Respuesta :
Answer:
U_2 = 578.359 m /s
d_e = 1.4924 cm
Explanation:
Given:
Length of nozzle L = 25 cm
Inlet diameter d_i = 5 cm
Nozzle Entrance : T_1 = 325 C , P_1 = 700 KPa , U_1 = 30 m / s , H_1 = 3112.5 KJ / kg , V_1 = 388.61 cm^3 / g
Nozzle Exit : T_2 = 250 C , P_2 = 350 KPa , U_2, H_2 = 2945.7 KJ / kg , V_2 = 667.75 cm^3 / g
To Find:
a. Velocity at exit U_2
b. Exit Diameter d_e
a.
Energy Equation is given as:
ΔH + ΔU^2 / 2 + gΔz = Q + W
Q = W = Δz = 0
Substitute the values:
(H_2 - H_1 ) + (U^2_2 - U^2_1 ) / 2 = 0
U_2 = sqrt((2* (H_1 - H_2 ) ) + U^2_1)
U_2 = sqrt ( 2 * 10^3 * (3112.5 -2945.7) + 900)
U_2 = 578.359 m / s
b.
Mass Balance :
U_1 * A_1 / V_1 = U_2 * A_2 / V_2
A = pi*d^2 / 4
U_1 * d_i ^ 2 / V_1 = U_2 * d_e ^2 / V_2
d_e = d_i * sqrt ( (U_1 / U_2) * (V_2 / V_1))
d_e = 5 * sqrt ((30 / 578.359) * (667.75 / 388.61))
d_e = 1.4924 cm
The exit velocity and exit diameter for the steam flow are respectively; 578.36 m/s and 1.4924 cm
What is the exit velocity?
We are given;
Length of nozzle; L = 25 cm
Inlet diameter d_i = 5 cm
Entrance temperature; T₁ = 325 °C
Entrance Pressure; P₁ = 700 KPa
Entrance velocity; v₁ = 30 m/s
Entrance enthalpy; H₁ = 3112.5 KJ/kg
Entrance volume; V₁ = 388.61 cm³/g
Exit temperature; T₂ = 250 °C
Exit Pressure; P₂ = 350 KPa
Exit enthalpy; H₂ = 2945.7 KJ/kg
Exit volume; V₂ = 667.75 cm³/g
a) From energy Equation, we have;
(H₂ - H₁) + (v₂² - v₁²)/2 = 0
v₂ = √(2*( H₁ - H₂)) + v₁²)
v₂ = √(2 * 10³ * (3112.5 - 2945.7) + 900)
v₂ = 578.36 m/s
B) From mass Balance, we know that;
v₁ * A₁/V₁ = v₂ * A₂/V₂
where;
A = π*d²/4
v₁ * d₁²/V₁ = v₂ * d₂²/V₂
d₂ = d₁√((v₁/v₂) × (V₂/V₁))
d₂ = 5√((30/578.359) × (667.75 / 388.61))
Exit diameter is; d₂ = 1.4924 cm
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