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Heat transfer is a fundamental principle in water engineering, influencing the design and operation of water treatment plants, pumping stations, and distribution systems in Kenya. Efficient management of heat affects water quality, energy consumption, and equipment longevity. This chapter applies mathematical principles to quantify heat transfer through conduction, convection, radiation, and evaporative cooling, enabling water engineers to optimize system performance.
Conduction is the transfer of heat through a solid material without movement of the material itself, occurring due to temperature gradients. In water engineering, conduction affects heat loss in pipelines and storage tanks, influencing water temperature control and energy efficiency. The governing equation for steady-state one-dimensional conduction is Fourier’s law:
$$ Q = -k A \frac{dT}{dx} $$
where \(Q\) is heat transfer rate (W), \(k\) is thermal conductivity (W/m·K), \(A\) is cross-sectional area (m²), and \(\frac{dT}{dx}\) is temperature gradient (K/m).
Example 1:
A steel pipe carrying hot water has a wall thickness of 0.01 m, thermal conductivity \(k = 50\) W/m·K, and surface area \(A = 2\) m². The temperature difference across the pipe wall is 40°C. Calculate the heat loss by conduction.
Given:
\(k = 50\) W/m·K,
\(A = 2\) m²,
\(\Delta T = 40\) K,
thickness \(L = 0.01\) m.
Formula:$$Q = k A \frac{\Delta T}{L}$$
Substitution:$$Q = 50 \times 2 \times \frac{40}{0.01}$$
Calculation:$$Q = 50 \times 2 \times 4000$$$$Q = 400,000 \text{ W}$$
Answer: 400,000 W (400 kW)
Example 2:
A concrete water storage tank wall is 0.2 m thick with thermal conductivity \(k = 1.7\) W/m·K. The inside surface area is 10 m², and the temperature difference between inside and outside is 15°C. Find the heat loss by conduction.
Given:
\(k = 1.7\) W/m·K,
\(A = 10\) m²,
\(\Delta T = 15\) K,
\(L = 0.2\) m.
Formula:$$Q = k A \frac{\Delta T}{L}$$
Substitution:$$Q = 1.7 \times 10 \times \frac{15}{0.2}$$
Calculation:$$Q = 1.7 \times 10 \times 75$$$$Q = 1275 \text{ W}$$
Answer: 1275 W (1.275 kW)
Example 3:
A copper pipe with diameter 0.05 m and thickness 0.005 m carries water at 60°C while the ambient temperature is 25°C. The pipe length is 5 m. Given copper’s thermal conductivity \(k = 400\) W/m·K, calculate the heat loss.
Given:
\(k = 400\) W/m·K,
\(L = 0.005\) m (thickness),
\(d = 0.05\) m,
length \(l = 5\) m,
\(\Delta T = 35\) K.
Calculate surface area:$$A = \pi d l = 3.1416 \times 0.05 \times 5 = 0.7854 \text{ m}^2$$
Formula:$$Q = k A \frac{\Delta T}{L}$$
Substitution:$$Q = 400 \times 0.7854 \times \frac{35}{0.005}$$
Calculation:$$Q = 400 \times 0.7854 \times 7000$$$$Q = 400 \times 5497.8$$$$Q = 2,199,120 \text{ W}$$
Answer: 2,199,120 W (2.2 MW)
Convection is heat transfer due to fluid motion, critical in water engineering for cooling or heating water through pipes, tanks, or natural water bodies. It involves combined conduction and fluid movement effects. The convective heat transfer rate is given by Newton’s law of cooling:
$$ Q = h A (T_s - T_\infty) $$
where \(h\) is the convective heat transfer coefficient (W/m²·K), \(A\) is surface area (m²), \(T_s\) is surface temperature (K), and \(T_\infty\) is fluid temperature away from the surface (K).
Example 1:
Water flows over a heated flat plate of area 3 m². The plate surface temperature is 80°C and the water temperature is 30°C. If \(h = 100\) W/m²·K, find the heat transferred to the water.
Given:
\(h = 100\) W/m²·K,
\(A = 3\) m²,
\(T_s = 80^\circ C\),
\(T_\infty = 30^\circ C\).
Formula:$$Q = h A (T_s - T_\infty)$$
Substitution:$$Q = 100 \times 3 \times (80 - 30)$$
Calculation:$$Q = 100 \times 3 \times 50$$$$Q = 15,000 \text{ W}$$
Answer: 15,000 W (15 kW)
Example 2:
A water tank surface area of 5 m² is exposed to ambient air at 25°C. The tank surface temperature is 45°C. If the convective heat transfer coefficient is 15 W/m²·K, calculate the heat loss.
Given:
\(h = 15\) W/m²·K,
\(A = 5\) m²,
\(T_s = 45^\circ C\),
\(T_\infty = 25^\circ C\).
Formula:$$Q = h A (T_s - T_\infty)$$
Substitution:$$Q = 15 \times 5 \times (45 - 25)$$
Calculation:$$Q = 15 \times 5 \times 20$$$$Q = 1,500 \text{ W}$$
Answer: 1,500 W (1.5 kW)
Example 3:
Water at 10°C flows inside a pipe with internal surface temperature 70°C. The pipe length is 10 m with internal diameter 0.1 m. If \(h = 500\) W/m²·K, calculate the heat transferred to water.
Given:
\(h = 500\) W/m²·K,
length \(l = 10\) m,
diameter \(d = 0.1\) m,
\(T_s = 70^\circ C\),
\(T_\infty = 10^\circ C\).
Calculate surface area:$$A = \pi d l = 3.1416 \times 0.1 \times 10 = 3.1416 \text{ m}^2$$
Formula:$$Q = h A (T_s - T_\infty)$$
Substitution:$$Q = 500 \times 3.1416 \times (70 - 10)$$
Calculation:$$Q = 500 \times 3.1416 \times 60$$$$Q = 500 \times 188.496$$$$Q = 94,248 \text{ W}$$
Answer: 94,248 W (94.25 kW)
Thermal radiation is the transfer of heat by electromagnetic waves, significant in water engineering for solar heating of water and heat loss from exposed surfaces. Radiative heat transfer depends on surface emissivity, temperature, and surrounding environment. The Stefan-Boltzmann law describes radiative heat transfer:
$$ Q = \epsilon \sigma A (T_s^4 - T_\infty^4) $$
where \(\epsilon\) is emissivity (dimensionless), \(\sigma = 5.67 \times 10^{-8}\) W/m²·K⁴ is Stefan-Boltzmann constant, \(A\) is surface area (m²), \(T_s\) and \(T_\infty\) are absolute temperatures (K).
Example 1:
A black-painted water tank (emissivity \(\epsilon = 0.95\)) with surface area 8 m² is at 350 K. Surrounding temperature is 300 K. Calculate the radiative heat loss.
Given:
\(\epsilon = 0.95\),
\(A = 8\) m²,
\(T_s = 350\) K,
\(T_\infty = 300\) K,
\(\sigma = 5.67 \times 10^{-8}\) W/m²·K⁴.
Formula:$$Q = \epsilon \sigma A (T_s^4 - T_\infty^4)$$
Calculate temperatures to fourth power:$$350^4 = 1.5 \times 10^{10}$$$$300^4 = 8.1 \times 10^{9}$$
Substitution:$$Q = 0.95 \times 5.67 \times 10^{-8} \times 8 \times (1.5 \times 10^{10} - 8.1 \times 10^{9})$$
Calculation:$$Q = 0.95 \times 5.67 \times 10^{-8} \times 8 \times 6.9 \times 10^{9}$$$$Q = 0.95 \times 5.67 \times 8 \times 6.9 \times 10^{1}$$$$Q = 0.95 \times 5.67 \times 8 \times 69$$$$Q = 0.95 \times 5.67 \times 552$$$$Q = 0.95 \times 3,131.84$$$$Q = 2,975.25 \text{ W}$$
Answer: 2,975 W (2.98 kW)
Example 2:
A water reservoir surface of 20 m² has an emissivity of 0.85 and temperature 310 K. The night sky temperature is 280 K. Calculate the net radiative heat loss.
Given:
\(\epsilon = 0.85\),
\(A = 20\) m²,
\(T_s = 310\) K,
\(T_\infty = 280\) K.
Formula:$$Q = \epsilon \sigma A (T_s^4 - T_\infty^4)$$
Calculate powers:$$310^4 = 9.2 \times 10^{9}$$$$280^4 = 6.1 \times 10^{9}$$
Substitution:$$Q = 0.85 \times 5.67 \times 10^{-8} \times 20 \times (9.2 \times 10^{9} - 6.1 \times 10^{9})$$
Calculation:$$Q = 0.85 \times 5.67 \times 10^{-8} \times 20 \times 3.1 \times 10^{9}$$$$Q = 0.85 \times 5.67 \times 20 \times 31$$$$Q = 0.85 \times 5.67 \times 620$$$$Q = 0.85 \times 3,515.4$$$$Q = 2,988.1 \text{ W}$$
Answer: 2,988 W (2.99 kW)
Example 3:
A solar water heater panel has area 4 m² and emissivity 0.9. The panel surface temperature is 360 K, and ambient temperature is 300 K. Calculate radiative heat loss.
Given:
\(\epsilon = 0.9\),
\(A = 4\) m²,
\(T_s = 360\) K,
\(T_\infty = 300\) K.
Formula:$$Q = \epsilon \sigma A (T_s^4 - T_\infty^4)$$
Calculate powers:$$360^4 = 1.68 \times 10^{10}$$$$300^4 = 8.1 \times 10^{9}$$
Substitution:$$Q = 0.9 \times 5.67 \times 10^{-8} \times 4 \times (1.68 \times 10^{10} - 8.1 \times 10^{9})$$
Calculation:$$Q = 0.9 \times 5.67 \times 10^{-8} \times 4 \times 8.7 \times 10^{9}$$$$Q = 0.9 \times 5.67 \times 4 \times 87$$$$Q = 0.9 \times 5.67 \times 348$$$$Q = 0.9 \times 1,974.4$$$$Q = 1,777 \text{ W}$$
Answer: 1,777 W (1.78 kW)
Evaporative cooling is heat loss due to water evaporation, important in cooling towers and natural water bodies. The process removes latent heat, lowering temperature. Heat loss by evaporation depends on evaporation rate and latent heat of vaporization:
$$ Q = m \times L_v $$
where \(Q\) is heat lost (J), \(m\) is mass of evaporated water (kg), and \(L_v\) is latent heat of vaporization (J/kg).
Example 1:
A cooling tower evaporates 0.5 kg of water per second. Given latent heat of vaporization \(L_v = 2.26 \times 10^6\) J/kg, calculate heat loss.
Given:
\(m = 0.5\) kg/s,
\(L_v = 2.26 \times 10^6\) J/kg.
Formula:$$Q = m \times L_v$$
Substitution:$$Q = 0.5 \times 2.26 \times 10^6$$
Calculation:$$Q = 1.13 \times 10^6 \text{ W}$$
Answer: 1,130,000 W (1.13 MW)
Example 2:
A pond loses 1000 kg of water by evaporation over 24 hours. Calculate the average heat loss rate. Use \(L_v = 2.26 \times 10^6\) J/kg.
Given:
\(m = 1000\) kg,
time \(t = 24 \times 3600 = 86400\) s,
\(L_v = 2.26 \times 10^6\) J/kg.
Formula (power):$$Q = \frac{m \times L_v}{t}$$
Substitution:$$Q = \frac{1000 \times 2.26 \times 10^6}{86400}$$
Calculation:$$Q = \frac{2.26 \times 10^9}{86400}$$$$Q = 26157 \text{ W}$$
Answer: 26,157 W (26.16 kW)
Example 3:
In a water treatment plant, 2 kg/s of water evaporates from open tanks. Calculate the heat loss due to evaporation. Use \(L_v = 2.26 \times 10^6\) J/kg.
Given:
\(m = 2\) kg/s,
\(L_v = 2.26 \times 10^6\) J/kg.
Formula:$$Q = m \times L_v$$
Substitution:$$Q = 2 \times 2.26 \times 10^6$$
Calculation:$$Q = 4.52 \times 10^6 \text{ W}$$
Answer: 4,520,000 W (4.52 MW)
A cast iron water pipe of thickness 0.015 m and thermal conductivity 55 W/m·K has an external surface area of 4 m². If the temperature difference across the pipe wall is 35°C, calculate the heat loss by conduction. (5 marks)
Water at 20°C flows inside a pipe with internal surface temperature 65°C. The pipe internal diameter is 0.08 m and length 8 m. Given \(h = 400\) W/m²·K, calculate the heat transferred to the water by convection. (5 marks)
A black-painted water reservoir of surface area 15 m² has temperature 320 K, while ambient temperature is 290 K. Calculate the radiative heat loss if emissivity is 0.9. (5 marks)
A cooling pond evaporates 1500 kg of water per day. Calculate the average heat loss, given latent heat of vaporization is \(2.26 \times 10^6\) J/kg. (5 marks)
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Create a free accountThis chapter explored the fundamental concepts and terms related to heat transfer, including conduction, convection, thermal radiation, and evaporative cooling as the primary mechanisms. It detailed the various modes by which heat moves through different materials and environments. The chapter then examined how to quantify heat, focusing on the measurement and calculation of heat quantities involved in physical processes. It also covered temperature change and how it relates to the energy exchanged during heating or cooling. The discussion extended to changes in physical state, explaining the processes involved when materials transition between solid, liquid, and gaseous phases. Latent heat was introduced to describe the energy absorbed or released during phase changes without temperature variation. Additionally, specific heat capacity was analyzed as the amount of heat required to change a substance's temperature-linked to the kinetic theory that explains heat in terms of particle motion. Overall, the chapter provided a comprehensive understanding of heat principles essential for practical applications.
A steel pipe used in a water treatment plant has a thickness of 2 cm, thermal conductivity \(k = 50 \, \text{W/m} \cdot \text{K}\), length 1 m, and temperature difference between the inner and outer surface of 40°C. Calculate the rate of heat conduction through the pipe wall. (Easy, 5 marks)
Water is heated in a tank by a convection process. If the convective heat transfer coefficient is \(h = 100 \, \text{W/m}^2 \cdot \text{K}\), the surface area is 2 m², and the temperature difference between the water surface and surrounding air is 15°C, find the heat transfer rate. (Easy, 5 marks)
Water treatment plant pipelines made of steel have a thickness of 20 mm and length of 5 m. If the temperature difference across the pipe wall is 30°C and the thermal conductivity of steel is 50 W/m°C, calculate the rate of heat conduction through the pipe wall. (4 marks)
Explain how convection heat transfer affects cooling of water in an open reservoir exposed to the atmosphere. (4 marks)
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