Question 6.12: A 1.5 m³ tank contains 500 kg of liquid water in equilibrium...

Choose the tank as the control volume. There is no work, and kinetic- and potentialenergy changes are assumed negligible. Equation (2.28) therefore is written:

\frac{d(m U)_{\mathrm{cv}}}{d t}+\Delta(H \dot{m})_{\mathrm{fs}}=\dot{Q}+\dot{W}       (2.28)

\frac{d(m U)_{\operatorname{tank}}}{d t}-H^{\prime} \dot{m}^{\prime}=\dot{Q}

where the prime denotes the state of the inlet stream. The mass balance, \dot{m}^{\prime}= d  m_{tank} / dt , may be combined with the energy balance to yield:

\frac{d(m U)_{\text {tank }}}{d t}-H^{\prime} \frac{d m_{\text {tank }}}{d t}=\dot{Q}

Multiplication by dt and integration over time (with H^{\prime} constant) give:

 Q = Δ ( mU )_{tank}   − H^{\prime} Δ m_{tank}

The definition of enthalpy may be applied to the entire contents of the tank to give:

 Δ ( mU )_{tank} = Δ ( mH ) _{tank}  −  Δ ( PmV )_{tank}

Because total tank volume mV and P are constant, Δ(PmV)_{tank} = 0. Then, with
\Delta(m H)_{\operatorname{tank}}=\left(m_2 H_2\right)_{\operatorname{tank}}-\left(m_1 H_1\right)_{\operatorname{tank}} , the two preceding equations combine to yield:

Q = ( m _2 H_2 ) tank  −  ( m_1 H_1 ) tank  −  H^{\prime} Δ m tank      (A)

where Δm_{tank}  is the 750 kg of water bled into the tank, and subscripts 1 and 2 refer to conditions in the tank at the beginning and end of the process. At the end of the process the tank still contains saturated liquid and saturated vapor in equilibrium at 100°C and 101.33 kPa. Hence, m_1H_1  and  m_2H_2  each consist of two terms, one for the liquid phase and one for the vapor phase. The numerical solution makes use of the following enthalpies taken from the steam tables:

H′ = 293.0  kJ⋅ kg^{−1} ; saturated liquid at 70° C

H^{ l}_{tank} = 419.1  kJ⋅ kg^{−1} ;  saturated liquid at  100° C

H^{ v}_{tank} = 2676.0  kJ⋅ kg^{−1} ;  saturated vapor at 100° C

The volume of vapor in the tank initially is 1.5 m³ minus the volume occupied by the 500 kg of liquid water. Thus,

m_1^v=\frac{1.5-(500)(0.001044)}{1.673}=0.772 \mathrm{~kg}

where 0.001044 and 1.673  m^{3}·kg^{−1} are the specific volumes of saturated liquid and saturated vapor at 100°C from the steam tables. Then,

At the end of the process, the masses of liquid and vapor are determined by a mass balance and by the fact that the tank volume is still 1.5 m³:

m_2=500+0.722+750=m_2^v+m_2^l

1.5 = 1.673  m_2^v + 0.001044  m_2^l

Solution gives:

m_2^l  = 1250.65  kg  and  m_2^v= 0.116  kg

Then, with H_2^l  = H_{1}^{l}  and  H_2^v = H_1^v

( m_2 H_2 )_{tank} = ( 1250.65 ) ( 419.1 ) + ( 0.116 ) ( 2676.0 ) = 524,458  kJ

Substitution of appropriate values into Eq. (A) gives

Q = 524,458 − 211,616 − ( 750 ) ( 293.0 ) = 93,092 kJ

Before conducting the rigorous analysis demonstrated here, one might make a reasonable estimate of the heat requirement by assuming that it is equal to the enthalpy change for heating 750 kg of water from 70°C to 100°C. This approach would give 94,575 kJ, which is slightly higher than the rigorous result, because it neglects the heat effect of condensing water vapor to accommodate the added liquid.