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Integration is a fundamental mathematical tool widely used in Refrigeration and Air Conditioning engineering to analyse systems involving rates of change, areas under curves, and accumulated quantities. This chapter focuses on applying integration techniques to solve practical problems such as calculating volumes of irregular shapes, determining work done by compressors, and evaluating heat transfer rates. Mastery of integration allows professionals to design and optimize components like evaporators, condensers, and ducting systems with precision and efficiency. The following topics will develop your ability to apply integration methods to real-world refrigeration and air conditioning scenarios.
Integration enables engineers in refrigeration and air conditioning to calculate quantities that are not directly measurable by simple formulas, such as the volume of complex-shaped refrigerant tanks or the work done in thermodynamic cycles. This section develops problem-solving skills using integration to analyze physical quantities relevant to the field.
Integration is used to find the area under performance curves such as pressure-volume diagrams or temperature-enthalpy charts, which are common in refrigeration cycle analysis. The area often represents work, heat transfer, or energy quantities that are essential for system design.
The fundamental formula for area under a curve \(y = f(x)\) between \(x = a\) and \(x = b\) is:
$$ A = \int_{a}^{b} f(x) \, dx $$
Example 1:
Calculate the area under the pressure-volume curve \(P(v) = 5v + 2\) kPa between volumes 1 m³ and 3 m³ for a refrigeration compressor.
Given: \(P(v) = 5v + 2\) kPa, \(v = 1\) to \(3\) m³
$$ A = \int_{1}^{3} (5v + 2) dv $$
$$ = [ \frac{5v^{2}}{2} + 2v ]_{1}^{3} $$
$$ = ( \frac{5 \times 3^{2}}{2} + 2 \times 3 ) - ( \frac{5 \times 1^{2}}{2} + 2 \times 1 ) $$
$$ = ( \frac{5 \times 9}{2} + 6 ) - ( \frac{5 \times 1}{2} + 2 ) $$
$$ = (22.5 + 6) - (2.5 + 2) = 28.5 - 4.5 = 24 \text{ kPa·m}^3 $$
Answer: 24 kPa·m³
Example 2:
Determine the total heat transferred when the heat transfer rate varies with time as \(Q(t) = 10t^{2} + 5\) kW over 0 to 4 seconds.
Given: \(Q(t) = 10t^{2} + 5\) kW, \(t = 0\) to \(4\) s
$$ Q_{total} = \int_{0}^{4} (10t^{2} + 5) dt $$
$$ = [ \frac{10t^{3}}{3} + 5t ]_{0}^{4} $$
$$ = ( \frac{10 \times 64}{3} + 20 ) - (0 + 0) = ( \frac{640}{3} + 20 ) $$
$$ = 213.33 + 20 = 233.33 \text{ kJ} $$
Answer: 233.33 kJ
Example 3:
Find the work done by a compressor if the pressure-volume relationship is \(P(v) = 3v^{2} + 4v + 1\) kPa between volumes \(v = 0\) and \(v = 2\) m³.
Given: \(P(v) = 3v^{2} + 4v + 1\) kPa, \(v = 0\) to \(2\) m³
$$ W = \int_{0}^{2} (3v^{2} + 4v + 1) dv $$
$$ = [ v^{3} + 2v^{2} + v ]_{0}^{2} $$
$$ = (8 + 8 + 2) - 0 = 18 \text{ kPa·m}^3 $$
Answer: 18 kPa·m³
Refrigeration systems often contain components such as evaporators or condensers with irregular shapes. Calculating their volume requires integration of cross-sectional areas along their length. This is essential for estimating refrigerant charge and system capacity.
The volume \(V\) of an object with cross-sectional area \(A(x)\) varying along length \(L\) is:
$$ V = \int_{0}^{L} A(x) \, dx $$
Example 1:
Calculate the volume of a condenser tube whose cross-sectional area varies as \(A(x) = 0.05x + 0.1\) m² along its 4 m length.
Given: \(A(x) = 0.05x + 0.1\) m², \(L = 4\) m
$$ V = \int_{0}^{4} (0.05x + 0.1) dx $$
$$ = [ 0.025x^{2} + 0.1x ]_{0}^{4} $$
$$ = (0.025 \times 16 + 0.4) - 0 = (0.4 + 0.4) = 0.8 \text{ m}^3 $$
Answer: 0.8 m³
Example 2:
Find the volume of a refrigerant receiver tank with circular cross-section radius varying with height \(r(h) = 0.2h + 0.5\) m, where \(h\) ranges from 0 to 3 m.
Given: \(r(h) = 0.2h + 0.5\) m, \(h = 0\) to \(3\) m
Cross-sectional area \(A(h) = \pi r(h)^{2} = \pi (0.2h + 0.5)^{2}\)
$$ V = \int_{0}^{3} \pi (0.2h + 0.5)^{2} dh $$
Expand the square:
$$ (0.2h + 0.5)^{2} = 0.04h^{2} + 0.2h + 0.25 $$
$$ V = \pi \int_{0}^{3} (0.04h^{2} + 0.2h + 0.25) dh $$
$$ = \pi [ \frac{0.04h^{3}}{3} + 0.1h^{2} + 0.25h ]_{0}^{3} $$
Calculate inside brackets:
$$ = \pi ( \frac{0.04 \times 27}{3} + 0.1 \times 9 + 0.25 \times 3 ) $$
$$ = \pi (0.36 + 0.9 + 0.75) = \pi \times 2.01 $$
$$ = 3.1416 \times 2.01 = 6.31 \text{ m}^3 $$
Answer: 6.31 m³
Example 3:
An evaporator coil has a cross-sectional area \(A(x) = 0.1 \sin(\pi x / 2)\) m² along its 2 m length. Calculate the volume.
Given: \(A(x) = 0.1 \sin(\frac{\pi x}{2})\) m², \(x = 0\) to \(2\) m
$$ V = \int_{0}^{2} 0.1 \sin(\frac{\pi x}{2}) dx $$
$$ = 0.1 \int_{0}^{2} \sin(\frac{\pi x}{2}) dx $$
Let \(u = \frac{\pi x}{2}\), then \(du = \frac{\pi}{2} dx \Rightarrow dx = \frac{2}{\pi} du\)
Change limits: when \(x=0, u=0\); when \(x=2, u=\pi\)
$$ V = 0.1 \times \int_{0}^{\pi} \sin u \times \frac{2}{\pi} du = \frac{0.2}{\pi} \int_{0}^{\pi} \sin u \, du $$
$$ = \frac{0.2}{\pi} [-\cos u]_{0}^{\pi} = \frac{0.2}{\pi} (-\cos \pi + \cos 0) $$
$$ = \frac{0.2}{\pi} (-(-1) + 1) = \frac{0.2}{\pi} (1 + 1) = \frac{0.4}{\pi} = 0.127 \text{ m}^3 $$
Answer: 0.127 m³
Work done by or on refrigerants during compression or expansion processes is often calculated by integrating pressure over volume changes. This is crucial for energy analysis and efficiency calculations of refrigeration cycles.
Work \(W\) done in a process changing volume from \(V_1\) to \(V_2\) with pressure \(P(V)\) is:
$$ W = \int_{V_1}^{V_2} P(V) \, dV $$
Example 1:
Calculate work done during isobaric compression where pressure \(P = 100\) kPa, volume changes from 0.5 m³ to 0.2 m³.
Given: \(P = 100\) kPa, \(V_1 = 0.5\) m³, \(V_2 = 0.2\) m³
$$ W = \int_{0.5}^{0.2} 100 \, dV = 100 \times (0.2 - 0.5) = 100 \times (-0.3) = -30 \text{ kPa·m}^3 $$
Negative sign indicates work done on the refrigerant.
Answer: -30 kPa·m³
Example 2:
Work done during compression where pressure varies as \(P(V) = 200 - 50V\) kPa, volume changes from 0.1 m³ to 0.4 m³.
Given: \(P(V) = 200 - 50V\) kPa, \(V_1 = 0.1\) m³, \(V_2 = 0.4\) m³
$$ W = \int_{0.1}^{0.4} (200 - 50V) dV $$
$$ = [ 200V - 25V^{2} ]_{0.1}^{0.4} $$
$$ = (200 \times 0.4 - 25 \times 0.16) - (200 \times 0.1 - 25 \times 0.01) $$
$$ = (80 - 4) - (20 - 0.25) = 76 - 19.75 = 56.25 \text{ kPa·m}^3 $$
Answer: 56.25 kPa·m³
Example 3:
Calculate work done during expansion with pressure-volume relationship \(P(V) = \frac{500}{V}\) kPa from volume 0.2 m³ to 1.0 m³.
Given: \(P(V) = \frac{500}{V}\) kPa, \(V_1 = 0.2\) m³, \(V_2 = 1.0\) m³
$$ W = \int_{0.2}^{1.0} \frac{500}{V} dV = 500 \int_{0.2}^{1.0} \frac{1}{V} dV $$
$$ = 500 [\ln V]_{0.2}^{1.0} = 500 (\ln 1.0 - \ln 0.2) = 500 (0 - (-1.609)) = 500 \times 1.609 $$
$$ = 804.7 \text{ kPa·m}^3 $$
Answer: 804.7 kPa·m³
Heat transfer rates in refrigeration systems can vary with time or position, requiring integration to find total heat transferred or energy consumption. This is applicable in transient heat exchanger analysis and dynamic system monitoring.
Total heat \(Q\) transferred over time interval \(t_1\) to \(t_2\) with heat transfer rate \(q(t)\) is:
$$ Q = \int_{t_1}^{t_2} q(t) \, dt $$
Example 1:
Heat transfer rate varies as \(q(t) = 15 + 3t\) kW over 0 to 5 seconds. Find total heat transferred.
Given: \(q(t) = 15 + 3t\) kW, \(t = 0\) to \(5\) s
$$ Q = \int_{0}^{5} (15 + 3t) dt = [ 15t + \frac{3t^{2}}{2} ]_{0}^{5} $$
$$ = (15 \times 5 + \frac{3 \times 25}{2}) - 0 = (75 + 37.5) = 112.5 \text{ kJ} $$
Answer: 112.5 kJ
Example 2:
Heat transfer rate in a evaporator varies as \(q(t) = 20 e^{-0.1 t}\) kW over 0 to 10 seconds. Calculate total heat transferred.
Given: \(q(t) = 20 e^{-0.1 t}\) kW, \(t = 0\) to \(10\) s
$$ Q = \int_{0}^{10} 20 e^{-0.1 t} dt = 20 \int_{0}^{10} e^{-0.1 t} dt $$
$$ = 20 [ \frac{-1}{0.1} e^{-0.1 t} ]_{0}^{10} = -200 ( e^{-1} - 1 ) $$
Calculate values:
$$ e^{-1} = 0.3679 $$
$$ Q = -200 (0.3679 - 1) = -200 (-0.6321) = 126.42 \text{ kJ} $$
Answer: 126.42 kJ
Example 3:
The power consumption of a refrigeration unit varies with time as \(P(t) = 100 + 50 \sin(\frac{\pi t}{12})\) W over 0 to 24 hours. Calculate total energy consumed.
Given: \(P(t) = 100 + 50 \sin(\frac{\pi t}{12})\) W, \(t = 0\) to \(24\) hours
Note: Energy in watt-hours (Wh)
$$ E = \int_{0}^{24} ( 100 + 50 \sin(\frac{\pi t}{12}) ) dt $$
Split the integral:
$$ = \int_{0}^{24} 100 dt + 50 \int_{0}^{24} \sin(\frac{\pi t}{12}) dt $$
Calculate first integral:
$$ = 100 \times 24 = 2400 \text{ Wh} $$
Calculate second integral:
Let \(u = \frac{\pi t}{12}\), then \(du = \frac{\pi}{12} dt \Rightarrow dt = \frac{12}{\pi} du\)
Change limits: \(t=0 \Rightarrow u=0\), \(t=24 \Rightarrow u=2\pi\)
$$ 50 \times \int_{0}^{2\pi} \sin u \times \frac{12}{\pi} du = \frac{600}{\pi} \int_{0}^{2\pi} \sin u \, du $$
$$ = \frac{600}{\pi} [ -\cos u ]_{0}^{2\pi} = \frac{600}{\pi} (-\cos 2\pi + \cos 0) = \frac{600}{\pi} (-1 + 1) = 0 $$
Total energy:
$$ E = 2400 + 0 = 2400 \text{ Wh} = 2.4 \text{ kWh} $$
Answer: 2.4 kWh
Calculate the area under the curve \(P(v) = 4v^{2} + 3v + 2\) kPa between volumes 0 and 3 m³. (5 marks)
A condenser tube has a cross-sectional area \(A(x) = 0.03x^{2} + 0.1\) m² along its 5 m length. Find its volume. (5 marks)
Determine the work done by a compressor if the pressure varies as \(P(V) = 150 - 20V\) kPa between volumes 0.2 m³ and 0.6 m³. (5 marks)
The heat transfer rate from an evaporator is given by \(q(t) = 25 + 5t\) kW for \(t = 0\) to 6 seconds. Calculate the total heat transferred. (5 marks)
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Create a free accountThis chapter focused on the application of integration to solve practical problems in engineering contexts. It began with basic applications of integration to determine areas and volumes, demonstrating how integrals can model real-world quantities. The use of algebraic substitution was then introduced to simplify integrals involving polynomial and rational functions, enabling easier evaluation. Following this, integration using trigonometric substitution was explored, providing techniques to handle integrals containing quadratic expressions under roots. The chapter proceeded to cover integration by parts, a method useful for integrating products of functions by breaking them down into simpler parts. Finally, the application of reduction formulae was examined, showing how complex integrals can be reduced step-by-step into simpler forms for easier calculation. Throughout, the chapter emphasized systematic approaches to integration that are essential for solving engineering mathematics problems efficiently.
Calculate the area under the curve \(y = 3x^2\) from \(x = 0\) to \(x = 4\). (2 marks)
Find the volume of the solid generated by revolving the curve \(y = x^3\) about the x-axis from \(x = 0\) to \(x = 2\). (3 marks)
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