Support Areas: Heat Transfer

Abstract

This chapter covers the subject matter of heat transfer and its specific applications for heating ventilation and air-conditioning (HVAC) systems. Section 2.1 describes the importance of heat transfer in HVAC design and calculations. Heat transfer principles and the modes of heat transfer – conduction, convection, and radiation are covered in Sect. 2.2. Section 2.3 is on conduction heat transfer with coverage of topics such as Fourier’s Law, multilayer heat conduction, and R-values of insulation and building materials. Comprehensive coverage of convection heat transfer, including the important topics of conduction-convection systems, overall resistance, and overall heat transfer coefficient is accomplished in Sect. 2.4. Section 2.5 covers the topics of design of heat exchangers and condensers.

Appendices

Practice Problems

Practice Problem 2.1

The cross-section of a partition wall consists of 0.80 in. thick wooden panel exposed to a temperature of 85 °F. Adjacent to the wooden panel is 3.25 in. thick fiberglass insulation followed by 0.75 in. thick particle board exposed to a conditioned space maintained at 70 °F. Only the conduction resistances are significant in this situation. Using the R-values given in Table 2.1, determine.

A. the combined R-value of the wall assembly.

B. the rate of heat gain by the conditioned space if the wall cross-section is 20 ft × 10 ft.

Practice Problem 2.2

A building wall consists of a 50 mm thick brick with thermal conductivity 0.67 W/m.K and an adjacent layer of 10 mm thick wall board with thermal conductivity 0.18 W/m.K. The brick is exposed to ambient air with heat transfer coefficient of 36 W/m².K and the heat transfer coefficient on the wall side is 11 W/m².K. During the summer months, the average ambient temperature is 42 °C. Typically, the inside is maintained at 21 °C. Determine

• A. the R-values of the brick and the wall board for the given thickness of each

• B. the rate of heat gain in the building per unit area of the wall

Practice Problem 2.3

A wall partition inside a building separates an unconditioned corridor space at 27 °C from a conditioned space maintained at 21 °C. The wall assembly consists of 20 mm thick gypsum wall board (R = 0.0793 m².K/W) exposed to ambient air on both sides and 50 mm of fiberglass batt insulation (R = 2.291 m².K/W) in between the exposed surfaces. Calculate the heat gained by the conditioned space per unit area of the wall surface after including the convection resistances at the exposed surfaces.

Solutions to Practice Problems

Practice Problem 2.1

• (Solution)

• A. The R-values given in Table 2.1 are per inch thickness. Therefore, to obtain the combined R-value of the wall assembly, multiply the table R-value of each component by its thickness and then add up the resulting component values as shown in the following table.

  Component	R-value (per in.), hr-ft2-°F/Btu	Thickness (in)	Resistance, R (hr-ft2-°F/Btu)
  Wooden panel	1.24	0.80	0.992
  Fiberglass insulation	3.67	3.25	11.927
  Particle board	1.06	0.75	0.795
  		∑Rcond, ua	13.71

• B. Calculate the heat gain through the wall using Eq. 2.8.

$$ {q}_{\mathrm{wall}}=\frac{A\Delta {T}_{\mathrm{overall}}}{\sum {R}_{\mathrm{cond},\mathrm{ua}}}=\frac{\left(20\ \mathrm{ft}\times 10\ \mathrm{ft}\right)\left({85}^{{}^{\circ}}\mathrm{F}-{70}^{{}^{\circ}}\mathrm{F}\right)}{13.71\frac{\mathrm{hr}\hbox{-} {\mathrm{ft}}^2{\hbox{-}}^0\mathrm{F}}{\mathrm{Btu}}}=218.82\ \mathrm{Btu}/\mathrm{hr} $$

Practice Problem 2.2

• (Solution)

• A. Subscript ‘wb’ represents the wall board. Calculate the R-values of the brick and the wall board using Eq. 2.10.

$$ {R}_{\mathrm{brick}}=\frac{\Delta {X}_{\mathrm{brick}}}{k_{\mathrm{brick}}}=\frac{50\ \mathrm{m}\mathrm{m}\times \frac{1\ \mathrm{m}}{1000\ \mathrm{m}\mathrm{m}}}{0.67\frac{\mathrm{W}}{\mathrm{m}\cdot \mathrm{K}}}=0.0746\ {\mathrm{m}}^2\cdot \mathrm{K}/\mathrm{W} $$

$$ {R}_{\mathrm{wb}}=\frac{\Delta {X}_{\mathrm{wb}}}{k_{\mathrm{wb}}}=\frac{10\ \mathrm{m}\mathrm{m}\times \frac{1\ \mathrm{m}}{1000\ \mathrm{m}\mathrm{m}}}{0.18\frac{\mathrm{W}}{\mathrm{m}\cdot \mathrm{K}}}=0.0556\ {\mathrm{m}}^2\cdot \mathrm{K}/\mathrm{W} $$

• B. Calculate the heat loss per unit cross-section area of the wall using Eq. 2.17.

$$ {\displaystyle \begin{array}{l}\frac{q}{A}=\frac{\Delta {T}_{\mathrm{o}\mathrm{verall}}}{\sum {R}_{\mathrm{th},\mathrm{ua}}}=\frac{\Delta {T}_{\mathrm{o}\mathrm{verall}}}{\frac{1}{h_{\mathrm{o}}}+\sum {R}_{\mathrm{cond},\mathrm{ua}}+\frac{1}{h_{\mathrm{i}}}}\Rightarrow \\ {}\frac{q}{A}=\frac{42^{{}^{\circ}}\mathrm{C}-{21}^{{}^{\circ}}\mathrm{C}}{\frac{1}{36\frac{\mathrm{W}}{{\mathrm{m}}^2\cdot \mathrm{K}}}+0.0746\frac{{\mathrm{m}}^2\cdot \mathrm{K}}{\mathrm{W}}+0.0556\frac{{\mathrm{m}}^2\cdot \mathrm{K}}{\mathrm{W}}+\frac{1}{11\frac{\mathrm{W}}{{\mathrm{m}}^2\cdot \mathrm{K}}}}\\ {}\kern0.8em =84.36\ \mathrm{W}/{\mathrm{m}}^2\end{array}} $$

Note that ΔT ° C = ΔT K.

Practice Problem 2.3

• (Solution)

Subscript ‘GWB’ represents gypsum wall board, and subscript ‘FG’ represents fiberglass insulation. From Table 2.2, for horizontal heat flow across a vertical wall, the convection R-value is R = 0.12 m².K/W. Calculate the R-value of the wall assembly (including the convection resistances) by adding up all the resistances as shown.

$$ {\displaystyle \begin{array}{l}\sum {R}_{\mathrm{th},\mathrm{ua}}=2\left({R}_{\mathrm{conv}.}\right)+2\left({R}_{\mathrm{GWB}}\right)+{R}_{\mathrm{FG}}\\ {}\kern3.7em =2\left(0.12\frac{{\mathrm{m}}^2\cdot \mathrm{K}}{\mathrm{W}}\right)+2\left(0.0793\frac{{\mathrm{m}}^2\cdot \mathrm{K}}{\mathrm{W}}\right)+2.291\frac{{\mathrm{m}}^2\cdot \mathrm{K}}{\mathrm{W}}\\ {}\kern3.7em =2.690\ {\mathrm{m}}^2\cdot \mathrm{K}/\mathrm{W}\end{array}} $$

Calculate the heat gained by the conditioned space per unit cross-section area of the partition wall using Eq. 2.17.

$$ \frac{q}{A}=\frac{\Delta {T}_{\mathrm{overall}}}{\sum {R}_{\mathrm{th},\mathrm{ua}}}=\frac{27^{{}^{\circ}}\mathrm{C}-{21}^{{}^{\circ}}\mathrm{C}}{2.690\frac{{\mathrm{m}}^2\cdot \mathrm{K}}{\mathrm{W}}}=2.230\ \mathrm{W}/{\mathrm{m}}^2 $$

Note that ΔT ° C = ΔT K.