Respuesta :
Answer:606 m/s
Explanation:
Given
Steam Inlet temperature [tex]400^{\circ} C[/tex] and pressure 800 KPa
[tex]h_1=3267.7 kJ/kg[/tex]
initial Velocity 10 m/s
Steam Outlet Velocity is [tex]300^{\circ} C[/tex] and pressure is 200 KPa
[tex]h_2=3072.1 kJ/kg[/tex]
[tex]\nu _2=1.31632[/tex]
From steam table
Heat loss 25 KW
inlet area 800 cm^2
applying Steady Flow Energy Equation
[tex]h_1+\frac{1}{2}v_1^2++Q=h_2+\frac{1}{2} v_2^2+W[/tex]
[tex]3267.7+\frac{1}{2000}10^2-25=3072.1+\frac{1}{2000}v_2^2[/tex]
[tex]3267.7-3072.1+0.05-25=\frac{1}{2}v_2^2[/tex]
[tex]v_2=\sqrt{195.65\times 2}=606 m/s[/tex]
and volume flow rate is [tex]m=\dot{m}\mu _2=2.082\times 1.31623=2.74 m^3/s[/tex]
For an inlet area of 800cm², The veloctiy of the steam at nozzle exit is 305.939 m/s while The flow rate of the steam at nozzle exit is 2.740 m^3/s
Steady flow energy equation?
Steady state flow is when the property of fluid does not change with time such properties include pressure, temperature, velocity.
Steady flow energy equation is the equation representing teh steady state flow represented as:
Q - W = m ( h2 - h1 ) + m ( c2^2 - c1^2 ) / 2000 + mg ( z2 - z1 )
where
Q = heat transfer kW
H = work transfer kW
h = specific enthalpy kj/kg
m = mass flow rate kg/s
c = velocity m/s
z = elevation m
g = acceleration due to gravity m/s^2
1 = inlet conditions
2 = outlet conditions
from steam table
T1 = 400 C
P1 = 800 kPa
gives h1 = 3267.5613 kj/kg and ρ1 = 2.6023.789 kg/m^3
T2 = 300 C
P2 = 200 kPa
gvies h2 = 3072.089 kj/kg and ρ2 = 0.789 kg/m^3
ρ = density
Q + m ( h1 + c1^2 / 2000 ) = W + m ( h2 +c2^2 / 2000 )
-25 + m ( 3267.5613 + 10^2 / 2000 ) = 0 + m ( 3072.089 +c2^2 / 2000 )
m = ρ1 * c1 * A1
m = 0.789 * 10 * 0.08
m = 2.0818 kg/s
substituting m into the equation gives
c2 = 305.939 m/s
flow rate of the steam at the exit Q2
Q2 = m / ρ2
Q2 = 2.0818 / 0.7597
Q2 = 2.740 m^3/s
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