Example Heat Recovery Calculation - Mitsubishi Electric Lossnay LGH-15RX3 Technical Manual

CHAPTER 3 G General Technical Considerations

2. Example Heat Recovery Calculation

(1) Setting of conditions
(Note: Tokyo Power, industrial power 6 kV supply)
Operation time
Electricity fee
Capacity per
1 kW of electricity
Energy unit cost
G
Return air volume (RA) = 7,200 m
G
Air conditions
Season
Dry bulb temp. Wet bulb temp. Relative humidity Absolute humidity
Item
DB [°C]
Outdoors 0
Indoors 20
(2) Selection of Lossnay model (select from treatment air volume catalogue)
Model name: LU-160 with combination of LU-1605 × 1 unit
G
G
Processing air volume per unit RA = 7,200 m
G
Heat recovery efficiency : Heat recovery efficiency = 73%, Enthalpy recovery efficiency (cooling) = 62%,
G
Static pressure loss (unit-type) RA = 156.9 Pa, OA = 186.3 Pa (Note: Each with filters)
G
Power consumption (pack-type) = none because of unit type
(3) State of indoor supply air
Temperature [°C]
Enthalpy
[kJ/kg (DA)]
Numerical value obtained
from above equation and
psychometric chart
(4) Outdoor air load and heat recovered
Fresh air load without
Lossnay (q )
1
Outdoor air load with
Lossnay (q )
2
Heat recovered (q )
3
(%) outdoor air load
(5) Recovered money (power rates)
Yearly saved money
Remarks
24
Units When Heating
10h/day × 26 days/mo. × 5 mo./yr. = 1,300 h/yr
(h/yr)
(yen/kWh)
(kW/kW)
(yen/kWh) 16.22/3.1 = 5.23
3 G
/Hr Outdoor air volume (OA) = 8,000 m
Winter heating
× [kg/kg (DA)]
WB [°C] RH [%]
–2.7 50 0.0018
13.8 50 0.0072
Enthalpy recovery efficiency (heating) = 67%
Heating
= { 20 (Indoor temperature) – 0 (outdoor air temperature)} ×
0.73 (heat recovery efficiency) + 0 (outdoor air temperature)
= 14.6
= {38.5 (Indoor enthalpy) – 5.0 (outdoor air enthalpy)} ×
0.67 (enthalpy recovery efficiency) + 5.0(outdoor air enthalpy)
= 27.4
• Dry-bulb temperature = 14.6 °C • Wet-bulb temperature = 9.2 °C
• Relative humidity = 49% • Absolute humidity = 0.005 kg/kg (DA)
• Enthalpy = 27.4 kJ/kg (DA)
Heating
= 1.2 (Air specific gravity) × 8,000 (outdoor air volume) ×
{ 38.5 (indoor enthalpy) – 5.0 (outdoor air enthalpy) }
= 321,600 kJ/h = 89.3 kW
) ×
= 89.3 (Outdoor air load) (q
1
{ 1 – 0.67 (enthalpy recovery efficiency)}
= 29.5kW
or
= Air specific gravity × outdoor air volume × (indoor enthalpy – indoor blow enthalpy)
= q – q
1 2
= 89.3 – 29.5
= 59.8
or
) × enthalpy recovery efficiency
= Outdoor air load (q
1
• Outdoor air load = 89.3 kW = 100%
• Outdoor air load with Lossnay = 29.5 kW = 33%
• Heat recovered = 59.8 kW = 67%
Heating
= Heat recovered: kW × Unit price yen/W × operation time Hr/year = Heat recovered: kW × Unit price yen/W × operation time Hr/year
= 59.8 kW × 5.232 yen/kWh × (1,300hr/year)
= 406,580 yen
If recovered heat is converted to electricity : heating = 59.8 kW/3.1 = 19.3 kW/h
10h/day × 26 days/mo. × 4 mo./yr. = 1,040 h/yr
16.22
3.1
Enthalpy h Dry bulb temp. Wet bulb temp. Relative humidity Absolute humidity
[kJ/kg (DA)] DB [°C]
5.0 (1.2) 33
38.5 (9.2) 26
3 3
/Hr, OA = 8,000 m Air volume ratio (RA/OA) = 0.9
= 33 (Outdoor air temperature) – { 33 (outdoor air temperature) –
26 (indoor temperature)} × 0.73 (heat recovery efficiency)
= 27.89
= 85 (Outdoor air enthalpy) – { 85 (outdoor air enthalpy) –
53.2 (indoor enthalpy)} × 0.62 (enthalpy recovery efficiency)
= 65.3
• Dry-bulb temperature = 27.89 °C • Wet-bulb temperature = 22.4 °C
• Relative humidity = 62% • Absolute humidity = 0.0146 kg/kg (DA)
• Enthalpy = 65.3 kJ/kg (DA)
Caution: See the psychrometric chart on the next page.
= 1.2 (Air specific gravity) × 8,000 (outdoor air volume) ×
{ 85.0 (outdoor air enthalpy) – 53.2 (indoor enthalpy) }
= 305,280 kJ/h = 84.8 kW
= 84.8 (Outdoor air load) (q
{ 1 – 0.62 (enthalpy recovery efficiency) }
= 32.2 kW
or
= Air specific gravity × outdoor air volume × (indoor enthalpy – indoor blow enthalpy)
= q – q
1 2
= 84.8 – 32.2
= 56.2 kW
or
= Outdoor air load (q
• Outdoor air load =84.8 kW = 100%
• Outdoor air load with Lossnay = 32.2 kW = 38%
• Heat recovered = 52.6 kW = 62%
= 52.6 kW × 6.86 yen × (1,040hr/year)
= 375,269 yen
When Cooling
17.84
2.6
17.84/2.6 = 6.86
3 G
/Hr Air volume ratio (RA/OA) = 0.9
Summer cooling
× [kg/kg (DA)]
WB [°C] RH [%]
27.1 63 0.0202
18.7 50 0.0105
Cooling
Cooling
) ×
1
) × enthalpy recovery efficiency
1
Cooling
cooling = 52.6 kW/2.6 = 20.2 kW/h
Enthalpy h
[kJ/kg (DA)]
85.0 (20.3)
53.0 (12.7)