Section 4.4.10.1: Material and energy balance for a tank

Assume water properties.

Equation name

Equation

  

Variables (units)
Balances      
Material balance $ \frac{dM}{dt} = F_1 - F_2 $ M holdup (kg)
    F1 feed flow (kg/s)
    F2 outlet flow (kg/s)
Energy balance $ \frac{dU}{dt} = h_1 F_1 - h_2 F_2$ U internal energy of contents (kJ)
(note 1)   h1 feed specific enthalpy (kJ/kg)
    h2 outlet specific enthalpy (kJ/kg)
Tank Model      
gravity flow model $vF_2 - 0.005 \sqrt{L} = 0$ v specific volume ( m3/kg)
    L level of contents (m)
`well mixed' model T - T2 = 0 T tank temperature (0C)
    T2 outlet temperature (0C)
mass - volume rel'n M v - V = 0 V volume of contents (m3)
volume - level rel'n V - 0.2 ((1 + 0.8 L)3 - 1) = 0    
(note 2)      
Thermo and definitions    
energy - enthalpy rel'n u - h + pv = 0 u specific internal energy (kJ/kg)
    h specific enthalpy (tank) (kJ/kg)
    p pressure (kPa) (note 3)
specific energy def'n U - M u = 0    
Phys props      
enthalpy correl'n h1 - 4.1846 T1 = 0 T1 feed temperature (0C)
(feed)      
enthalpy correl'n h - 4.1846 T = 0    
(tank)      
enthalpy correl'n h2 - 4.1846 T2 = 0    
(outlet)      
density correl'n (998.23 - 0.255 (T - 20)) v - 1 = 0    
Specifications      
feed flow spec $F_1 - \left(2 + \frac{0.02 t}{1 + 0.01 t}\right) = 0 $    
feed temp. spec T1 - 20 = 0    
pressure spec p - 100 = 0    

Notes:

1.
rate of change of energy holdup = net enthalpy flow (in - out)
2.
tank assumed to have sloping sides (radius a linear function of level).
3.
u = h - pv: assume p is in kPa (1 bar = 100 kPa) and v in m3/kg; then units of u and h are consistent (kJ/kg).

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