Electric Heating of Air in a House
The electric heating systems used in many houses consist of a simple duct with resistance heaters. Air is heated as it flows over resistance wires. Consider a 15-kW electric heating system. Air enters the heating section at 100 kPa and 17°C with a volume flow rate of 150 m³/min. If heat is lost from the air in the duct to the surroundings at a rate of 200 W, determine the exit temperature of air.
The electric heating system of a house is considered. For specified electric power consumption and air flow rate, the air exit temperature is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus \Delta m_{ CV }=0 \text { and } \Delta E_{ CV }=0 . 2 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values. 3 The kinetic and potential energy changes are negligible, \Delta ke \cong \Delta pe \cong 0 . 4 Constant specific heats at room temperature can be used for air.
Analysis We take the heating section portion of the duct as the system
(Fig. 5–41). This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus \dot{m}_{1}=\dot{m}_{2}=\dot{m} . Also, heat is lost from the system and electrical work is supplied to the system.
At temperatures encountered in heating and air-conditioning applications, Δh can be replaced by c_{p} \Delta T \text { where } c_{p}=1.005 kJ / kg · °C—the value at room temperature—with negligible error (Fig. 5–42). Then the energy balance for this steady-flow system can be expressed in the rate form as
\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}}=\underbrace{d E_{\text {system }} /d t^{\nearrow ^{0(steady)}}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc., energies } \end{array}}=0
\begin{aligned} \dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \\ \dot{W}_{e, \text { in }}+\dot{m} h_{1} &=\dot{Q}_{\text {out }}+\dot{m} h_{2} \quad(\text { since } \Delta ke \cong \Delta pe \cong 0) \\ \dot{W}_{e, \text { in }}-\dot{Q}_{\text {out }} &=\dot{m} c_{p}\left(T_{2}-T_{1}\right) \end{aligned}
From the ideal-gas relation, the specific volume of air at the inlet of the duct is
v_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left(0.287 kPa \cdot m ^{3} / kg \cdot K \right)(290 K )}{100 kPa }=0.832 m ^{3} / kg
The mass flow rate of the air through the duct is determined from
\dot{m}=\frac{\dot{V}_{1}}{v_{1}}=\frac{150 m ^{3} / min }{0.832 m ^{3} / kg }\left(\frac{1 min }{60 s }\right)=3.0 kg / s
Substituting the known quantities, the exit temperature of the air is determined to be
\begin{aligned} (15 kJ / s )-(0.2 kJ / s ) &=(3 kg / s )\left(1.005 kJ / kg \cdot{ }^{\circ} C \right)\left(T_{2}-17\right){ }^{\circ} C \\ T_{2} &=21.9^{\circ} C \end{aligned}
Discussion Note that heat loss from the duct reduces the exit temperature of air.
