then 880 kW.

At 6 lit/sec water flow, the increase of water temperature in the flash tank is about 15 degr. C, which may be too much. In any case, the feed temperature to the process tanks must be controllable, for which reason there is a bypass lead, that through the retainer control valve RCV allows the flow of water through the heating coils of the process tanks and through the flash tank to be larger than that through the diesel engine (10 degr. C. temp. difference in the flash tank, requires 9 lit/sec). The more water is bypassed, the lower the feed temperature to the heating coils will be.

The RCV may also influence the flow of cooling water through the engine, for which reason temperature sensors on the process tanks and the diesel engine are needed to adjust and to keep the RCV on a balanced setting. An alternative solution would be to apply a fixed setting for the RCV and having such a high waterflow that the feed temperature to the heating coils always will be too low. The required temperature can then be adjusted with an additional burner. This will hardly cost more energy because the efficiency of the flash tank increases in this manner ( higher heat-transfer temp. difference - lower flash gas/vapor temperature). Anyway, the existing burners must remain in operational condition to take over, in case the diesel engine would brake down for a longer time.

 

3) A switch valve V1 is placed in the original cooling circuit of the diesel engine. When it is closed, this circuit is restored. If required, waste-heat from the exhaust gases can still be recoverd with the flow through the bypass and RCV. When V1 is open, the radiator is fully bypassed and all the water flows through the boost-pump P2 to the flash tank. Back valves V2 and V3 prevent undesired back flows - an additional back valve in the bypass may be useful.

ENERGY & COSTS BALANCE
If 5% energy-loss is assumed in piping and other equipment:
Recovered from the engine's cooling circuit: 475 kW (500)
Recovered from the exhaust gases: 360 kW (380)
Unrecovered waste heat: 120 kW
Engine shaft power: 200kW

Fuel (diesel) need 1200 kW equals to 3160 liter per 24 hrs continuous operation.
Fuel costs at 0.35 US$/liter: 1100 US$ per day.
Energy cost = 0.23 $/kWh , at 17% overall efficiency.

With waste-heat recovery, the energy cost is 0.044 $/kWh , at 86% overall efficiency
Savings per 24 hr period: 760 US$


Pay-off time for every 25,000 US$ investment is then app. 1 month ! With these figures, further assessment can be made as to determine type and size of pumps, valves, ducting and dimensioning of the flash tank.

I can be a consulting engineer for you