Question

Saturated steam at 232.1 kPa and 80% quality is being used to heat a liquid food by direct steam injection. The product enters the heating system at 200 kg/min and 158C, and leaves at 1058C. The specific heat of the product is 3.85 kJ/kgK and does not change significantly during the heating process. Estimate the steam requirement.

Answer 1

To develop this problem we will apply the thermodynamic concepts of the first law, in which we will make a balance of the first law. Then look for me in the thermodynamic tables under the conditions given the values for the enthalpy (Considering the given quality). We will calculate from this the enthalpy of the system and finally we will find the steam requirement

$C_p = 3.85kJ/kg$

$\dot{m_p} = 200kg/min$

$T_1 = 15\°C$

$T_2 = 105\°C$

Seam thermodynamics table,

$P = 232.1kPa = 2.321bar$

$\mu = 0.8$

$h_c = 524.99kJ/kg$

$h_v = 2713.5kJ/kg$

Enthalpy system,

$h_s = x(h_v-h_c) +H_c$

$h_s = (0.8)(2713.79-524.99)+524.99$

$h_s = 2275.79kJ/kg$

From energy balance

$\dot_{m_A}C_{pa}(T_A-0)+(\dot{m_s}-\dot{m_A})h_s = \dot{m_s}C_{pa}(T_B-0)$

$(200)(3.85)(15-0)+(\dot{m_s}-200)(2275.79)=\dot{m_s}(3.85)(105-0)$

$\dot{m_s} = 237.03kg/min$

$S = 237.03-200$

$S = 37.03 kg/min$