Refrigeration and air conditioning systems rely heavily on precise mathematical calculations to optimize performance and ensure safety. Differentiation, a fundamental concept in calculus, is essential for analyzing rates of change in temperature, pressure, and other variables critical to system design and operation. This chapter introduces differentiation and demonstrates its application in solving real-world problems faced by refrigeration and air conditioning professionals in Kenya.

3.1 Introduction to differentiation

Differentiation is the process of finding the rate at which a function changes at any point, which is crucial in understanding dynamic behaviors of physical systems in refrigeration and air conditioning. It enables engineers to analyze how variables such as temperature or pressure vary over time or distance, thus facilitating system optimization and fault detection. This section covers the fundamental definition of differentiation, basic rules, and their application through worked examples relevant to refrigeration and air conditioning contexts.

3.1.1 Concept and Definition of Differentiation

Differentiation measures how a function \(f(x)\) changes as its input \(x\) changes, giving the instantaneous rate of change or slope of the function at any point. The derivative of \(f(x)\) with respect to \(x\) is defined as the limit of the average rate of change over an interval as the interval approaches zero.

The derivative \(f'(x)\) is given by:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Worked Examples

Example 1: The temperature \(T\) in a refrigeration chamber varies with time \(t\) in minutes according to \(T(t) = 5t^2 + 3t + 20\). Find the rate of change of temperature at \(t = 4\) minutes.

Given: \(T(t) = 5t^2 + 3t + 20\), \(t=4\) min

$$ \frac{dT}{dt} = \lim_{h \to 0} \frac{5(t+h)^2 + 3(t+h) + 20 - (5t^2 + 3t + 20)}{h} $$

Calculate derivative using power rule:

$$ \frac{dT}{dt} = 10t + 3 $$

Substitute \(t=4\):

$$ \frac{dT}{dt} = 10 \times 4 + 3 = 40 + 3 = 43 $$

Answer: The rate of change of temperature at 4 minutes is 43 °C/min

Example 2: The pressure \(P\) in a compressor varies with volume \(V\) as \(P(V) = \frac{100}{V}\). Find the rate of change of pressure when \(V = 5 \, m^3\).

Given: \(P(V) = 100 V^{-1}\), \(V=5 \, m^3\)

$$ \frac{dP}{dV} = \lim_{h \to 0} \frac{100 (V + h)^{-1} - 100 V^{-1}}{h} $$

Using the power rule for negative exponents:

$$ \frac{dP}{dV} = -100 V^{-2} $$

Substitute \(V=5\):

$$ \frac{dP}{dV} = -100 \times 5^{-2} = -100 \times \frac{1}{25} = -4 $$

Answer: The rate of change of pressure at volume 5 m³ is -4 kPa/m³

Example 3: The heat transfer rate \(Q\) in a condenser is given by \(Q(x) = 3x^3 - 2x^2 + 4\), where \(x\) is the thickness of the condenser plate in cm. Find the rate of change of heat transfer rate at \(x=2\) cm.

Given: \(Q(x) = 3x^3 - 2x^2 + 4\), \(x=2\) cm

$$ \frac{dQ}{dx} = 9x^2 - 4x $$

Substitute \(x=2\):

$$ \frac{dQ}{dx} = 9 \times 2^2 - 4 \times 2 = 9 \times 4 - 8 = 36 - 8 = 28 $$

Answer: The rate of change of heat transfer rate at 2 cm thickness is 28 units/cm

Example 4: The efficiency \(\eta\) of an air conditioning unit depends on the ambient temperature \(T\) as \(\eta(T) = 50 - 0.5T^2\). Find the rate of change of efficiency at \(T = 6^{\circ}C\).

Given: \(\eta(T) = 50 - 0.5T^2\), \(T=6^\circ C\)

$$ \frac{d\eta}{dT} = -0.5 \times 2T = -T $$

Substitute \(T=6\):

$$ \frac{d\eta}{dT} = -6 $$

Answer: The rate of change of efficiency at 6°C is -6 %/°C

Example 5: The volume \(V\) of refrigerant gas in a cylinder varies with pressure \(p\) as \(V(p) = 20 \ln(p + 1)\). Find the rate of change of volume at \(p=4 \, atm\).

Given: \(V(p) = 20 \ln(p + 1)\), \(p=4\)

$$ \frac{dV}{dp} = 20 \times \frac{1}{p+1} $$

Substitute \(p=4\):

$$ \frac{dV}{dp} = 20 \times \frac{1}{5} = 4 $$

Answer: The rate of change of volume at 4 atm is 4 m³/atm

3.1.2 Basic Differentiation Rules

The differentiation process is simplified by applying standard rules that allow quick calculation of derivatives for many functions common in refrigeration and air conditioning engineering.

The main rules are:

• Power Rule: \(\frac{d}{dx} x^n = n x^{n-1}\)
• Constant Rule: \(\frac{d}{dx} c = 0\)
• Constant Multiple Rule: \(\frac{d}{dx} [c f(x)] = c \frac{d}{dx} f(x)\)
• Sum Rule: \(\frac{d}{dx} [f(x) + g(x)] = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)\)

Worked Examples

Example 1: Find the derivative of the function \(f(x) = 4x^3 - 2x + 7\).

Given: \(f(x) = 4x^3 - 2x + 7\)

Apply power and constant rules:

$$ \frac{df}{dx} = 4 \times 3 x^{3-1} - 2 \times 1 x^{1-1} + 0 $$

$$ \frac{df}{dx} = 12x^2 - 2 $$

Answer: \(\frac{df}{dx} = 12x^2 - 2\)

Example 2: The velocity \(v(t)\) of refrigerant flow is given by \(v(t) = 6t^4 + 5\). Find \(\frac{dv}{dt}\).

Given: \(v(t) = 6t^4 + 5\)

$$ \frac{dv}{dt} = 6 \times 4 t^{4-1} + 0 = 24 t^3 $$

Answer: \(\frac{dv}{dt} = 24 t^3\)

Example 3: The pressure \(P(x) = 8x^2 + 3x - 5\). Find \(\frac{dP}{dx}\).

Given: \(P(x) = 8x^2 + 3x - 5\)

$$ \frac{dP}{dx} = 8 \times 2 x^{2-1} + 3 \times 1 x^{1-1} + 0 = 16x + 3 $$

Answer: \(\frac{dP}{dx} = 16x + 3\)

Example 4: Find the derivative of \(Q(t) = 10 - 4t^5 + 7t\).

Given: \(Q(t) = 10 - 4t^5 + 7t\)

$$ \frac{dQ}{dt} = 0 - 4 \times 5 t^{5-1} + 7 \times 1 t^{1-1} = -20 t^4 + 7 $$

Answer: \(\frac{dQ}{dt} = -20 t^4 + 7\)

Example 5: The temperature function is \(T(x) = 3x^3 + 2x^2 + x + 1\). Find \(\frac{dT}{dx}\).

Given: \(T(x) = 3x^3 + 2x^2 + x + 1\)

$$ \frac{dT}{dx} = 3 \times 3 x^{3-1} + 2 \times 2 x^{2-1} + 1 \times 1 x^{1-1} + 0 = 9x^2 + 4x + 1 $$

Answer: \(\frac{dT}{dx} = 9x^2 + 4x + 1\)

3.1.3 Differentiation of Common Functions in Refrigeration and Air Conditioning

Many functions encountered in refrigeration and air conditioning involve exponential, logarithmic, and trigonometric forms. Knowing how to differentiate these is vital for system analysis.

Key differentiation formulas:

• Exponential: \(\frac{d}{dx} e^{ax} = a e^{ax}\)
• Natural Logarithm: \(\frac{d}{dx} \ln x = \frac{1}{x}\)
• Sine: \(\frac{d}{dx} \sin x = \cos x\)
• Cosine: \(\frac{d}{dx} \cos x = -\sin x\)

Worked Examples

Example 1: Find \(\frac{d}{dt} e^{0.1t}\) where \(t\) is time in minutes.

Given: \(f(t) = e^{0.1t}\)

$$ \frac{df}{dt} = 0.1 e^{0.1t} $$

Answer: \(\frac{df}{dt} = 0.1 e^{0.1t}\)

Example 2: Differentiate \(P(x) = \ln(3x + 2)\) with respect to \(x\).

Given: \(P(x) = \ln(3x + 2)\)

$$ \frac{dP}{dx} = \frac{1}{3x + 2} \times 3 = \frac{3}{3x + 2} $$

Answer: \(\frac{dP}{dx} = \frac{3}{3x + 2}\)

Example 3: Find the derivative of \(Q(\theta) = 5 \sin \theta\), where \(\theta\) is in radians.

Given: \(Q(\theta) = 5 \sin \theta\)

$$ \frac{dQ}{d\theta} = 5 \cos \theta $$

Answer: \(\frac{dQ}{d\theta} = 5 \cos \theta\)

Example 4: Differentiate \(R(\theta) = 7 \cos (2\theta)\).

Given: \(R(\theta) = 7 \cos (2\theta)\)

$$ \frac{dR}{d\theta} = 7 \times (-\sin (2\theta)) \times 2 = -14 \sin (2\theta) $$

Answer: \(\frac{dR}{d\theta} = -14 \sin (2\theta)\)

Example 5: The refrigerant pressure varies as \(P(t) = 50 e^{-0.05 t}\). Find \(\frac{dP}{dt}\) at \(t=10\) seconds.

Given: \(P(t) = 50 e^{-0.05 t}\), \(t=10\)

$$ \frac{dP}{dt} = 50 \times (-0.05) e^{-0.05 t} = -2.5 e^{-0.05 t} $$

Substitute \(t=10\):

$$ \frac{dP}{dt} = -2.5 e^{-0.5} = -2.5 \times 0.6065 = -1.516 $$

Answer: The rate of change of pressure at 10 seconds is approximately -1.52 kPa/s

3.1.4 Practical Application: Rate of Change in Temperature and Pressure in Refrigeration Systems

In refrigeration, understanding how temperature and pressure change over time or distance within components is critical for system efficiency and safety. Differentiation allows calculation of these rates, enabling engineers to predict system behavior and make necessary adjustments.

The rate of change of a variable \(y\) with respect to \(x\) is given by the derivative \(\frac{dy}{dx}\). For example, if temperature \(T(t)\) changes with time \(t\), \(\frac{dT}{dt}\) represents how quickly temperature varies at any instant.

Worked Examples

Example 1: In a refrigeration evaporator, temperature varies as \(T(t) = 25 - 3 e^{-0.2 t}\). Find the rate of temperature change at \(t=5\) minutes.

Given: \(T(t) = 25 - 3 e^{-0.2 t}\), \(t=5\)

$$ \frac{dT}{dt} = -3 \times (-0.2) e^{-0.2 t} = 0.6 e^{-0.2 t} $$

Substitute \(t=5\):

$$ \frac{dT}{dt} = 0.6 e^{-1} = 0.6 \times 0.3679 = 0.221 $$

Answer: The temperature is increasing at 0.221 °C/min at 5 minutes

Example 2: Pressure inside a compressor varies over time as \(P(t) = 100 - 15 \ln(t + 1)\). Find \(\frac{dP}{dt}\) at \(t=4\) minutes.

Given: \(P(t) = 100 - 15 \ln(t + 1)\), \(t=4\)

$$ \frac{dP}{dt} = -15 \times \frac{1}{t+1} = -\frac{15}{t+1} $$

Substitute \(t=4\):

$$ \frac{dP}{dt} = -\frac{15}{5} = -3 $$

Answer: Pressure decreases at 3 kPa/min at 4 minutes

Example 3: The heat flow rate \(Q(x)\) through a pipe varies with distance \(x\) as \(Q(x) = 4x^3 - 6x^2 + 2x\). Find the rate of change of heat flow at \(x=3\) meters.

Given: \(Q(x) = 4x^3 - 6x^2 + 2x\), \(x=3\)

$$ \frac{dQ}{dx} = 12x^2 - 12x + 2 $$

Substitute \(x=3\):

$$ \frac{dQ}{dx} = 12 \times 9 - 12 \times 3 + 2 = 108 - 36 + 2 = 74 $$

Answer: Heat flow rate increases by 74 units per meter at 3 meters

Example 4: The refrigerant volume \(V(t)\) changes with time as \(V(t) = 7 \sin(0.5 t)\). Find \(\frac{dV}{dt}\) at \(t= \pi\) seconds.

Given: \(V(t) = 7 \sin(0.5 t)\), \(t= \pi\)

$$ \frac{dV}{dt} = 7 \times 0.5 \cos(0.5 t) = 3.5 \cos(0.5 t) $$

Substitute \(t= \pi\):

$$ \frac{dV}{dt} = 3.5 \cos(0.5 \times \pi) = 3.5 \cos(\frac{\pi}{2}) = 3.5 \times 0 = 0 $$

Answer: The rate of change of volume at \(t=\pi\) seconds is 0 m³/s

Example 5: The temperature of a refrigerant decreases with time according to \(T(t) = 40 e^{-0.1 t} + 5\). Calculate the rate of temperature change at \(t=10\) minutes.

Given: \(T(t) = 40 e^{-0.1 t} + 5\), \(t=10\)

$$ \frac{dT}{dt} = 40 \times (-0.1) e^{-0.1 t} = -4 e^{-0.1 t} $$

Substitute \(t=10\):

$$ \frac{dT}{dt} = -4 e^{-1} = -4 \times 0.3679 = -1.4716 $$

Answer: Temperature is decreasing at approximately 1.47 °C/min at 10 minutes

Chapter Summary

This chapter introduced the concept of differentiation, focusing on its fundamental principles and techniques for finding the derivative of functions. It explored various applications of differentiation such as calculating rates of change and solving practical engineering problems. The chapter then covered differentiation of parametric equations, providing methods to find derivatives when variables are expressed in terms of a third parameter. Implicit differentiation was explained to handle functions defined implicitly rather than explicitly. Logarithmic differentiation was presented as an effective tool for differentiating complex functions involving products, quotients, or powers. The concept of total differential was introduced along with its use in estimating small changes and understanding rates of change in multivariable contexts. Finally, the chapter examined the determination of maxima, minima, and saddle points for functions of two variables, essential for optimization problems in engineering.