HVAC Systems and Equipment

Abstract

This chapter on HVAC systems and equipment has seven sections with several sub-sections in each section. After a brief introduction in Sect. 8.1, basic HVAC systems are covered in Sect. 8.2, including direct expansion (DX) cooling systems, chilled-water cooling systems, and steam-based heating systems. Section 8.3 covers chilled-water systems in detail including water-cooled, air-cooled, and evaporative condensers. Section 8.3 also includes coverage of different types of compressors, such as scroll, screw, and centrifugal compressors used in chilled-water systems. Energy recovery systems including heat recovery ventilators (HRV), and energy recovery ventilators (ERV) are covered in Sect. 8.4. Different energy recovery devices such as plate and frame heat exchangers, run-around coils, sensible wheels, heat recovery wheels, energy recovery wheels, enthalpy wheels, and their analysis for both summer and winter operations are explained in detail in Sect. 8.4, with the use of several illustrative examples and practice problems. Acoustics and noise control are covered in Sect. 8.5, including sources of noise and noise mitigation strategies. Section 8.6 covers vibration and vibration control in HVAC equipment and systems, including vibration isolation mechanisms such as elastomer pads, floor springs, and spring hangers. Constant air volume (CAV) and variable air volume (VAV) systems are covered in Sect. 8.7, with both single-zone and multiple-zone applications. Illustrative examples to compare CAV and VAV systems are presented in Sect. 8.7.

Appendices

Practice Problems

Practice Problem 8.1

A sensible heat recovery run-around coil is used in preheating 10,000 cfm of outside winter air at 40 °F DB and 50% rh to a temperature of 60 °F DB. It is estimated that the sensible heat exchange effectiveness of the run around coil is 70%. If an equivalent mass flow rate of air from the facility is to be ventilated, determine

• A. the temperature of the air in the facility,

• B. the actual heat transfer achieved by the system,

• C. the gpm of the circulating water required if the temperature increase of the water is limited to 10 °F,

• D. cfm of return / exhaust air if the facility is maintained at 60% rh.

Practice Problem 8.2

An enthalpy wheel with a moisture effectiveness of 70% is used in processing 2000 cfm of cold, dry winter air at 40 °F and 30% rh to a pre-heated air temperature of 55 °F. 2000 cfm of return / exhaust air at 68 °F and 50% rh enters the wheel. Determine

• A. the humidity ratio and the relative humidity of pre-heated air supplied to the air handling unit,

• B. the kW of heating achieved by the wheel,

• C. the conditions of the exhaust air leaving the wheel.

Practice Problem 8.3

0.8 m³/s of winter air at 6 °C DB and 3 °C WB is preheated using an equivalent volume flow of facility return / exhaust air at 20 °C and 50% rh. The heat exchange is accomplished in an ERV (enthalpy wheel). The total and sensible effectiveness of the ERV are equal at 85%. Determine

• A. the conditions (DB temperature, enthalpy, relative humidity, and humidity ratio) of both the air streams exiting the ERV.

• B. the moisture effectiveness of the ERV,

• C. the annual savings in energy costs due to the use of ERV if power costs are $1.03 per kWH and the annual usage of the heating system is estimated to be 1000 hours.

Solutions to Practice Problems

Practice Problem 8.1

• (Solution)

Determine the specific volume of inlet outside air by locating its state point (OA, 40 °F DB, 50% rh) on the psychrometric chart as shown in the excerpt of the chart below.

$$ {v}_{\mathrm{OA}}=12.7\ {\mathrm{ft}}^3/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a} $$

$$ {\dot{m}}_{\mathrm{OA}}=\frac{{\dot{V}}_{\mathrm{OA}}}{v_{\mathrm{OA}}}=\frac{10000\frac{{\mathrm{ft}}^3}{\min }}{12.7\frac{{\mathrm{ft}}^3}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}}=787.40\ \mathrm{lbm}\ \mathrm{d}\ \mathrm{a}/\min $$

The mass flow rates of the outside air and exhaust air from the facility are equal. Therefore,

$$ {\dot{m}}_{\mathrm{EA}}={\dot{m}}_{\mathrm{OA}}=787.40\ \mathrm{lbm}\ \mathrm{d}\ \mathrm{a}/\min $$

• A. Since the mass flow rates of the outside air (OA) and exhaust air (EA) are equal,

$$ {\displaystyle \begin{array}{l}{\dot{m}}_{EA}{c}_{pa}={\dot{m}}_{OA}{c}_{pa}\Rightarrow \\ {}{C}_{EA}={C}_{OA}={C}_{\mathrm{min}}=\left(787.40\frac{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}{\min}\right)\left(0.24\frac{\mathrm{Btu}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}{\hbox{-}}^0\mathrm{F}}\right)\\ {}\kern7.4em =188.98\ \mathrm{Btu}/\min {\hbox{-}}^0\mathrm{F}\end{array}} $$

• Calculate the temperature of air from the facility entering the HRV using Eq. 8.8.

• B. Calculate the actual heat transfer achieved by the system using Eq. 8.6.

$$ {\displaystyle \begin{array}{l}{q}_{HRV,W}={\dot{m}}_{OA}{c}_{pa}\left({T}_{OA,e}-{T}_{OA,i}\right)={C}_{OA}\left({T}_{OA,e}-{T}_{OA,i}\right)\\ {}\kern11em =\left(188.98\frac{\mathrm{Btu}}{\min \hbox{-} {}^{{}^{\circ}}\mathrm{F}}\right)\left({60}^{{}^{\circ}}\mathrm{F}-{40}^{{}^{\circ}}\mathrm{F}\right)\\ {}\kern11em =3779.6\ \mathrm{Btu}/\min \end{array}} $$

• C. Energy balance for the run-around coil where it transfers heat to cold outside air results in the following set of equations.

• q_{HRV, W}= heat gained by the outside air = heat lost by the circulating water =

$$ {\displaystyle \begin{array}{l}{q}_{HRV,W}={\dot{m}}_w{c}_{pw}\varDelta {T}_w\Rightarrow \\ {}{\dot{m}}_w=\frac{q_{HRV,W}}{c_{pw}\varDelta {T}_w}=\frac{3779.6\frac{\mathrm{Btu}}{\min }}{\left(1.0\frac{\mathrm{Btu}}{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}\hbox{-} {}^{{}^{\circ}}\mathrm{F}}\right)\left({10}^{{}^{\circ}}\mathrm{F}\right)}=377.96\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\min \end{array}} $$

• Convert the mass flow rate of water to gpm using the density of water.

$$ {\dot{V}}_w=\frac{{\dot{m}}_w}{\rho_w}=\frac{377.96\frac{\mathrm{lbm}}{\min }}{8.34\frac{\mathrm{lbm}}{\mathrm{gal}}}=45.32\ \mathrm{gpm} $$

• D. Determine the specific volume of return / exhaust air by locating its state point (EA, 69 °F DB, 60% rh) on the psychrometric chart as shown. From the excerpt of the psychrometric chart shown,

$$ {v}_{\mathrm{EA}}=13.52\ {\mathrm{ft}}^3/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a} $$

$$ {\displaystyle \begin{array}{l}{\dot{V}}_{\mathrm{EA}}={\dot{m}}_{\mathrm{EA}}{v}_{\mathrm{EA}}=\left(787.40\frac{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}{\min}\right)\left(13.52\frac{{\mathrm{ft}}^3}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}\right)\\ {}\kern5.9em =10,646\ {\mathrm{ft}}^3/\min \left(\mathrm{cfm}\right)\end{array}} $$

Practice Problem 8.2

• (Solution)

Locate the state points of inlet outside air (OAi, 40 °F DB, 30% rh), and inlet exhaust air (EAi, 68 °F DB, 50% rh) on the psychrometric chart as shown.

From the excerpt of the chart shown,

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{OA},\mathrm{i}}=11.3\ \mathrm{Btu}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a},\kern0.75em {v}_{\mathrm{OA},\mathrm{i}}=12.65\ {\mathrm{ft}}^3/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{\mathrm{OA},\mathrm{i}}=0.0016\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\\ {}{h}_{\mathrm{EA},\mathrm{i}}=24.3\ \mathrm{Btu}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a},\kern1em {v}_{\mathrm{EA},\mathrm{i}}=13.46\ {\mathrm{ft}}^3/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{\mathrm{EA},\mathrm{i}}=0.0073\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\end{array}} $$

Calculate the mass flow rates of the outside air and exhaust air using specific volumes.

$$ {\dot{m}}_{\mathrm{OA}}=\frac{{\dot{V}}_{\mathrm{OA},\mathrm{i}}}{v_{\mathrm{OA},\mathrm{i}}}=\frac{2000\frac{{\mathrm{ft}}^3}{\min }}{12.65\frac{{\mathrm{ft}}^3}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}}=158.10\ \mathrm{lbm}\ \mathrm{d}\ \mathrm{a}/\min $$

$$ {\dot{m}}_{\mathrm{EA}}=\frac{{\dot{V}}_{\mathrm{EA},\mathrm{i}}}{v_{\mathrm{EA},\mathrm{i}}}=\frac{2000\frac{{\mathrm{ft}}^3}{\min }}{13.45\frac{{\mathrm{ft}}^3}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}}=148.70\ \mathrm{lbm}\ \mathrm{d}\ \mathrm{a}/\min $$

Since ${\dot{m}}_{\mathrm{EA}}$< ${\dot{m}}_{\mathrm{OA}}$, ${\dot{m}}_{\mathrm{EA}}={\dot{m}}_{\mathrm{min}}=148.70\ \mathrm{lbm}\ \mathrm{d}\ \mathrm{a}/\min$

• A. Using the given moisture effectiveness for winter operation, determine the humidity ratio of the exit outside air (supply air to AHU) using Eq. 8.22.

$$ {\displaystyle \begin{array}{l}{\varepsilon}_{moist.,W}=\frac{{\dot{m}}_{OA}\left({\omega}_{OA,e}-{\omega}_{OA,i}\right)}{{\dot{m}}_{\mathrm{min}}\left({\omega}_{EA,i}-{\omega}_{OA,i}\right)}\Rightarrow \\ {}0.70=\frac{\left(158.10\frac{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}{\min}\right)\left({\omega}_{OA,e}-0.0016\frac{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}\right)}{\left(148.70\frac{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}{\min}\right)\left(0.0073\frac{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}-0.0016\frac{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}\right)}\\ {}\Rightarrow {\omega}_{OA,e}=0.0053\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\end{array}} $$

• On the psychrometric chart, locate the state point of the exit outside air (supply air to AHU) at the intersection of

$$ {T}_{OA,e}={55}^{{}^{\circ}}\mathrm{F},\mathrm{and}\ {\omega}_{OA,e}=0.0053\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a} $$

• At the state point of the exit outside air (supply air to AHU), the relative humidity is rh_OA,e = 60%

• B. At the state point of the exit outside air (supply air to AHU), the enthalpy is

$$ {h}_{OA,e}=19.0\ \mathrm{Btu}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a} $$

• Calculate the total heat transfer achieved using Eq. 8.16.

$$ {\displaystyle \begin{array}{l}{q}_{T,W}={\dot{m}}_{OA}\left({h}_{OA,e}-{h}_{OA,i}\right)\\ {}\kern1.8em =\left(158.10\frac{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}{\min}\right)\left(19\frac{\mathrm{Btu}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}-11.3\frac{\mathrm{Btu}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}\right)\\ {}\kern1.8em =1217.4\ \mathrm{Btu}/\min \end{array}} $$

• Convert the total heat transfer achieved to kW of heating.

$$ {q}_{T,W}=\left(1217.4\ \mathrm{Btu}/\min \right)\left(\frac{0.0176\ \mathrm{kW}}{1\ \mathrm{Btu}/\min}\right)=21.43\ \mathrm{kW} $$

• C. Calculate the enthalpy of the exit exhaust air using the total heat transfer achieved and Eq. 8.16.

$$ {\displaystyle \begin{array}{l}{q}_{T,W}={\dot{m}}_{EA}\left({h}_{EA,i}-{h}_{EA,e}\right)\Rightarrow \\ {}1217.4\frac{\mathrm{Btu}}{\min }=\left(148.70\frac{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}{\min}\right)\left(24.30\frac{\mathrm{Btu}}{\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}}-{h}_{EA,e}\right)\Rightarrow \\ {}{h}_{EA,e}=16.11\ \mathrm{Btu}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\end{array}} $$

• Using the given moisture effectiveness for winter operation, determine the humidity ratio of the exit exhaust air using Eq. 8.22. Note that ${\dot{m}}_{EA}={\dot{m}}_{\mathrm{min}}$.

• On the psychrometric chart, locate the state point of the exit exhaust air at the intersection of

$$ {h}_{EA,e}=16.11\ \mathrm{Btu}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a},\mathrm{and}\ {\omega}_{EA,e}=0.0033\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a} $$

• From the psychrometric chart, the conditions of the exit exhaust air, EAe (leaving the enthalpy wheel) are:

$$ {\displaystyle \begin{array}{l}{T}_{\mathrm{DB},\mathrm{EA},\mathrm{e}}={52}^0\mathrm{F},{T}_{\mathrm{WB},\mathrm{EA},\mathrm{e}}={42}^0\mathrm{F},{h}_{EA,e}=16.11\ \mathrm{Btu}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{EA,e}=0.0033\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{d}\ \mathrm{a},\kern1.5em {\mathrm{rh}}_{\mathrm{EA},\mathrm{e}}=40\%\end{array}} $$

Practice Problem 8.3

• (Solution)

Locate the state points of inlet outside air (OAi, 6 °C DB, 3 °C WB), and inlet exhaust air (EAi, 20 °C DB, 50% rh) on the psychrometric chart as shown.

From the excerpt of the chart shown,

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{OA},\mathrm{i}}=15\ \mathrm{kJ}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a},\kern0.75em {v}_{\mathrm{OA},\mathrm{i}}=0.795\kern0.40em {\mathrm{m}}^3/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{\mathrm{OA},\mathrm{i}}=0.0035\ \mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}\\ {}{h}_{\mathrm{EA},\mathrm{i}}=39\ \mathrm{kJ}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a},\kern1em {v}_{\mathrm{EA},\mathrm{i}}=0.84\ {\mathrm{m}}^3/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{\mathrm{EA},\mathrm{i}}=0.0076\ \mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}\end{array}} $$

Calculate the mass flow rates of the outside air and exhaust air using specific volumes.

$$ {\dot{m}}_{\mathrm{OA}}=\frac{{\dot{V}}_{\mathrm{OA},\mathrm{i}}}{v_{\mathrm{OA},\mathrm{i}}}=\frac{0.8\frac{{\mathrm{m}}^3}{\mathrm{s}}}{0.795\frac{{\mathrm{m}}^3}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}}=1.006\ \mathrm{kg}\ \mathrm{d}\ \mathrm{a}/\mathrm{s} $$

$$ {\dot{m}}_{\mathrm{EA}}=\frac{{\dot{V}}_{\mathrm{EA},\mathrm{i}}}{v_{\mathrm{EA},\mathrm{i}}}=\frac{0.8\frac{{\mathrm{m}}^3}{\mathrm{s}}}{0.84\frac{{\mathrm{m}}^3}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}}=0.9524\ \mathrm{kg}\ \mathrm{d}\ \mathrm{a}/\mathrm{s} $$

• Since ${\dot{m}}_{\mathrm{EA}}$< ${\dot{m}}_{\mathrm{OA}}$, ${\dot{m}}_{\mathrm{EA}}={\dot{m}}_{\mathrm{min}}=0.9524\ \mathrm{kg}\ \mathrm{d}\ \mathrm{a}/\mathrm{s}$

• A. Using the given total effectiveness of the enthalpy wheel and Eq. 8.21, calculate the enthalpy of the exit outside air, that is, h_{OA, e}.

$$ {\displaystyle \begin{array}{l}{\varepsilon}_{T, ERV,W}=\frac{{\dot{m}}_{OA}\left({h}_{OA,e}-{h}_{OA,i}\right)}{{\dot{m}}_{\mathrm{min}}\left({h}_{EA,i}-{h}_{OA,i}\right)}\Rightarrow \\ {}0.85=\frac{\left(1.006\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left({h}_{OA,e}-15\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}\right)}{\left(0.9524\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(39\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}-15\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}\right)}\Rightarrow \\ {}{h}_{OA,e}=34.3\ \mathrm{kJ}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}\end{array}} $$

• Using the given sensible effectiveness of the enthalpy wheel and an equation similar to Eq. 8.21, calculate the dry bulb temperature of the exit outside air, that is, T_OA,e.

$$ {\displaystyle \begin{array}{l}{\varepsilon}_{Sen, ERV,W}=\frac{{\dot{m}}_{OA}{c}_{pa}\left({T}_{OA,e}-{T}_{OA,i}\right)}{{\dot{m}}_{\mathrm{min}}{c}_{pa}\left({T}_{EA,i}-{T}_{OA,i}\right)}\Rightarrow \\ {}0.85=\frac{\left(1.006\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(1.0\frac{\mathrm{kJ}}{\mathrm{kg}\cdot {}^{\circ}\mathrm{C}}\right)\left({T}_{OA,e}-{6}^{\circ}\mathrm{C}\right)}{\left(0.9524\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(1.0\frac{\mathrm{kJ}}{\mathrm{kg}\cdot {}^{\circ}\mathrm{C}}\right)\left({20}^{\circ}\mathrm{C}-{6}^{\circ}\mathrm{C}\right)}\Rightarrow \\ {}{T}_{OA,e}={17.3}^{\circ}\mathrm{C}\end{array}} $$

• On the excerpt of the psychrometric chart locate the state point of exit outside air at the intersection of its enthalpy and dry bulb temperature as shown.

$$ {\mathrm{OA}}_{\mathrm{e}}:{h}_{OA,e}=34.3\ \mathrm{kJ}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a},{T}_{OA,e}={17.3}^{{}^{\circ}}\mathrm{C} $$

• From the psychrometric chart, the conditions of the exit outside air, OAe (leaving the ERV) are:

$$ {\displaystyle \begin{array}{l}{T}_{\mathrm{DB},\mathrm{OA},\mathrm{e}}={17.3}^{{}^{\circ}}\mathrm{C},\kern0.75em {h}_{OA,e}=34.3\ \mathrm{kJ}/\mathrm{kj}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{OA,e}=0.0062\ \mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a},\kern0.5em {\mathrm{rh}}_{\mathrm{OA},\mathrm{e}}=52\%\end{array}} $$

• Calculate the total heat transfer achieved during winter operation using the first part of Eq. 8.16.

$$ {\displaystyle \begin{array}{l}{q}_{T,W}={\dot{m}}_{OA}\left({h}_{OA,e}-{h}_{OA,i}\right)\\ {}\kern1.75em =\left(1.006\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(34.3\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}-15\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}\right)=19.42\ \mathrm{kW}\end{array}} $$

• Calculate the enthalpy of the exit exhaust air using the second part of Eq. 8.16.

$$ {\displaystyle \begin{array}{l}{q}_{T,W}={\dot{m}}_{EA}\left({h}_{EA,i}-{h}_{EA,e}\right)\Rightarrow \\ {}19.42\ \mathrm{kW}=\left(0.9524\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(39\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}-{h}_{EA,e}\right)\Rightarrow \\ {}{h}_{EA,e}=18.6\ \mathrm{kJ}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}\end{array}} $$

• Calculate the dry bulb temperature of the exit exhaust air using Eq. 8.17

$$ {\displaystyle \begin{array}{l}{q}_{Sen,W}={\dot{m}}_{OA}{c}_{pa}\left({T}_{OA,e}-{T}_{OA,i}\right)={\dot{m}}_{EA}{c}_{pa}\left({T}_{EA,i}-{T}_{EA,e}\right)\Rightarrow \\ {}\left(1.006\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(1.0\frac{\mathrm{kJ}}{\mathrm{kg}\cdot {}^{\circ}\mathrm{C}}\right)\left({17.3}^{\circ}\mathrm{C}-{6}^{\circ}\mathrm{C}\right)\\ {}\kern5.25em =\left(0.9524\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(1.0\frac{\mathrm{kJ}}{\mathrm{kg}\cdot {}^{\circ}\mathrm{C}}\right)\left({20}^{\circ}\mathrm{C}-{T}_{EA,e}\right)\\ {}\Rightarrow {T}_{EA,e}={8.1}^{\circ}\mathrm{C}\end{array}} $$

• On the psychrometric locate the state point of exit outside air at the intersection of its enthalpy and dry bulb temperature.

$$ {\mathrm{EA}}_{\mathrm{e}}:{h}_{EA,e}=18.6\ \mathrm{kJ}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a},\kern1em {T}_{EA,e}={8.1}^{{}^{\circ}}\mathrm{C} $$

• From the psychrometric chart, the conditions of the exit exhaust air, EA,e (leaving the ERV) are:

$$ {\displaystyle \begin{array}{l}{T}_{\mathrm{DB},\mathrm{EA},\mathrm{e}}={8.1}^{{}^{\circ}}\mathrm{C},\kern0.75em {h}_{EA,e}=18.60\ \mathrm{kJ}/\mathrm{kj}\ \mathrm{d}\ \mathrm{a}\\ {}{\omega}_{EA,e}=0.0042\ \mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a},\kern0.5em {\mathrm{rh}}_{\mathrm{EA},\mathrm{e}}=62\%\end{array}} $$

• B. Calculate the moisture effectiveness of the ERV using Eq. 8.22.

$$ {\displaystyle \begin{array}{l}{\varepsilon}_{moist.,W}=\frac{{\dot{m}}_{OA}\left({\omega}_{OA,e}-{\omega}_{OA,i}\right)}{{\dot{m}}_{\mathrm{min}}\left({\omega}_{EA,i}-{\omega}_{OA,i}\right)}\\ {}\kern3em =\frac{\left(1.006\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(0.0062\frac{\mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}-0.0035\frac{\mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}\right)}{\left(0.9524\frac{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}{\mathrm{s}}\right)\left(0.0076\frac{\mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}-0.0035\frac{\mathrm{kg}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}\right)}\\ {}\kern3em =0.7\left(70\%\right)\end{array}} $$

• C. In the solution to part A, the total heat transfer achieved by the ERV was determined to be 19.42 kW. This is the equivalent heat input saved by the usage of the ERV. Therefore, the annual savings due to the usage of ERV is

$$ \mathrm{Annual}\ \mathrm{Savings}=\left(19.42\ \mathrm{kW}\right)\left(\frac{1000\ \mathrm{h}}{\mathrm{yr}}\right)\left(\frac{\$1.03}{\mathrm{kWh}}\right)=\$20,003/\mathrm{yr} $$