Question 7.51: A 0.5 m^3 tank contains carbon dioxide at 300 K, 150 kPa is ...

A 0.5 m ^{3} tank contains carbon dioxide at 300 K, 150 kPa is now filled from a supply of carbon dioxide at 300 K, 150 kPa by a compressor to a final tank pressure of 450 kPa. Assume the whole process is adiabatic and reversible. Find the final mass and temperature in the tank and the required work to the compressor.

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C.V. The tank and the compressor.
Continuity Eq.4.15:            m _{2}- m _{1}= m _{ in }

Energy Eq.4.16:                  m _{2} u _{2}- m _{1} u _{1}={ }_{1} Q _{2}-{ }_{1} W _{2}+ m _{ in } h _{ in }

Entropy Eq.7.13:                m _{2} s _{2}- m _{1} s _{1}=\int d Q / T +{ }_{1} S _{2  gen }+ m _{ in } s _{ in }

Process:         Adiabatic    { }_{1} Q _{2}=0 ,         Process ideal  { }_{1} S _{2 \text { gen }}=0, \quad s _{1}= s _{ in }

\Rightarrow m _{2} s _{2}= m _{1} s _{1}+ m _{ in } s _{ in }=\left( m _{1}+ m _{ in }\right) s _{1}= m _{2} s _{1} \Rightarrow s _{2}= s _{1}

Constant s so since the temperatures are modest use Eq.6.23

T _{2}= T _{1}\left(\frac{ P _{2}}{ P _{1}}\right)^{\frac{ k -1}{ k }}=300   K (450 / 150)^{0.2242}= 3 8 3 . 7 9   K

 

m _{1}= P _{1} V _{1} / RT _{1}=\frac{150   kPa \times 0.5   m ^{3}}{0.1889   kJ / kg – K \times 300   K }=1.3235   kg

 

m _{2}= P _{2} V _{2} / RT _{2}=450   kPa \times 0.5   m ^{3} /(0.1889 \times 383.79)  kJ / kg = 3 . 1 0 3 5   kg

 

\Rightarrow   m _{ in }=1.78   kg

 

\begin{aligned}{ }_{1} W _{2} &= m _{ in } h _{ in }+ m _{1} u _{1}- m _{2} u _{2}= m _{ in } RT _{ in }+ m _{1}\left( u _{1}- u _{ in }\right)- m _{2}\left( u _{2}- u _{ in }\right) \\&=1.78(0.1889 \times 300)+ m _{1}(0)-3.1035 \times 0.653(383.79-300) \\&=- 6 8 . 9 3   kJ \quad \text { (work must come in })\end{aligned}

 

…………………………………….

Eq.4.15 : 1=\frac{\dot{m}_{1}}{\dot{m}_{3}}+\frac{\dot{m}_{2}}{\dot{m}_{3}}

Eq.4.16 : 0=\frac{\dot{m}_{1}}{\dot{m}_{3}} h_{1}+\frac{\dot{m}_{2}}{\dot{m}_{3}} h_{2}-h_{3}+\dot{Q} / \dot{m}_{3}

Eq.7.13 : \left(m_{2} s_{2}-m_{1} s_{1}\right)_{ c.v. }=\sum m_{i} s_{i}-\sum m_{e} s_{e}+\int_{0}^{t} \frac{\dot{Q}_{ c.v. }}{T} d t+{ }_{1} S_{2 gen }

Eq.6.23 : \frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}

 

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