Answer:
a) 358.8 KJ/kg
b) 0.0977 KJ/K- kg
c) 83.28%
Explanation:
N2 at 300 k. ( use the properties of N2 at 300 k (T1) )
Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk , R = 0.1297 KJ/kgk , y = 1.4 ,
Given data:
T2 = 645 k
P1 = 1 bar , P2 = 10.5 bar
<u>a)Determine the work input in KJ/Kg of N2 flowing </u>
Winput = h2 - h1 = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg
<u>b) Determine the rate of entropy in KJ/K- kg of N2 flowing </u>
Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1
= 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )
= 0.0977 KJ/K- kg
<u>c) Determine isentropic compressor efficiency </u>
Isentropic compressor efficiency = 83.28%
calculated using the relation below
( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )
where T'2 = 587.314
Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.
Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:
The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:
= moles of a gas / total number moles of gas
The rigid tank has total pressure of 1MPa.
molar mass = 14g/mol
mass in the tank = 2000g
number of moles in the tank: = 142.85mols
molar mass = 44g/mol
mass in the tank = 4000g
number of moles in the tank: = 90.91mols
Total number of moles: 142.85 + 90.91 = 233.76 mols
To calculate partial pressure:
For Nitrogen gas:
= 0.6
For Carbon Dioxide:
0.4
Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.
Answer:
The initial temperature is 649 K (376 °C).
The final pressure is 0.965 MPa
Explanation:
From the ideal gas equation
PV = nRT
P is the initial pressure of water = 2 MPa = 2×10^6 Pa
V is intial volume = 150 L = 150/1000 = 0.15 m^3
n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol
R is gas constant = 8.314 m^3.Pa/mol.K
T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C
From pressure law,
P1/T1 = P2/T2
P2 (final pressure) = P1T2/T1
T2 (final temperature) = 40 °C = 40 + 273 = 313 K
P1 (initial pressure) = 2 MPa
T1 (initial temperature) = 649 K
P2 = 2 × 313/649 = 0.965 MPa
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