some basic hvac. first law of thermodynamics h = u + pv enthalpy second law: entropy hvac - cooling...
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Cooling Rates
1. evTT cooling flow rate= evQBtu/h, cooling flow / 12000 Tons to Btu/h conversion
2. evQ Btu/h cooling flow rate = acF´ kW elec Power to AC * e * b * 3412. * COP see Spread Sheet
3. kpH´Btu/h Heat equ. Power to Ko = acF´ kW elec Power to AC * e * b * 3412.
4. kpF´ kW elec Power to Komp = acF´ kW elec Power to AC * e * b
5. kpH´ Btu/h Heat equ. Power to Komp = kpF´ kW elec Power to Komp * 3412electrical HV
6. COP = evQBtu/h, Rate of heat absorbed / kpH´Btu/h, work to compresssor practical COP (heat loss incl)
7. COPH = cdQBtu/h, condenser / kpH´Btu/h Btu/h COP HEATPUMP defined
8. EERBtu/Wh = evQBtu/h, Rate of heat absorbed / kpF´WW, Total Power input EER defined
9. EERBtu/Wh = COP * 3.412Btu/h / W EER in terms of COP
10. kpF´kw = kpF´Ww / 1000
11. kW|T = acF´kW elec Power to AC / evTTons cooling rate produced by AC kw per Ton
12. kW|T = 12000Btu/h/T / (3412Btu/h/kW * COP) kW per Ton as f of COP
13. kWh|Th = 12000. / (3412. * COP) kWh per Tonh
14. kW|T = 12k Btu/h/T / EERBtu/Wh kW per Ton as f of EER
15. abH´Btu/h, equ. Heat to generator - = mstlb/h of steam, * hfgBtu/lb, 955 1-stage, 860 2-stage Absorption thermal Power to Generator (Absorption chiller)
Nipuna
COP = evQ / kpH’ = (h1-h4) / (h2-h1)
HVAC - Cooling
Saturation humidity line: Wet bulb temperature lines Relative humidity line
Specific volume lines Enthalpy lines
Psychrometric Chart
HVAC - Cooling
The energy efficiency rating (EER) of an air conditioner is its BTU/h rating over its Wattage.
Example: window air conditioner Rating: 10,000-BTU/h Power Consumption: 1,200 watts EER = 10,000 BTU/h/1,200 watts = 8.3 Btu/Wh
Normally a higher EER is accompanied by a higher price.
HVAC - Cooling
Choice between two 10,000-BTU/h units
1. EER of 8.3, consumes 1,200 watts2. EER of 10, consumes 1000 watts.
Price difference is $100. Usage: 4 months a year, 6 hours a day. Electricity Cost: $0.10/kWh. ===========================================4 mo. x 30 days/mo. x 6 hr/day = 720 hours
(720 h x .2 kW) x $0.10/kWh = $14.40 Savings
Since the EER 10 unit costs $100 more, it will take about seven years for this more expensive unitto break even
HVAC - Cooling
FIRETUBE BOILERS WATERTUBE BOILERS
Disadvantages of the Watertube design include:High initial capital cost Cleaning is more difficult due to the design No commonality between tubes Physical size may be an issue
Disadvantages of Firetube Boilers include:Not suitable for high pressure applications 250 psig and above Limitation for high capacity steam generation
HVAC -Heating
Thermostatic Steam Traps
Mechanical Steam Traps
HVAC -Heating
Heat Flow
1. hQBtu/h, W Heat flow = Aft2, m2 * TF, C / R(F*ft2*h)/(Btu) , (C*m2/W) Heat Flow as function of Thermal Resistance R
2. hQ Btu/h, W Heat flow = Aft2, m2 * C Btu/(h*ft2*F), W/(m2 *C) * TF, C Heat Flow (loss) in terms of Conductance C
3. hQBtu/h, W = Aft2, m2 * (kBtu/(h*ft*F, W/ (m*C) / Lft, m) Btu/(h*ft2*F) , W/(m2 *C) * TF, C Heat Loss (flow) in terms of conductivity k
Heat Flux
4. hqBtu/(h*ft2), W/m2 Heat flux = TF, C / R(F*ft2*h)/(Btu) , (C*m2/W) Heat flux as function of R
5. hq Btu/h*ft2), W/m2 Heat flux = C Btu/(h*ft2*F), W/(m2 *C) Conductance * TF, C Heat flux as function of C
6. hqBtu/h*ft2), , W/m2 Heat flux = k Btu/(h*ft*F),conductivity, W/(m*C) / Lft,m * TF, C Heat flux as function of k
R – C – k
7. R (F*ft2*h)/(Btu) , (C*m2/W) = 1 / C Btu/(h*ft2*F) , W/(m2 *C) R –C Relationship
8. C Btu/(h*ft2*F), W/(m2 *C) = k Btu/(h*ft*F), W/(m*C) / Lft, m C – k Relationship
9. R(F*ft2*h)/(Btu) , (C*m2/W) = Lft, m / k Btu/(h*ft*F) , W/(m*C) R – k Relationship
Convenience Conductivity k´, per inch, per cm
10. hQBtu/h, W = Aft2, cm2 * (k´Btu*in/(h*ft2*F), W/cm*C / L´in, cm) Btu/(h*ft2*F), W/(cm2*C) * TF C Heat flow (loss) in terms of convenience k
11. hqBtu/h*ft2), Heat flux = k´ Btu*in/(h*ft2*F),conductivity / L´in * TF Heat flux and convenience k
12. C Btu/(h*ft2*F), = k´ Btu*in/(h*ft2*F) / L´in C – convenience k Relationship
13. R(F*ft2*h)/(Btu) = L´in / k´ Btu*in/(h*ft2*F) R – convenience kRelationship
Convenience Resistance R´ – R per inch, cm
14. R´ (h*ft2*F) / (Btu * in) = (h*ft2*F) /Btu / in , (cm * C) / W = 1 / k´ Btu*in/(h*ft2*F), W/cm*C R´ per inch, cm = R / L´ = (L / k ) / L´ , L´ cm, in
15. R(F*ft2*h)/(Btu), (C*m2/W) = R´ (h*ft2*F) / (Btu * in), (cm * C) / W * L´in, cm R = R´ * L´
Building Envelope Nipuna en:p:ÙN:
HVAC - Envelope