Answer :
Answer:
a) The inlet velocity, v₁ = 60.95 m/s
b) The exit temperature, [tex]T_2 = 687.02[/tex]
Explanation:
a) Determine the inlet velocity
The gas considered is [tex]CO_2[/tex]
The inlet Area of the nozzle, [tex]A_1 = 40 cm^2[/tex] = 0.004 m²
Gas constant for [tex]CO_2[/tex], R = 0.1889 kJ/kg-K
Input Pressure, P₁ = 1 MPa = 1000 kPa
Output Pressure, P₂ = 100 kPa
Input temperature, T₁ = 500°C = 500 + 273 = 773 K
Mass flow rate,
[tex]\dot{m} = 6000 kg/hr\\\dot{m} = \frac{6000}{3600} = 1.67 kg/s[/tex]
The relationship between mass flow rate and Inlet velocity is given by:
[tex]\dot{m} = \rho_1 A_1 v_1[/tex]..............(*)
It is necessary to first get the input density, [tex]\rho_1[/tex] using the relationship below:
[tex]P_1 = \rho_1 R_1 T_1\\1000 = \rho_1 * 0.1889 * 773\\\rho_1 = \frac{1000}{0.1889 * 773} \\\rho_1 = 6.85 kg/m^2[/tex]
Insert the necessary parameters into equation (*) to determine the inlet velocity:
1.67 = 6.85 * 0.004 * v₁
1.67 = 0.0274 v₁
v₁ = 1.67/0.0274
v₁ = 60.95 m/s
b) The exit temperature
At [tex]T_1 = 773 K[/tex], [tex]c_p = 1.156 kJ/kg-K = 1156 J/kg-K[/tex]
Exit Velocity, v₂ = 450 m/s
In an adiabatic process, there is no exchange of heat
i.e energy expended = work done = 0, q = w = 0
There is no change in potential energy
For a steady state process:
[tex]q - w = c_p (T_2 - T_1) + 0.5(v_2^2 - v_1^2) + g(z_2 - z_1)[/tex]
Considering all the assumptions for an an adiabatic process, the equation above reduces to:
[tex]c_p (T_2 - T_1) = -0.5(v_2^2 - v_1^2)\\1156(T_2 - 500) = - 0.5(450^2 - 60.95^2)\\1156(T_2 - 500) = -99392.55\\T_2 - 500 = -99392.55/1156\\T_2 - 500 = -85.98\\T_2 = -85.98 + 500\\T_2 = 414.02^0C\\T_2 = 414.02 + 273\\T_2 = 687.02 K[/tex]
A) The inlet velocity is : 60.95 m/s
B) The exit temperature is : 687.02 K
Given data :
Inlet Temperature ( T₁ ) = 500°C = 773 K
Inlet pressure ( P₁ ) = 1 MPa = 10³ kPa
Mass flow rate = 6000 kg/h = 1.67 kg/s
outlet pressure ( P₂ ) = 100 kPa
outlet velocity ( V₂ ) = 450 m/s
Inlet area of nozzle = 40 cm² = 0.004 m²
A) Determine the inlet velocity ( V₁ )
R ( gas constant ) of CO₂ = 0.1889 kJ / kg.k
First step : Determine the value of ρ₁ ( density of CO₂ )
ρ₁ = [tex]\frac{P}{RT}[/tex] --- ( 1 )
where : P = 1000 , R = 0.1889, T = 773
ρ₁ = 1000 / ( 0.1889 * 773 )
= 6.85 kg/m³
Next step : Determine the inlet velocity ( V₁ )
V₁ = mas flow rate / ( ρ₁ * inlet area of nozzle )
= 1.67 / ( 6.85 * 0.004 )
= 60.95 m/s
B) Determine the exit temperature
For an adiabatic process
Cp ( T1 - T2 ) = - 0.5 ( V₂² - V₁² ) ---- ( 2 )
where : V₂ = 450 m/s, Cp = 1156 kJ/kg.k, V₁ = 60.95 m/s, V₂ = 450 m/s, T₁ = 500°C
Back to equation ( 2 )
T₂ - 500 = -99392.55 / 1156
∴ T₂ = -85.98 + 500
= 414.02°C = 687.02 K
Hence we can conclude that The inlet velocity is : 60.95 m/s and The exit temperature is : 687.02 K
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