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IgorC [24]
2 years ago
7

A compressor receives air at 290 K, 95 kPa and shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate o

f 0.01 kg/s air to a pipeline. Find the maximum possible exit pressure of the compressor.
Engineering
1 answer:
aev [14]2 years ago
3 0

Answer:

P2 = 3.9 MPa

Explanation:

Given that

T₁ = 290 K

P₁ = 95 KPa

Power P = 5.5 KW

mass flow rate  = 0.01 kg/s

solution

with the help of table A5

here air specific heat and adiabatic exponent is

Cp = 1.004 kJ/kg K

and k = 1.4

so

work rate will be

W = m × Cp × (T2 - T1)              ..........................1

here T2 = W ÷ ( m × Cp) + T1    

so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290

T2 = 838 k

so final pressure will be here

P2 = P1 × (\frac{T2}{T1})^\frac{k}{k-1}        ..............2

P2 = 95 × (\frac{838}{290})^\frac{1.4}{1.4-1}

P2 = 3.9 MPa

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A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect
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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:

P_{total} = P_{1}+P_{2}+...

The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:

X_{i} = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

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molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: n=\frac{2000}{14} = 142.85mols

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mass in the tank = 4000g

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Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

P_{i}=P_{total}.X_{i}

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P_{N_{2}}=1.\frac{142.85}{233.76}

P_{N_{2}} = 0.6

For Carbon Dioxide:

P_{total}=P_{N_{2}}+P_{CO_{2}}

P_{CO_{2}} = P_{total}-P_{N_{2}}

P_{CO_{2}}=1-0.6

P_{CO_{2}}= 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

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