Answer to Question #177045 in Mechanical Engineering for Micheal Vaughn

Question

Simple two stage (effect) parallel flow multi-effect evaporation unit with vapour compression A two-effect thermal vapor-compression system is designed at the following operating conditions: Motive steam ms: Ps: Ts mm; Pm Tm Feed water mf 1 mt2 Cooling Sea water md2 mev: Pev: Tev Demiater (1) (2) Intake Heating Sea water Steam mm md1 Down condenser md2-mev md Distillate Condensate mb 1 mb2 mb Rejected brine mev The areas of the exchangers are equal: Ae1=Ae2 - Boiling temperatures, T61= 75°C; Tb2, of 70°C; Feed water temperature; T= 65°C. - Motive steam pressure, Pm = 800 kPa. - Brine reject concentration, Xb= 60,000 ppm; Intake seawater salinity, Xp = 30,000 ppm - Intake seawater temperature, Tcw = 15°C System capacity, ma= 1 kg/s Determine the specific heat transfer area, the entrainment ratio, and the flow rates of: entrainment vapor; motive steam; supply steam and cooling water, and the performance ratio.

Answer 1

As per the data

$Q_s=mc_p \triangle T$

Stream mass flow rate $\frac{Q_s}{C_p \triangle T}= \frac{8000}{4.2 \times(75-70)}=142.85 kg/s$

$Q_s=Q_k$

$mC_p \triangle T=mC_p \triangle T$

Specific heat of sea water = 3.85 kg/s

$m_s C_{p_s} \triangle T=m_cC_{p_c} \triangle T$

$m_c= \frac{m_s C_{p_s} \triangle T}{C_{p_c} \triangle T}= \frac{142.85 \times4.2 \times (75-70)}{3.85 \times(65-15)}=15.58 kg/s$

$Q_s=3000kW ; T_{cw}=15^OC; T_{b1}=75^oC; T_{b2}=70^oC; T_F=65^oC$

$x_f=35000, x_b=30000$

$A_{e1}=A_{e2}$

$U_{e1}=U_{e2}=2.25 kW/m^{2 \space o}C$

$U_{c}=1.9 kW/m^{2 \space o}C$

$C_p=4.2$

$\triangle T_1=75-15=60^oC$

$\triangle T_2=70-65=5^oC$

$\triangle TIn(\triangle T_1-\triangle T_2)= \frac{60.5}{In \frac{60}{5}}=22.16 ^oC$

$U=2.25 kW/m^{2 \space o}C=2250 W/m^{2 \space o}C$

$Q=UA\triangle TIn(\triangle T_1-\triangle T_2)$

$A = \frac{Q}{U\triangle TIn(\triangle T_1-\triangle T_2) }= \frac{8000}{2.250 \times 22.13}=60.25 m^2$