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

The exit temperature of the gas = 32° C

Explanation:

Solution

Given that:

Inlet temperature T₁ = 27°C ≈ 300.15 K

Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa

Volume flow rate , V = 15 m/s³

Diameter of the deduct, D = 500 mm = 0.5 m

Electric heater power, W heater = 130 kW = 130 * 10^3 W

The heat lost Q = 80 kW =  80 * 10^3 W

Now,

From the ideal gas law, density of the air at the inlet is given as :

ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300

=0.6667 kg/m³

The mass flow rate through the duct is computed below:

m = ρ₁ V = 0.6667 * 15 = 10 kg/s

Thus

Applying the first law of thermodynamics to the process is shown below:

Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)

So,

If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:

Q + m (h₁) = m (h₂) + W

or

Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)

Thus

h₂ - h₁ = Cp T₂ - T₁

Now by method of substitution the known values are:

(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)

Note: The heat transfer is  taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas

So,

Solving for T₂,

T₂ = 32° C

Therefore the exit temperature of the gas = 32° C

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