METI / ECCJ / ACE Training Program MTPEC06 (2007.01)

[IV] Case Examples of EE&C

Contents:

Ex.1 Adjustment of Combustion Air Ratio in Boiler
Ex.2 Heat Insulation to Un-insulated Steam Piping, etc.
Ex.3 Waste heat recovery from the tunnel kiln
Ex.4 Introduction of Inverter control to Aerator Blower Motor
Ex. 5 Improvement in Compressor Ope. Management

Food Processing Industry

Ex.1 Adjustment of Combustion Air Ratio in Boiler

Existing problem: Two 2t/h boilers, but the high air ratio operation O² concentration: 7.9% in exhaust gas (Air ratio = 1.6)
Improvement measures: Adjustment of air ratio, 1.2 based on standard of EC Law
Preconditions for estimation:

A fuel oil consumption		520kL/y
Properties of A fuel oil: }
	Lower calorific value Specific gravity Theoretical amount of air Theoretical amount of exhaust gas	Hu = 34,270kJ/L Sg = 0.83kg/L A₀ = 8.99m³N/L G₀ = 9.55m³N/L
Mean exhaust gas temperature: Atmospheric temperature: Mean specific heat of exhaust gas: Heat loss of exhaust gas at air ratio, m: Fuel reduction rate by air ratio improvement: Price of A fuel oil		Tg = 220ºC(Assumed constant) t0 = 20ºC cg = 1.38kJ/m³NK Lg = [G₀+(m-1)×A₀] ×cg ×(Tg-t₀)(kJ/L) Fr = (Lg₁-Lg₂) / (Hu – Lg₂) ×100 (%) ¥41/L

Ex.1 Adjustment of Combustion Air Ratio in Boiler
(Food Processing Industry)

4. Estimated energy saving
	Exhaust gas heat loss at m = 1.6: Lg₁ = 4,125kJ/L Exhaust gas heat loss at m = 1.2: Lg₂ = 3,132kJ/L Fuel reduction rate: Fr = 3.2% A fuel Oil reduction: 520kL/y ×0.032 = 16.6kL/y

5. Effect
	Reduction in crude oil equivalent
		= 16.6kL/y ×39.1MJ/L /38.76MJ/L = 16.75kL/y
	CO² reduction
		= 16.6kL/y ×39.1GJ/kL × 0.0189tC/GJ ×44/12 = 45t/y
	Cost saving
		= 16.6kL/y × \41,000/kL} = k\680/y (= US$5,920/y)

Metallic Products Manufacturing

Ex.2 Heat Insulation to Un-insulated Steam Piping, etc.

１．Existing problem:	There are many places of un-insulated piping for steam, drain and hot water, etc.
2. Improvement measures:	Enhancement of heat insulation to steam and drain piping (Drain is fed to boiler.)

3. Preconditions for estimation:
Insulation places & bare piping equivalent length (EQ):

Fluid temp.	Pipe size	Piping length	Valves No. & EQ	Flanges No. & EQ	Total bare piping EQ
Steam: 150ºC	50A	3m	6×1.28m	20×0.49m	20.48m
Drain: 90ºC	25A	5m	4×1.21m	10×0.54m	15.24m

Unit radiation heat:

Insulation efficiency:
Boiler operation rate:
Operating time:
LPG: Low calorific value:
Specific weight:
Price:

Steam piping (150ºC, 50A):400W/m
:Drain piping (90ºC, 25A):100W/m
90%
80% (Average)
10h/day ×270day/y = 2,700h/y
97,000kJ/m³_N
2.07kg/m³_N
¥114.4/m³_N

Ex.2 Heat Insulation to Un-insulated Steam Piping, etc. (Metallic Products Manufacturing)

4. Estimated energy saving
	Radiation heat loss before improvement: 400W/m × 20.48m + 100W/m ×15.24m = 9,716W = 34,978kJ/h Reduction of radiation heat loss after improvement: 34,978kJ/h × 0.9 = 31,480kJ/h LPG saving volume: 31,480kJ/h × 2,700h/y / (97,000kJ/m²_N × 0.8) = 1,095.3m³_N/y = 2,267kg/y
5. Effect: Reduction in crude oil equivalent = 2.267t/y ×50.2GJ/t /38.76GJ/kL
	= 2.94kL/y (Crude oil: 38.76MJ/L) CO² reduction = 2.267t/y ×50.2GJ/t ×0.0163tC/GJ ×44/12 = 6.8t/y Cost saving = 1,095.3m³_N /y × \114.4/m³_N = k\125/y (= US$1,087/y)

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