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

Question

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 m[latex]^3[/latex]/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.

Accepted Answer

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 [latex]\Delta m_{\mathrm{CV}}=0[/latex] and [latex]\Delta E_{\mathrm{CV}}[/latex]=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, [latex]\Delta k e \cong[/latex] [latex]\Delta \mathrm{pe} \cong 0[/latex]. 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 [latex]\dot{m}_{1}=\dot{m}_{2}=\dot{m}[/latex]. 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 [latex]c_{p} \Delta T[/latex] where [latex]c_{p}=1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}[/latex]-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

[latex]\underbrace{E_{ ext {in }}-E_{ ext {out}}}_{\begin{array}{c} ext { Rate of net energy transfer} \ ext { by heat, work, and mass }\end{array}} =\underbrace{{dE_{ ext {system }}}/dt earrow ^{0( ext{steady})}}_{\begin{array}{c} ext {Rate of Change in internal, kinetic, } \ ext { potential, etc., energies }\end{array}}=0[/latex]

[latex]\dot{E}_{ ext {in }} =\dot{E}_{ ext {out }}[/latex]

[latex]\dot{W}_{ ext {e.in }}+\dot{m} h_{1}=\dot{Q}_{ ext {out }}+\dot{m} h_{2} \quad( ext { since } \Delta \mathrm{ke} \cong \Delta \mathrm{pe} \cong 0) [/latex]

[latex]\dot{W}_{e, ext { in }}-\dot{Q}_{ ext {out }}=\dot{m} c_{p}\left(T_{2}-T_{1} ight)[/latex]

From the ideal-gas relation, the specific volume of air at the inlet of the duct is

[latex]v_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K} ight)(290 \mathrm{~K})}{100 \mathrm{kPa}}=0.832 \mathrm{~m}^{3} / \mathrm{kg}[/latex]

The mass flow rate of the air through the duct is determined from

[latex]\dot{m}=\frac{\dot{V}_{1}}{v_{1}}=\frac{150 \mathrm{~m}^{3} / \mathrm{min}}{0.832 \mathrm{~m}^{3} / \mathrm{kg}}\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}} ight)=3.0 \mathrm{~kg} / \mathrm{s}[/latex]

Substituting the known quantities, the exit temperature of the air is determined to be

[latex](15 \mathrm{~kJ} / \mathrm{s})-(0.2 \mathrm{~kJ} / \mathrm{s}) =(3 \mathrm{~kg} / \mathrm{s})\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C} ight)\left(T_{2}-17 ight)^{\circ} \mathrm{C} [/latex]

[latex]T_{2} =21.9^{\circ} \mathrm{C}[/latex]

Discussion Note that heat loss from the duct reduces the exit temperature of air.

Question 5.11: Electric Heating of Air in a House The electric heating syst...

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