By the end of this chapter, you will be able to:
These skills will help you analyze real-world engineering data accurately, making your work more reliable and efficient in the trade.
Refrigeration and Air Conditioning professionals in Kenya rely heavily on data analysis to optimize system performance, diagnose faults, and ensure energy efficiency. Understanding statistical measures such as mean, median, mode, and standard deviation is essential to interpret temperature readings, pressure variations, and system reliability data. These statistics provide insight into typical operating conditions and variability, which is critical for maintenance planning and quality control in refrigeration systems.
The mean is the average value of a data set, representing the central tendency of measurements such as temperature readings or pressure values over time. It is widely used in refrigeration and air conditioning to summarize system performance data.
The formula for the mean of \( n \) data points \( x_1, x_2, ..., x_n \) is
$$ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i $$
Example 1: A technician records compressor discharge pressures (in kPa) as 350, 360, 340, 355, and 365 during five tests. Find the mean pressure.
Given: \( x_1 = 350 \), \( x_2 = 360 \), \( x_3 = 340 \), \( x_4 = 355 \), \( x_5 = 365 \), \( n = 5 \)
$$ \bar{x} = \frac{1}{5} (350 + 360 + 340 + 355 + 365) $$$$ = \frac{1}{5} (1770) $$$$ = 354 \text{ kPa} $$
Answer: 354 kPa
Example 2: Temperature readings (°C) inside a cold room are 2.5, 3.0, 2.8, 3.2, 2.9, and 3.1. Calculate the mean temperature.
Given: \( x_1 = 2.5 \), \( x_2 = 3.0 \), \( x_3 = 2.8 \), \( x_4 = 3.2 \), \( x_5 = 2.9 \), \( x_6 = 3.1 \), \( n = 6 \)
$$ \bar{x} = \frac{1}{6} (2.5 + 3.0 + 2.8 + 3.2 + 2.9 + 3.1) $$$$ = \frac{1}{6} (17.5) $$$$ = 2.9167 \text{ °C} $$
Answer: 2.92 °C
Example 3: A refrigeration engineer measures the voltage (V) supplied to a compressor motor over 8 hours as 220, 222, 219, 221, 223, 220, 218, and 224 volts. Find the mean voltage.
Given: \( x = \{220, 222, 219, 221, 223, 220, 218, 224\} \), \( n=8 \)
$$ \bar{x} = \frac{1}{8} (220 + 222 + 219 + 221 + 223 + 220 + 218 + 224) $$$$ = \frac{1}{8} (1767) $$$$ = 220.875 \text{ V} $$
Answer: 220.88 V
Example 4: In a maintenance check, the refrigerant pressure (kPa) in 10 units is recorded as 800, 790, 805, 795, 810, 785, 800, 795, 790, and 805. Calculate the mean pressure.
Given \( n=10 \), data as above.
$$ \bar{x} = \frac{1}{10} (800 + 790 + 805 + 795 + 810 + 785 + 800 + 795 + 790 + 805) $$$$ = \frac{1}{10} (7975) $$$$ = 797.5 \text{ kPa} $$
Answer: 797.5 kPa
Example 5: A technician measures the airflow velocity (m/s) at six points in a duct: 4.5, 4.7, 4.6, 4.8, 4.9, 4.4. Determine the mean airflow velocity.
Given \( n=6 \), data as above.
$$ \bar{x} = \frac{1}{6} (4.5 + 4.7 + 4.6 + 4.8 + 4.9 + 4.4) $$$$ = \frac{1}{6} (27.9) $$$$ = 4.65 \text{ m/s} $$
Answer: 4.65 m/s
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Create a free accountA refrigeration technician recorded the temperatures (in °C) inside a cold room over 5 days as follows: 2, 3, 2, 4, 3. Calculate the mean temperature. (3 marks)
The following pressures (in kPa) were measured at different points in an air conditioning system: 100, 105, 98, 102, 107, 101. Find the median pressure. (3 marks)
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