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Cooling Load with Dehumidification Calculator

Cooling Load with Dehumidification

1. What is a Cooling Load with Dehumidification Calculator?

Definition: This calculator computes the cooling load (\( \dot{q} \)) in a cooling and dehumidification process, accounting for both sensible heat removal (temperature drop) and latent heat removal (moisture condensation).

Purpose: It is used in HVAC systems to determine the total cooling required to both lower the air temperature and remove moisture, aiding in the design and sizing of air conditioning systems.

2. How Does the Calculator Work?

The calculator uses the following formula for the cooling load:

Cooling Load: \[ \dot{q} = \dot{m}_a \left( (h_1 - h_2) - (W_1 - W_2) h_w \right) \]

Where:

  • \( \dot{q} \): Cooling load (Btu/hr, convertible to kW)
  • \( \dot{m}_a \): Mass flow rate of dry air (lb_m/hr, kg/hr)
  • \( h_1 \): Enthalpy of entering air (Btu/lb_m, kJ/kg)
  • \( h_2 \): Enthalpy of leaving air (Btu/lb_m, kJ/kg)
  • \( W_1 \): Entering humidity ratio (lb_v/lb_da, kg_v/kg_da)
  • \( W_2 \): Leaving humidity ratio (lb_v/lb_da, kg_v/kg_da)
  • \( h_w \): Enthalpy of condensate water (Btu/lb_v, kJ/kg)

Unit Conversions:

  • Mass Flow Rate (\( \dot{m}_a \)): lb_m/hr, kg/hr (1 kg/hr = 2.20462 lb_m/hr)
  • Enthalpies (\( h_1 \), \( h_2 \), \( h_w \)): Btu/lb_m or Btu/lb_v, kJ/kg (1 kJ/kg = 0.429923 Btu/lb_m or Btu/lb_v)
  • Humidity Ratios (\( W_1 \), \( W_2 \)): lb_v/lb_da, kg_v/kg_da (numerically the same, label changes for clarity)
  • Cooling Load (\( \dot{q} \)): Btu/hr, kW (1 Btu/hr = 0.000293071 kW)

Steps:

  • Enter the mass flow rate (\( \dot{m}_a \)), entering and leaving enthalpies (\( h_1 \), \( h_2 \)), entering and leaving humidity ratios (\( W_1 \), \( W_2 \)), and the enthalpy of condensate water (\( h_w \)), and select their units.
  • Convert mass flow rate to lb_m/hr, enthalpies to Btu/lb_m (or Btu/lb_v for \( h_w \)), and ensure humidity ratios are consistent.
  • Calculate the cooling load using the formula.
  • Convert the result to the selected unit (Btu/hr or kW).
  • Display the result with 5 decimal places, or in scientific notation if the value is greater than 10,000 or less than 0.00001.

3. Importance of Cooling Load with Dehumidification Calculation

Calculating the cooling load in a cooling and dehumidification process is crucial for:

  • HVAC Design: Determines the total cooling required to both lower air temperature and remove moisture, aiding in the sizing of cooling and dehumidification equipment.
  • Energy Efficiency: Helps optimize energy usage by ensuring the system is appropriately sized for both sensible and latent loads.
  • System Performance: Ensures the system can achieve the desired temperature and humidity levels for comfort and process requirements.

4. Using the Calculator

Examples:

  • Example 1: For \( \dot{m}_a = 21687 \, \text{lb_m/hr} \), \( h_1 = 31.4 \, \text{Btu/lb_m} \), \( h_2 = 22.2 \, \text{Btu/lb_m} \), \( W_1 = 0.0112 \, \text{lb_v/lb_da} \), \( W_2 = 0.0082 \, \text{lb_v/lb_da} \), \( h_w = 23.0 \, \text{Btu/lb_v} \), cooling load in Btu/hr:
    • \( \dot{q} = 21687 \times ( (31.4 - 22.2) - (0.0112 - 0.0082) \times 23.0 ) \)
    • \( \dot{q} = 21687 \times ( 9.2 - 0.003 \times 23.0 ) \)
    • \( \dot{q} = 21687 \times ( 9.2 - 0.069 ) \approx 21687 \times 9.131 \approx 198023.99700 \)
    • Since 198023.99700 > 10000, use scientific notation: \( 1.98024e+5 \)
  • Example 2: For \( \dot{m}_a = 1000 \, \text{kg/hr} \), \( h_1 = 70 \, \text{kJ/kg} \), \( h_2 = 50 \, \text{kJ/kg} \), \( W_1 = 0.015 \, \text{kg_v/kg_da} \), \( W_2 = 0.010 \, \text{kg_v/kg_da} \), \( h_w = 50 \, \text{kJ/kg} \), cooling load in kW:
    • Convert: \( \dot{m}_a = 1000 \times 2.20462 = 2204.62 \, \text{lb_m/hr} \), \( h_1 = 70 \times 0.429923 = 30.0946 \, \text{Btu/lb_m} \), \( h_2 = 50 \times 0.429923 = 21.4962 \, \text{Btu/lb_m} \), \( h_w = 50 \times 0.429923 = 21.4962 \, \text{Btu/lb_v} \)
    • \( \dot{q} = 2204.62 \times ( (30.0946 - 21.4962) - (0.015 - 0.010) \times 21.4962 ) \)
    • \( \dot{q} = 2204.62 \times ( 8.5984 - 0.005 \times 21.4962 ) \)
    • \( \dot{q} = 2204.62 \times ( 8.5984 - 0.107481 ) \approx 2204.62 \times 8.490919 \approx 18720.54538 \, \text{Btu/hr} \)
    • Convert to kW: \( 18720.54538 \times 0.000293071 = 5.48641 \, \text{kW} \)
    • Since 5.48641 < 10000 and > 0.00001, display with 5 decimal places: \( 5.48641 \)
  • Example 3: For \( \dot{m}_a = 0.01 \, \text{lb_m/hr} \), \( h_1 = 30 \, \text{Btu/lb_m} \), \( h_2 = 20 \, \text{Btu/lb_m} \), \( W_1 = 0.012 \, \text{lb_v/lb_da} \), \( W_2 = 0.008 \, \text{lb_v/lb_da} \), \( h_w = 20 \, \text{Btu/lb_v} \), cooling load in Btu/hr:
    • \( \dot{q} = 0.01 \times ( (30 - 20) - (0.012 - 0.008) \times 20 ) \)
    • \( \dot{q} = 0.01 \times ( 10 - 0.004 \times 20 ) \)
    • \( \dot{q} = 0.01 \times ( 10 - 0.08 ) \approx 0.01 \times 9.92 = 0.09920 \)
    • Since 0.09920 < 0.00001, use scientific notation: \( 9.92000e-2 \)

5. Frequently Asked Questions (FAQ)

Q: What is a cooling and dehumidification process in HVAC?
A: A cooling and dehumidification process in HVAC involves lowering the air temperature and removing moisture, typically using a cooling coil where air is cooled below its dew point, causing condensation.

Q: Why does the formula account for both sensible and latent heat?
A: The formula accounts for sensible heat (temperature drop, \( h_1 - h_2 \)) and latent heat (moisture removal, \( (W_1 - W_2) h_w \)) because both contribute to the total cooling load in a dehumidification process.

Q: How do I determine the enthalpy and humidity ratio values if I don’t have a psychrometric chart?
A: Enthalpy (\( h_1 \), \( h_2 \)) and humidity ratios (\( W_1 \), \( W_2 \)) can be calculated using psychrometric equations or software, based on temperature, pressure, and humidity conditions, or approximated using standard air properties.

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