Answer:
Given,
Temperature;
T = 393;;K
Convert to Celcius;
T = (393-273) degrees
T = 120°C
Using Table A-4 (Saturated water - Temperature table), at T = 120 C;
vf = 0.001060 m³/kg
vg = 0.89133 m³/kg
Quality is given as;
75% = 0.75
Specific volume is given as;
v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)
v= 0.66876 m³/kg
We know;
v = V/m
0.66876 = 100/m
m = 149.53 kg
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:
The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)
Explanation:
Since the given volume flow rate is gallons per minute.
We know that 1 gallon = 3.785 liters and
1 minute = 60 seconds
Let the flow rate be 
Now replacing the gallon and the minute by the above values we get

Thus 
Now since we know that 1 liter = 
Using this in above relation we get

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.
A.
It would be released without a doubt so be careful!
Hope this helps :)