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)

Answer: The exit temperature of the gas in deg C 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 kJ = 1000 J)

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

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

We know that for an ideal gas the mass flow rate will be calculated as follows.

     [tex]P_{1}V_{1} = mRT_{1}[/tex]

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

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

                = 10 kg/s

Now, according to the steady flow energy equation:

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

[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})C_{p} = \frac{130 - 80}{10}[/tex]

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

[tex]T_{2}[/tex] = 5 K + 300 K

[tex]T_{2}[/tex] = 305 K

           = (305 K - 273 K)

           = [tex]32^{o}C[/tex]

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

(a) Design a low-pass filter with a cutoff frequency of 12 kHz. Use0.15 uF capacitor and an appropriate resistor. B) Sketch and label the circuit. C) What is the gain of the filter at a frequencyof 20 kHz? Give your answer both as ratio (Vout/Vin) and in dB