Question 7.236E: A rigid 35 ft^3 tank contains water initially at 250 F, with...

C.V. Tank and walls out to the source. Neglect storage in walls. There is flow out and no boundary or shaft work.

Continuity Eq.4.15:        m _{2}- m _{1}=- m _{ e }

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

Entropy Eq.7.12:          m _{2} s _{2}- m _{1} s _{1}=- m _{ e }s_{e}+\int dQ / T +{ }_{1} S _{2  gen }

State 1: 250 F, Table F.7.1                v _{ f1 }=0.017, \quad v _{ g 1}=13.8247   ft ^{3} / lbm

state 2:

From the energy equation we get

……………………………………..

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.7.12 : \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} \sum_{ c.s. } \frac{\dot{Q}_{ c.v. }}{T} d t+{ }_{1} S_{2 gen }