Paragraph summary on airplane history
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Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k
Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance = = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = = 240 / 0.11 = 2181.82 ohms