Math Problem Statement

Examine the economics of night operation of the primary air system of an office fan coil air conditioning system to reduce the refrigeration energy. Assume that by operating the extract fan without the supply fan running, but with the air handling unit mixing dampers set on full fresh air, it is possible to ventilate the space at the rate of 11 s⁻¹ m⁻² (treated floor area) and when operating in this mode, the extract fan operates against a total pressure of 0.5 kPa and at an efficiency of 60 percent.

Solution

The image provided shows Example 1.1, which involves analyzing the economics of night operation of an air conditioning system in an office. The goal is to reduce the need for refrigeration energy by operating an extract fan under certain conditions. Here's a breakdown of the mathematical steps included in this example:

Assumptions:

  1. The air system operates at a rate of 11 s⁻¹ m⁻² (flow rate per unit treated floor area) against a total pressure of 0.5 kPa with an efficiency of 60%.
  2. The mechanical cooling plant requires 1 W of electrical input to provide 2 W of cooling during the day.

Problem Breakdown:

  1. Electrical Power to Operate the Fan: The power required is given by: Electrical Power=1.0W×0.5kPa0.6=0.83W m2\text{Electrical Power} = \frac{1.0 \, \text{W} \times 0.5 \, \text{kPa}}{0.6} = 0.83 \, \text{W m}^{-2}

  2. Heat Removed by Night Fan Operation (per degree rise): The formula used here is: Heat Removed=11s1×1C×358273+23C=1.22W m2K1\text{Heat Removed} = \frac{11 \, \text{s}^{-1} \times 1^\circ C \times 358}{273 + 23^\circ C} = 1.22 \, \text{W m}^{-2} \text{K}^{-1} This equation estimates the heat removed per degree of air temperature rise.

  3. Electrical Power to Operate Cooling Plant: To remove 1.22 W m⁻² of heat, the electrical power required is: Power=1.22W m22=0.61W m2\text{Power} = \frac{1.22 \, \text{W m}^{-2}}{2} = 0.61 \, \text{W m}^{-2}

  4. Air Temperature Rise for Precooling: The air temperature rise required for viable night operation is calculated using: Temperature Rise=0.83W m20.61W m2=1.36K\text{Temperature Rise} = \frac{0.83 \, \text{W m}^{-2}}{0.61 \, \text{W m}^{-2}} = 1.36 \, \text{K} This value is rounded to 1.5 K for practical purposes.

Key Insights:

  • The calculations show how much energy the system saves by using night cooling via fan operation rather than active refrigeration.
  • Efficiency and temperature rise are critical factors in determining the overall energy consumption.

Would you like a deeper explanation of any of these steps or further details on the physics behind the calculations?

Here are 5 related questions to expand on this:

  1. How does the efficiency of the fan affect the overall energy savings in cooling systems?
  2. Can this model be applied to other HVAC systems for different types of buildings?
  3. How would changes in external air temperature impact the performance of night cooling?
  4. What other methods could be used to optimize energy efficiency in HVAC systems?
  5. How can night cooling contribute to long-term cost savings in energy consumption?

Tip: When working with HVAC systems, always consider external conditions (temperature, humidity) and equipment efficiency, as these factors can drastically influence energy savings.

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Math Problem Analysis

Mathematical Concepts

Physics (Air conditioning energy systems)
Power and energy efficiency
Heat transfer

Formulas

Electrical power = (1.0 W × 0.5 kPa) / 0.6 = 0.83 W m⁻²
Heat removed = (11 s⁻¹ × 1°C × 358) / (273 + 23°C) = 1.22 W m⁻² K⁻¹
Electrical power for cooling = 1.22 W m⁻² (heat) = 0.61 W m⁻²
Air temperature rise = 0.83 W m⁻² / 0.61 W m⁻² = 1.36 K

Theorems

Principles of energy conservation
Heat transfer laws
Work-energy principle

Suitable Grade Level

Undergraduate (Engineering, Physics, Environmental Science)