3.2 Chart Reading & Psychrometric Equations at Sea Level vs. Altitude (5,000 ft)

3.2 Chart Reading & Psychrometric Equations at Sea Level vs. Altitude (5,000 ft)

In standard HVAC design, engineers frequently rely on standard sea-level psychrometric charts and rule-of-thumb thermal equations. However, on the PE Mechanical exam, questions involving high altitude (such as 5,000 ft in Denver, CO) are classic test items designed to verify that candidates understand the underlying physics rather than blindly plugging numbers into standard formulas.

1. Atmospheric Pressure Variation with Altitude

The U.S. Standard Atmosphere model defines atmospheric barometric pressure $P$ as a function of altitude $Z$ (in feet) above sea level up to $36,000\text{ ft}$:

$P = 14.696 \left( 1 - 6.8754 \times 10^{-6} Z \right)^{5.2559} \quad [\text{psia}]$

$\text{or} \quad P = 29.921 \left( 1 - 6.8754 \times 10^{-6} Z \right)^{5.2559} \quad [\text{in. Hg}]$

Barometric Pressures and Densities Across Key Elevations

2. Impact of Altitude on Psychrometric Properties

When altitude increases and total barometric pressure $P$ decreases:

1. Humidity Ratio ($W$): $W = 0.62198 \frac{p_v}{P - p_v}$ For a given dew-point temperature (fixed partial vapor pressure $p_v$), as $P$ decreases, the denominator $(P - p_v)$ becomes smaller. Therefore, the humidity ratio $W$ is significantly higher at altitude for the same dry-bulb and dew-point temperatures.

2. Specific Volume ($v$): $v = \frac{R_{da} T}{P - p_v}$ As $P$ decreases, specific volume $v$ expands (air becomes less dense). At $5,000\text{ ft}$, $v$ is approximately $20%$ greater than at sea level for the same temperature.

3. Specific Enthalpy ($h$): $h = 0.240 T_{db} + W (1061 + 0.444 T_{db})$ Because $W$ is higher at altitude for the same relative humidity and dry bulb, the specific enthalpy of moist air is higher at altitude.

4. Saturation Curve Shift: On an altitude psychrometric chart (such as ASHRAE Chart #2 for $5,000\text{ ft}$), the saturation curve shifts upward and to the left compared to the sea-level chart (Chart #1).

3. Derivation of Heat Transfer Rate Equations: Sea Level vs. Altitude

All continuous-flow thermal energy equations originate from First Law mass and energy rate balances:

$\dot{q} = \dot{m}_{da} \Delta h = (\rho \times 60 \times \text{CFM}) \times \Delta h$

Sensible Heat Transfer Rate ($\dot{q}_s$)

$\dot{q}_s = \dot{m}_{da} c_{p,ma} \Delta T = (60 \times \rho \times \text{CFM}) \times c_{p,ma} \times \Delta T$

• At Sea Level Standard Conditions:

  • $\rho_{std} = 0.0750\text{ lb}_m/\text{ft}^3$
  • $c_{p,ma} = c_{p,da} + W c_{p,v} \approx 0.240 + (0.010)(0.444) = 0.244\text{ Btu}/(\text{lb}_m\cdot^\circ\text{F})$
  • Constant $C_s = 60 \times 0.0750 \times 0.240 = 1.08\text{ Btu}/(\text{hr}\cdot\text{CFM}\cdot^\circ\text{F})$ (or $1.10$ using $0.244$) $\dot{q}_s = 1.08 \times \text{CFM} \times (T_2 - T_1) \quad [\text{Btu/hr}]$

• At 5,000 ft Altitude (Denver):

  • $\rho_{alt} = 0.0624\text{ lb}_m/\text{ft}^3$
  • Constant $C_{s,alt} = 60 \times 0.0624 \times 0.240 = 0.8986 \approx 0.90\text{ Btu}/(\text{hr}\cdot\text{ACFM}\cdot^\circ\text{F})$ $\dot{q}_s = 0.90 \times \text{ACFM} \times (T_2 - T_1) \quad [\text{Btu/hr}]$

Latent Heat Transfer Rate ($\dot{q}_l$)

$\dot{q}_l = \dot{m}_{da} h_{fg} \Delta W = (60 \times \rho \times \text{CFM}) \times h_{fg} \times \Delta W$

• At Sea Level Standard Conditions ($h_{fg} \approx 1060\text{ Btu/lb}_w$):

  • Constant $C_l = 60 \times 0.0750 \times 1060 = 4770 \approx 4840\text{ Btu}\cdot\text{lb}_{da}/(\text{hr}\cdot\text{CFM}\cdot\text{lb}_w)$ $\dot{q}_l = 4840 \times \text{CFM} \times (W_2 - W_1) \quad [\text{Btu/hr}, W \text{ in lb/lb}]$
  • In grains of moisture ($1\text{ lb} = 7000\text{ grains}$): $\dot{q}_l = \frac{4840}{7000} \times \text{CFM} \times \Delta W_{\text{grains}} = 0.69 \times \text{CFM} \times \Delta W_{\text{grains}} \quad [\text{Btu/hr}]$

• At 5,000 ft Altitude:

  • Constant $C_{l,alt} = 60 \times 0.0624 \times 1060 = 3968 \approx 4000\text{ Btu}\cdot\text{lb}_{da}/(\text{hr}\cdot\text{ACFM}\cdot\text{lb}_w)$ $\dot{q}_l = 4000 \times \text{ACFM} \times (W_2 - W_1) \quad [\text{Btu/hr}, W \text{ in lb/lb}]$ $\dot{q}_l = 0.57 \times \text{ACFM} \times \Delta W_{\text{grains}} \quad [\text{Btu/hr}]$

Total Heat Transfer Rate ($\dot{q}_t$)

$\dot{q}_t = \dot{m}_{da} \Delta h = (60 \times \rho \times \text{CFM}) \times \Delta h$

• At Sea Level: $C_t = 60 \times 0.0750 = 4.5\text{ lb}_m/(\text{hr}\cdot\text{CFM})$ $\dot{q}_t = 4.5 \times \text{CFM} \times (h_2 - h_1) \quad [\text{Btu/hr}]$

• At 5,000 ft Altitude: $C_{t,alt} = 60 \times 0.0624 = 3.74\text{ lb}_m/(\text{hr}\cdot\text{ACFM})$ $\dot{q}_t = 3.74 \times \text{ACFM} \times (h_2 - h_1) \quad [\text{Btu/hr}]$

4. Airflow Conversion: Standard CFM (SCFM) vs. Actual CFM (ACFM)

In HVAC equipment specifications, airflow can be expressed in two ways:

• Standard CFM (SCFM): Volumetric flow referenced to standard sea-level air density ($\rho_{std} = 0.075\text{ lb/ft}^3$). SCFM is directly proportional to mass flow rate: $\dot{m}_{da} = 60 \times 0.075 \times \text{SCFM} = 4.5 \times \text{SCFM} \text{ [lb/hr]}$.
• Actual CFM (ACFM): The physical volume of air moving through the duct or fan per minute at actual operating pressure and temperature: $\dot{m}{da} = 60 \times \rho{actual} \times \text{ACFM}$.

Equating dry air mass flow rates:

$\text{ACFM} = \text{SCFM} \times \left( \frac{\rho_{std}}{\rho_{actual}} \right) = \text{SCFM} \times \left( \frac{P_0}{P} \right) \times \left( \frac{T}{T_0} \right)$

Key Exam Rule of Thumb: For a given sensible cooling load ($q_s$) and supply air temperature differential ($\Delta T$), a fan at 5,000 ft must deliver 20% MORE actual CFM (ACFM) than at sea level to achieve the same cooling capacity: $\frac{\text{ACFM}_{5000}}{\text{CFM}_{sl}} = \frac{1.08}{0.90} = 1.20$

5. Step-by-Step Worked Example: High-Altitude Cooling Coil Sizing

Problem Statement

A commercial building located in Denver, Colorado (elevation $5,000\text{ ft}$, barometric pressure $P = 12.23\text{ psia}$, air density $\rho = 0.0624\text{ lb/ft}^3$) has a calculated space sensible cooling load of $q_s = 180,000\text{ Btu/hr}$ and a space latent cooling load of $q_l = 40,000\text{ Btu/hr}$.

The indoor design condition is $75.0^\circ\text{F}$ DB and $50%\text{ RH}$. The supply air temperature leaving the AHU is $55.0^\circ\text{F}$ DB.

Calculate:

1. Required supply airflow in Actual CFM (ACFM) at $5,000\text{ ft}$
2. What airflow would have been required if the identical building were located at sea level
3. The total cooling coil load in tons of refrigeration at $5,000\text{ ft}$, assuming entering coil air enthalpy $h_1 = 33.2\text{ Btu/lb}{da}$ and leaving coil air enthalpy $h_2 = 23.1\text{ Btu/lb}{da}$

Solution Steps

Step 1: Calculate sensible heat constant at 5,000 ft $C_{s,alt} = 60 \times \rho_{alt} \times c_p = 60 \times 0.0624 \times 0.240 = 0.8986 \approx 0.90\text{ Btu}/(\text{hr}\cdot\text{ACFM}\cdot^\circ\text{F})$

Step 2: Calculate required supply airflow at 5,000 ft $\Delta T = T_{room} - T_{supply} = 75.0^\circ\text{F} - 55.0^\circ\text{F} = 20.0^\circ\text{F}$ $\text{ACFM} = \frac{q_s}{C_{s,alt} \times \Delta T} = \frac{180,000}{0.8986 \times 20.0} = \frac{180,000}{17.971} = 10,016\text{ ACFM} \approx 10,000\text{ ACFM}$

Step 3: Calculate required airflow at sea level $\text{CFM}_{sl} = \frac{q_s}{1.08 \times \Delta T} = \frac{180,000}{1.08 \times 20.0} = \frac{180,000}{21.60} = 8,333\text{ CFM}$ $\text{Airflow Ratio} = \frac{10,016}{8,333} = 1.202 \quad (+20.2\%\text{ increase})$

Step 4: Calculate total cooling coil capacity at 5,000 ft $C_{t,alt} = 60 \times \rho_{alt} = 60 \times 0.0624 = 3.744\text{ lb}_m/(\text{hr}\cdot\text{ACFM})$ $\dot{q}_{total} = C_{t,alt} \times \text{ACFM} \times (h_1 - h_2)$ $\dot{q}_{total} = 3.744 \times 10,016 \times (33.2 - 23.1) = 3.744 \times 10,016 \times 10.1 = 378,749\text{ Btu/hr}$ $\text{Tons of Refrigeration} = \frac{378,749\text{ Btu/hr}}{12,000\text{ Btu/hr}\cdot\text{ton}} = 31.56\text{ tons}$ (Note: If calculated using standard sea level 4.5 factor incorrectly: $4.5 \times 10,016 \times 10.1 = 455,227\text{ Btu/hr} = 37.9\text{ tons}$, resulting in a $20%$ equipment oversizing error!)