2.4 Mass, Energy & Combustion Balances for HVAC Equipment

2.4 Mass, Energy & Combustion Balances for HVAC Equipment

Practical HVAC design synthesizes mass and energy balances across integrated equipment loops: hydronic water loops, refrigeration refrigerant loops, cooling tower condenser loops, air distribution ducts, and fuel combustion systems. On the PE exam, mastery of rapid conservation equations, cooling tower water balances, and combustion chemistry is essential.

---

1. Fundamental HVAC Heat Transfer "Shortcut" Equations

In standard design practice, steady-flow energy balances for air and water are simplified by evaluating fluid density ($\rho$) and specific heat ($c_p$) at standard atmospheric conditions:

Hydronic Water Loops ($60^\circ\text{F} - 180^\circ\text{F}$)

$\dot{q} = \dot{m} c_p \Delta T = \left(\text{GPM} \times 8.33 \frac{\text{lb}_m}{\text{gal}} \times 60 \frac{\text{min}}{\text{hr}}\right) \times \left(1.0 \frac{\text{Btu}}{\text{lb}_m\cdot^\circ\text{F}}\right) \times \Delta T$

$\dot{q} = 500 \times \text{GPM} \times \Delta T \quad (\text{Btu/hr})$

$\text{Cooling Tons} = \frac{500 \times \text{GPM} \times \Delta T}{12,000} = \frac{\text{GPM} \times \Delta T}{24}$

Note: If fluid is an ethylene or propylene glycol mixture, the constant $500$ must be adjusted for specific gravity ($SG$) and specific heat ($c_p$): $C = 500 \times SG \times c_p$.

Air Distribution Ducts (Standard Air: $\rho = 0.075 \text{ lb}_m/\text{ft}^3$, $c_p = 0.240 \text{ Btu/lb}_m\cdot^\circ\text{F}$)

$\text{Sensible Heat: } \dot{q}_s = \left(\text{CFM} \times 60 \times 0.075\right) \times 0.240 \times \Delta T = 1.08 \times \text{CFM} \times \Delta T \quad (\text{Btu/hr})$

$\text{Latent Heat: } \dot{q}_l = \left(\text{CFM} \times 60 \times 0.075\right) \times 1,061 \times \Delta W = 4,840 \times \text{CFM} \times \Delta W \quad (\text{Btu/hr})$

$\text{Total Heat: } \dot{q}_t = \left(\text{CFM} \times 60 \times 0.075\right) \times \Delta h = 4.5 \times \text{CFM} \times \Delta h \quad (\text{Btu/hr})$

Where $\Delta T$ is dry-bulb temperature difference ($^\circ\text{F}$), $\Delta W$ is humidity ratio difference ($\text{lb}{water}/\text{lb}{dry air}$), and $\Delta h$ is enthalpy difference ($\text{Btu/lb}_{dry air}$).

---

2. Chiller & Cooling Tower Energy Balances

A water chiller acts as an energy combiner: heat absorbed from the building chilled water loop plus the electrical power input to the compressor must be rejected to the condenser water loop.

   +-------------------------------------------------------------+
   |                      WATER CHILLER                          |
   |                                                             |
   |   [Evaporator]  <--- Q_evap = 500 * GPM_chw * Delta_T_chw   |
   |        |                                                    |
   |        V                                                    |
   |   [Compressor]  <--- W_comp = 3,412.14 * kW_in              |
   |        |                                                    |
   |        V                                                    |
   |   [Condenser]   ---> Q_cond = Q_evap + W_comp               |
   +-----------------------------|-------------------------------+
                                 |
                                 V
   +-------------------------------------------------------------+
   |                      COOLING TOWER                          |
   |                                                             |
   |   Q_tower = 500 * GPM_cw * (T_cwr - T_cws) = Q_cond         |
   |   Evaporation: E = Q_tower / h_fg                           |
   |   Makeup: M = E + B + D                                     |
   +-------------------------------------------------------------+

The "Standard Cooling Tower Ton"

• Refrigeration Ton: $1 \text{ ton} = 12,000 \text{ Btu/hr}$ of cooling effect at the evaporator.
• Tower Heat Rejection: For electric centrifugal chillers operating at nominal full-load efficiency ($\sim 0.60 \text{ kW/ton}$), the heat of compression adds $0.60 \times 3,412.14 \approx 2,047 \text{ Btu/hr}$ per ton. The total heat rejected is $12,000 + 2,047 \approx 14,047 \text{ Btu/hr}$.
• Standard Rating Benchmark: Industry standards (CTI/ASHRAE) rate cooling towers at $15,000 \text{ Btu/(hr}\cdot\text{ton)}$ (a $25%$ heat of compression factor). At standard $3.0 \text{ GPM/ton}$ condenser water flow, this corresponds to a $10^\circ\text{F}$ temperature range ($85^\circ\text{F}$ entering, $95^\circ\text{F}$ leaving):

$\dot{q}_{tower} = 500 \times 3.0 \text{ GPM} \times 10^\circ\text{F} = 15,000 \frac{\text{Btu}}{\text{hr}\cdot\text{ton}}$

---

3. Cooling Tower Mass & Water Treatment Balances

Evaporative cooling towers reject heat by evaporating water into the ambient airstream. As pure water evaporates, dissolved solids (minerals, calcium, silica) remain behind, concentrating in the sump basin. Water must be intentionally bled off (blowdown) and replaced with fresh city makeup water.

                      Evaporation (E) [Pure H2O Vapor]
                            ^
                            |
  Makeup (M) -----------> [Cooling Tower Sump] -----------> Blowdown (B) [High TDS]
  [Low TDS]                 |                  -----------> Drift (D) [Mist]
                            V
                   Circulation (GPM_cw)

Conservation of Mass Equations

$\text{Total Water Mass Balance: } M = E + B + D$

Where $M$ is makeup rate, $E$ is evaporation rate, $B$ is blowdown rate, and $D$ is drift loss (typically negligible in modern drift eliminators: $D < 0.005% \times \text{GPM}_{circ}$). Neglecting drift ($D \approx 0$):

$M = E + B$

Dissolved Solids (TDS) Balance & Cycles of Concentration ($C$)

$M \times \text{TDS}_{makeup} = B \times \text{TDS}_{tower}$

$\text{Cycles of Concentration } (C) = \frac{\text{TDS}_{tower}}{\text{TDS}_{makeup}} = \frac{\text{Chlorides}_{tower}}{\text{Chlorides}_{makeup}} = \frac{M}{B}$

Substituting $M = E + B$ into the cycles of concentration equation:

$C = \frac{E + B}{B} = \frac{E}{B} + 1 \implies \frac{E}{B} = C - 1$

$\text{Blowdown Flow Rate: } B = \frac{E}{C - 1}$

$\text{Makeup Flow Rate: } M = E \times \left(\frac{C}{C - 1}\right)$

Evaporation Rate Approximation Formula

$E = \frac{\dot{q}_{tower}}{h_{fg}} \approx \frac{\dot{q}_{tower}}{1,050 \text{ Btu/lb}_m}$

In terms of circulation flow rate and tower temperature range:

$E (\text{GPM}) = 0.0008 \times \text{GPM}_{circ} \times (T_{cwr} - T_{cws}) = 0.0008 \times \text{GPM}_{circ} \times \Delta T_{range}$

---

4. Combustion Chemistry, Excess Air & Boiler Efficiency

Combustion is the rapid exothermic chemical reaction between fuel hydrocarbons and oxygen in atmospheric air.

Stoichiometric (Theoretical) Combustion of Methane ($\text{CH}_4$)

Atmospheric air is approximately $21% \text{ O}_2$ and $79% \text{ N}_2$ by volume (mole ratio $\frac{79}{21} = 3.76 \text{ moles N}_2 / \text{mole O}_2$).

$\text{CH}_4 + 2 \left(\text{O}_2 + 3.76 \text{ N}_2\right) \longrightarrow \text{CO}_2 + 2 \text{ H}_2\text{O} + 7.52 \text{ N}_2 + \text{Heat}$

For 1 mole (or $1 \text{ ft}^3$) of methane gas:

• Theoretical $\text{O}_2$ Required: $2.0 \text{ moles}$
• Theoretical Air Required: $2.0 \times 4.76 = 9.52 \text{ moles}$ (or $9.52 \text{ ft}^3$ of air per $\text{ft}^3$ of $\text{CH}_4$).

Percent Excess Air

To ensure complete combustion and prevent hazardous carbon monoxide ($\text{CO}$) and soot generation, commercial boilers operate with excess air ($10%$ to $25%$ excess air for natural gas):

$\% \text{ Excess Air} = \left(\frac{\text{Actual Air} - \text{Theoretical Air}}{\text{Theoretical Air}}\right) \times 100\% = \left(\frac{\text{Actual } \text{O}_2 - \text{Theoretical } \text{O}_2}{\text{Theoretical } \text{O}_2}\right) \times 100\%$

$\text{Theoretical Air-to-Fuel Ratio by Mass: } AF_{mass} = \frac{m_{air}}{m_{fuel}}$

Higher Heating Value (HHV) vs. Lower Heating Value (LHV)

$\text{HHV} = \text{LHV} + m_{H_2O} \times h_{fg,H_2O}$

• HHV (Gross Calorific Value): Assumes all water vapor formed during combustion is fully condensed back to liquid at $60^\circ\text{F}$ or $77^\circ\text{F}$, releasing its latent heat of condensation ($\sim 1,050 \text{ Btu/lb}_m$ of water). Natural gas HHV $\approx 1,000 - 1,050 \text{ Btu/ft}^3$ ($\sim 23,800 \text{ Btu/lb}_m$).
• LHV (Net Calorific Value): Assumes all water leaves the boiler chimney as vapor, without recovering latent heat. Natural gas LHV $\approx 900 - 950 \text{ Btu/ft}^3$.

Condensing vs. Non-Condensing Boiler Mechanics

Thermal Efficiency (%)
  100% +--------------------------------------------/ (Latent Heat Recovery Zone)
       |                                           /
   95% +------------------------------------------/    Condensing Boilers
       |                                         /     (Eff = 90% - 98%)
   90% +----------------------------------------/  
       |               Flue Gas Dew Point (~130°F)
   85% +---------------------+------------------ (Sensible-Only Cooling Zone)
       |                    /                    Non-Condensing Boilers (Eff = 80% - 85%)
   80% +-------------------/--------------------
       +-------------------+--------------------+------> Water Return Temperature (°F)
                          130°F                180°F

• Non-Condensing Boilers: Designed with minimum return water temperature $>140^\circ\text{F}$ to keep flue gas temperature above the acid dew point ($\sim 130^\circ\text{F}$), preventing corrosive condensate from degrading steel/cast-iron heat exchangers. Peak thermal efficiency is limited to $80% - 85%$.
• Condensing Boilers: Constructed from 316L stainless steel or cast aluminum to resist acidic condensate ($\text{pH } 3.0 - 5.0$). When return water is below $120^\circ\text{F}$ (ideally $80^\circ\text{F} - 100^\circ\text{F}$), flue gas water vapor condenses, boosting overall thermal efficiency to $92% - 98%$.

---

5. Worked Example: Comprehensive Cooling Tower Makeup & Blowdown Analysis

Problem: A campus central plant operates a $1,500 \text{-ton}$ water-cooled centrifugal chiller operating at full load with an efficiency of $0.56 \text{ kW/ton}$. The condenser water is circulated through an induced-draft cooling tower at $3.0 \text{ GPM/ton}$ entering the tower at $95.0^\circ\text{F}$ and leaving at $85.0^\circ\text{F}$. Water analysis reveals that city makeup water contains $120 \text{ ppm}$ of total dissolved solids (TDS), while the cooling tower water chemistry must be maintained at a maximum of $600 \text{ ppm}$ TDS to prevent calcium carbonate scaling. Assume latent heat of vaporization $h_{fg} = 1,050 \text{ Btu/lb}_m$ and drift loss is negligible.

Find:

1. The total heat rejection rate ($\dot{Q}_{tower}$) to the cooling tower in $\text{Btu/hr}$.
2. The evaporation rate ($E$) in $\text{GPM}$.
3. The Cycles of Concentration ($C$).
4. The required continuous blowdown rate ($B$) and makeup water rate ($M$) in $\text{GPM}$.

Step-by-Step Solution:

Step 1: Calculate total condenser heat rejection rate $\dot{Q}_{tower}$. $\dot{Q}_{evap} = 1,500 \text{ tons} \times 12,000 \frac{\text{Btu}}{\text{hr}\cdot\text{ton}} = 18,000,000 \frac{\text{Btu}}{\text{hr}}$ $\dot{W}_{comp} = 1,500 \text{ tons} \times 0.56 \frac{\text{kW}}{\text{ton}} \times 3,412.14 \frac{\text{Btu}}{\text{hr}\cdot\text{kW}} = 840.0 \text{ kW} \times 3,412.14 = 2,866,198 \frac{\text{Btu}}{\text{hr}}$ $\dot{Q}_{tower} = \dot{Q}_{evap} + \dot{W}_{comp} = 18,000,000 + 2,866,198 = 20,866,198 \frac{\text{Btu}}{\text{hr}}$

(Check via water loop: $\text{GPM}_{cw} = 1,500 \times 3.0 = 4,500 \text{ GPM}$. $\dot{Q} = 500 \times 4,500 \times (95 - 85) = 22,500,000 \text{ Btu/hr}$ at standard $10^\circ\text{F}$ tower rating). Use the exact heat rejection of $20,866,198 \text{ Btu/hr}$ for evaporation calculation.

Step 2: Calculate evaporation rate ($E$). $E = \frac{\dot{Q}_{tower}}{h_{fg} \times \rho_{water}} = \frac{20,866,198 \text{ Btu/hr}}{1,050 \text{ Btu/lb}_m \times 8.33 \text{ lb}_m/\text{gal} \times 60 \text{ min}/\text{hr}}$ $E = \frac{20,866,198}{524,790} = 39.76 \text{ GPM}$

(Check via rule-of-thumb: $E = 0.0008 \times 4,500 \text{ GPM} \times 10 = 36.0 \text{ GPM}$ using nominal cooling). Using exact thermodynamic duty: $E = 39.8 \text{ GPM}$.

Step 3: Calculate Cycles of Concentration ($C$). $C = \frac{\text{TDS}_{tower}}{\text{TDS}_{makeup}} = \frac{600 \text{ ppm}}{120 \text{ ppm}} = 5.0$

Step 4: Calculate Blowdown rate ($B$) and Makeup rate ($M$). $B = \frac{E}{C - 1} = \frac{39.76 \text{ GPM}}{5.0 - 1} = \frac{39.76}{4} = 9.94 \text{ GPM}$ $M = E + B = 39.76 + 9.94 = 49.70 \text{ GPM}$

(Check: $M = E \times \frac{C}{C - 1} = 39.76 \times \frac{5}{4} = 49.70 \text{ GPM}$).

---

6. NCEES Reference Handbook Navigation & Exam Traps

• Section 1.3 & 1.4: Combustion & Heat Transfer: Review stoichiometric air-fuel tables, heating values, and cooling tower definitions.
• Tower Ton Confusion: Remember that an evaporative cooling tower ton is nominally rated at $15,000 \text{ Btu/hr}$, not $12,000 \text{ Btu/hr}$, because it must reject both the building evaporator load and the compressor power.
• Cycles of Concentration Denominator: The blowdown formula is $B = \frac{E}{C - 1}$. Do not divide by $C$; dividing by $C$ rather than $C-1$ is a standard NCEES exam distractor trap!