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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K)​

Respuesta :

Answer: The exit temperature of the gas is [tex]32^{o}C[/tex].

Explanation:

The given data is as follows.

[tex]C_{p}[/tex] = 1000 J/kg K,     R = 500 J/Kg K = 0.5 kJ/kg K (as 1 J = 0.001 kJ)

[tex]P_{1}[/tex] = 100 kPa,      [tex]V_{1} = 15 m^{3}/s[/tex]

[tex]T_{1} = 27^{o}C[/tex] = (27 + 273) K = 300 K

For the given gas we assume that it is an ideal gas with constant pressure, negligible change in kinetic and potential energy, constant specific heat. At inlet,

     [tex]P_{1}V_{1} = mRT_{1}[/tex]

or,      m = [tex]\frac{P_{1}V_{1}}{RT}[/tex]

             = [tex]\frac{100 \times 15}{0.5 \times 300}[/tex]

According to steady flow energy equation,

[tex]mh_{1} + Q = mh_{2} + W[/tex]

or,    [tex]h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}[/tex]

[tex]C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}[/tex]

[tex](T_{2} - T_{1}) \times C_{p} = \frac{130 - 80}{10}[/tex]

[tex](T_{2} - T_{1})[/tex] = 5 K

[tex]T_{2}[/tex] = (300 + 5) K = 305 K or [tex]32^{o}C[/tex]    (as [tex][305 - 273]^{o}C = 32^{o}C[/tex])

Thus, we can conclude that exit temperature of the gas is [tex]32^{o}C[/tex].