Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ = 
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ = 
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction, 
, is given as follows;
= T₀ × 
= T₀ × 
× (s₄ + s₂ - s₁ - s₃)
= T₀ × 
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output, 
= 
+ ≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.
Answer:
small guitar with no strings?
Explanation:
it would be fun to make i think
Answer:
51.4 Ohms
Explanation:
By applying voltage division rule
where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then
Answer:
The mass flow rate of steam m=5.4 Kg/s
Explanation:
Given:
At the inlet of turbine P=10 MPa ,T=500 C
AT the exit of turbine P=10 KPa ,x=0.9
Required power=5 MW
From steam table
<u> At 10 MPa and 500 C:</u>
h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)
<u>At 10 KPa:</u>
=2675.1 KJ/Kg, 
=417.51 KJ/Kg
= 7.3 KJ/Kg-K ,
=1.3 KJ/Kg-K
So enthalpy of steam at the exit of turbine
h= 
Now by putting the values
h= 417.51+0.9(2675.1- 417.51) KJ/Kg
h=2449.34 KJ/Kg
Lets take m is the mass flow rate of steam
So 
m=5.4 Kg/s
So the mass flow rate of steam m=5.4 Kg/s
