Answer:
a) 3581.15067 kw
b) 95.4%
Explanation:
<u>Given data:</u>
compressor efficiency = 85%
compressor pressure ratio = 10
Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K
At turbine inlet : pressure = 950 kPa, temperature = 1400k
Turbine efficiency = 88% , exit pressure of turbine = 100 kPa
A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW
attached below is a detailed solution to the given question
Answer:
False
Explanation:
No matter if something happened in the past year or so, it still should be included for safety reasons so it wont happen again
Answer:
V₂ = 20 V
Vt = 20 V
V₁ = 20 V
V₃ = 20 V
I₁ = 10 mA
I₃ = 3.33 mA
It = 18.33 mA
Rt = 1090.91 Ω
Pt = 0.367 W
P₁ = 0.2 W
P₂ = 0.1 W
P₃ = 0.067 W
Explanation:
Part of the picture is cut off. I assume there is a voltage source Vt there?
First, use Ohm's law to find V₂.
V = IR
V₂ = (0.005 A) (4000 Ω)
V₂ = 20 V
R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.
V₁ = 20 V
V₃ = 20 V
Vt = 20 V
Now we can use Ohm's law again to find I₁ and I₃.
V = IR
I = V/R
I₁ = (20 V) / (2000 Ω)
I₁ = 0.01 A = 10 mA
I₃ = (20 V) / (6000 Ω)
I₃ = 0.00333 A = 3.33 mA
The current It passing through Vt is the sum of the currents in each branch.
It = I₁ + I₂ + I₃
It = 10 mA + 5 mA + 3.33 mA
It = 18.33 mA
The total resistance is the resistance of the parallel resistors:
1/Rt = 1/R₁ + 1/R₂ + 1/R₃
1/Rt = 1/2000 + 1/4000 + 1/6000
Rt = 1090.91 Ω
Finally, the power is simply each voltage times the corresponding current.
P = IV
Pt = (0.01833 A) (20 V)
Pt = 0.367 W
P₁ = (0.010 A) (20 V)
P₁ = 0.2 W
P₂ = (0.005 A) (20 V)
P₂ = 0.1 W
P₃ = (0.00333 A) (20 V)
P₃ = 0.067 W
