Thermodynamics

The velocities involved in pipe and duct flow are relatively low, and thekinetic energy changes are usually insignificant. This is particularly true whenthe pipe or duct diameter is constant and the heating effects are negligible.Kinetic energy changes may be significant, however, for gas flow in ductswith variable cross-sectional areas especially when the compressibility effectsare significant. The potential energy term may also be significant when thefluid undergoes a considerable elevation change as it flows in a pipe or duct.Chapter 5 | 245EXAMPLE 5–11Electric Heating of Air in a HouseThe electric heating systems used in many houses consist of a simple ductwith resistance heaters. Air is heated as it flows over resistance wires. Considera 15-kW electric heating system. Air enters the heating section at 100kPa and 17°C with a volume flow rate of 150 m 3 /min. If heat is lost fromthe air in the duct to the surroundings at a rate of 200 W, determine the exittemperature of air.Solution The electric heating system of a house is considered. For specifiedelectric power consumption and air flow rate, the air exit temperature isto be determined.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0 and E CV 0. 2 Air is an ideal gassince it is at a high temperature and low pressure relative to its critical-pointvalues. 3 The kinetic and potential energy changes are negligible, ke pe 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 boundaryduring the process. We observe that there is only one inlet and one exitand thus ṁ 1 ṁ 2 ṁ. Also, heat is lost from the system and electricalwork is supplied to the system.At temperatures encountered in heating and air-conditioning applications,h can be replaced by c p T where c p 1.005 kJ/kg · °C—the valueat room temperature—with negligible error (Fig. 5–42). Then the energybalance for this steady-flow system can be expressed in the rate form asE # in E # out dE system >dt 0⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭Rate of net energy transferby heat, work, and massRate of change in internal, kinetic,potential, etc., energiesFrom the ideal-gas relation, the specific volume of air at the inlet of theduct isv 1 RT 1P 1E # in E # outW # e,in m # h 1 Q # out m # h 2 1since ¢ke ¢pe 02W # e,in Q # out m # c p 1T 2 T 1 2¡ 0 (steady) 10.287 kPa # m 3 /kg # K21290 K2100 kPa 0.832 m 3 /kgThe mass flow rate of the air through the duct is determined fromm # V# 1 150 m3 /min 1 mina b 3.0 kg/sv 1 0.832 m 3 /kg 60 sQ˙out = 200 W˙T 2 = ?T 1 = 17°CP 1 = 100 kPaV 1 = 150 m 3 /minFIGURE 5–41Schematic for Example 5–11.AIR–20 to 70°C∆h = 1.005 ∆T (kJ/kg)W˙e, in = 15 kWFIGURE 5–42The error involved in h c p T,where c p 1.005 kJ/kg · °C, is lessthan 0.5 percent for air in thetemperature range 20 to 70°C.