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Igoryamba
3 years ago
7

A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa and 400°C. It is cooled to 40°

C and 300 kPa. Calculate the heat transfer for the process.
Engineering
1 answer:
leonid [27]3 years ago
7 0

Answer:

heat transfer for the process is - 643.3 kJ

Explanation:

given data

mass m = 2 kg

pressure p1 = 500 kPa

temperature t1 = 400°C = 673.15 K

temperature t2 = 40°C = 313.15 K

pressure p2 = 300 kPa

to find out

heat transfer for the process

solution

we know here mass is constant so

m1 = m2

so by energy equation

m ( u2 - u1 ) = Q - W

Q is heat transfer

and in process  P =  A+ N that is linear spring

so

W = ∫PdV

= 0.5 ( P1+P2) ( V1 - V2)

so for case 1

P1V1 = mRT

put here value

500 V1 = 2 (0.18892) (673.15)

V1 = 0.5087 m³

and

for case 2

P2V2 = nRT

300 V2 = 2 (0.18892) (313.15)

V2 = 0.3944 m³

and

here W will be

W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )

W = -45.72 kJ

and

Q is here for Cv = 0.83 from ideal gas table

Q = mCv ( T2-T1 ) + W

Q = 2 × 0.83 ( 40 - 400 ) - 45.72

Q = - 643.3 kJ

heat transfer for the process is - 643.3 kJ

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A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
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Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

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P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

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N = 9.66×10²⁴

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The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

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3 years ago
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