Question 6.236E: Nitrogen at 90 lbf/in.^2, 260 F is in a 20 ft^3 insulated ta...

CV Both tanks + pipe + valve Insulated : Q = 0            Rigid: W = 0

Energy Eq.3.5:            m \left( u _{2}- u _{1}\right)=0-0 \quad \Rightarrow u _{2}= u _{1}= u _{ a 1}

Entropy Eq.6.37:        m \left( s _{2}- s _{1}\right)=\int dQ / T +{ }_{1} S _{2  gen} ={ }_{1} S _{2  gen}  \quad( dQ =0)

State 1:     P _{1}, T _{1},      V _{ a } \quad \Rightarrow      Ideal gas

m = PV/RT = (90 × 20 × 144)/ (55.15 × 720) = 6.528 lbm

State 2:    V _{2}= V _{ a }+ V _{ b } ;     \text { uniform final state } \quad v _{2}= V _{2} / m ; \quad u _{2}= u _{ a 1}

Ideal gas   u (T)  \Rightarrow u _{2}= u _{ a 1} \quad \Rightarrow T _{2}= T _{ a 1}= 7 2 0   R

From entropy equation and Eq.6.19 for entropy change

Irreversible due to unrestrained expansion in valve P ↓ but no work out.

If not a uniform final state then flow until P _{2 b }= P _{2 a } and valve is closed.

Assume no Q between A and B

Now we must assume m _{ a 2} went through rev adiabatic expansion

1) V _{2}= m _{ a 2} v _{ a 2}+ m _{ b 2} v _{ b 2};                    2) P _{ b 2}= P _{ a 2};

3) s _{ a 2}= s _{ a 1};                                    4) Energy eqs.

4 Eqs           4 unknowns :  P _{2}, T _{ a 2}, T _{ b 2}, x = m _{ a 2} / m _{ a 1}

Energy Eq.:    m_{a 2}\left(u_{a 2}-u_{a 1}\right)+m_{b 2}\left(u_{b 2}-u_{a 1}\right)=0

Vol constraint:      P _{2} V _{2} / m _{ a 1} R = x T _{ a 2}+(1- x ) T _{ b 2}= T _{ a 1}

Now we have final state in A

Substitute into energy equation

Comment: This shows less entropy generation as the mixing part is out compared with the previous solution.

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Eq.3.5: E_{2}-E_{1}={ }_{1} Q_{2}-{ }_{1} W_{2}

Eq.6.37 : S_{2}-S_{1}=\int_{1}^{2} d S=\int_{1}^{2} \frac{\delta Q}{T}+{ }_{1} S_{2 gen}

Eq. 6.19 : s_{2}-s_{1}=\left(s_{T 2}^{0}-s_{T 1}^{0}\right)-R \ln \frac{P_{2}}{P_{1}}