Cual es el costo del kwh
1 answer:
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Answer:
1) In series, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.
2) In parallel, the combined head and volume flow will move along the system curve from point 1 to point 3.
Explanation:
1) Pump in series:
When two or more pumps are connected in series, their resulting pump performance curve will be obtained by adding their respective heads at the same flow rate as shown in the first diagram attached.
In the first diagram, we have 3 curves namely:
- system curve
- single pump curve
- 2 pump in series curve
Also, we have points labeled 1, 2 and 3.
- Point 1 represents the point that the system operates with one pump running.
- Point 2 represents the point where the head of two identical pumps connected in series is twice the head of a single pump flowing at the same rate.
- Point 3 is the point where the system is operating when both pumps are running.
Now, since the flowrate is constant, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.
2) Pump in parallel:
When two or more pumps are connected in parallel, their resulting pump performance curve will be obtained by adding their respective flow rates at same head as shown in the second diagram attached.
In the second diagram, we have 3 curves namely:
- system curve
- single pump curve
- 2 pump in series curve
Also, we have points labeled 1, 2 and 3
- Point 1 represents the point that the system operates with one pump running.
- Point 2 represents the point where the flow rate of two identical pumps connected in series is twice the flow rate of a single pump.
- Point 3 is the point where the system is operating when both pumps are running.
In this case, the combined head and volume flow will move along the system curve from point 1 to point 3.
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.