Cours project by discipline «basic technological processes and devices» content

4. Technological calculation.
Design a single-body evaporating chamber an installation for concentrating G=26702 kg/h (7.417 kg/h) of aqueous ammonium nitrate solution from the initial mass concentration of x_initial=29% to x_exit=79% under the fallowing conditions:

1. 
   Heating is carried out by saturated water vapor pressure P_g.p=0.35 MPa.
   
2. 
   Absolute pressure in the vapor space of the evaporator P_b.k=0.018 MPa.
   
3. 
   The temperature of the solution entering the installation t₀=20˚C.
   
4. 
   The temperature of the solution entering the evaporator t₁=60 ˚C.
   
5. 
   The initial temperature of the cooling water t=12 ˚C.
   
6. 
   The temperature of the mixture of cooling water and condensate coming out of the barometric condenser is lower than condensation temperature by Δt=5 ˚C.
   
7. 
   Evaporating apparatus with a remote heating chamber type 2.
   

1. 
   Material balance.
   
   1. 
      The quantity of evaporated water is determined by the formula:
      

G_b=26702/3600=7.417 kg/sek
W=G_H*(1-x_initial/x_exit)=7.417*(1-29/79)=4.74 kg/h

2. 
   Quantity of the evaporated solution:
   

G_ox=G_b-W
G_ox=7.417-4.74=2.67 kg/h

2. 
   Thermal calculation.
   

2.1) Determination of the boiling point of the solution.
The boiling point of the solution is determined by the formual (6.4)
The temperature of the secondary steam coming from the evaporator in barometric condenser at P=0.0185 MPa, equal to 57.5 ˚C.
Based on practical recommendations, we accept hydraulic depression
Δ՜՜՜=1 ˚C. Than the temperature of the secondary steam in the evaporated apparate ravia:
T_n՜₌ 57.5+1=58.5
This temperature corresponds to the pressure P=0.0207MPa, heat of vaporization r=2362.55 kJ/kg.
Hydrostatic depression Δ՜՜ is defined as follows.
To select the value of the pipe height H, the heat transfer surface area of the evaporator is determined by orientation F_about.
F_about=w*r/q
When boiling aqueous solutions of salts, the specific heat load for devices with natural circulation is q=20000÷50000 W/m². We accept q=30000 W/m². Then:
F_about =4.74*2362.55*10³/30000=373.28 m²
According to GOST 11987-81, we accept a vaporizer with the following characteristics:
-heat exchange surface 355 m²;
-pipe length 5m;
-diameter of the tube 38x2 mm;
-pitch between pipes 48 mm;
-pipe material- steel X18H10T.
Approximately accept the boiling point of the solution in apparatus 20˚C more than the temperature of secondary steam t_boil≈80˚C. Consequently,
h_yp=[0.26+0.0014*(1289-972)]*5=2.47m
Where 1289 and 972 kg/m³ correspond to the density of the solution and water at 80°C.
Pressure increase in the middle layer of the solution:
𝛥P_г=(p(80˚C)*g*h)/2
𝛥P_г=1289*9.81*2.47/2= 156000 Pa= 0.022 MPa
Thus, the pressure in the middle layer of the boiling pipes:
P_г=0.0207+0.0156=0.0363 MPa
This pressure correspond to the boiling point of water t=75.4°C and the heat of vaporization r=2320 kJ/kg^.
Hence, the hydrostatic depression of ravia:
𝛥´´=75.4-58.5=16.9°C
Temperature depression is determined by the formula (6.75):
𝛥´=16.2*(273+75.4)²*15.26/2320*10³)=12.93˚C.
where 𝛥_atm=15.26 ˚C is the temperature depression of 77% solution at atmospheric pressure^.
Boiling point of the solution (formula 6.4):
t_s= t_{secondary steam}+ 𝛥= 57.5+12.93+16.9+1=87.33 ˚C.

2. 
   Useful temperature difference:
   

𝛥t_useful=T-t_boiling
𝛥t_пол=137.9-87.33=50.57 ˚C.
Here: 𝑃_𝑏=0.405MPa found from the vapor pressure of water heated to t=137.9⁰C.
2.3 Heating steam consumption G_h.s we determine from the thermal balance equation (formula 6.14).
G_{hot steam}=((4.74*2650*10³)+(2.67*2618.1*87.33)(7.417*3866*60))/0.95*2171*10³=4.89kg/sec

Where i=2650 kJ/kg is the enthalpy of secondary steam; c_i=3866J/kg*K specific of heat capacity of 10% solution; c_k=2618.1 J/kg*K is the specific heat capacity of the evaporated solution_[9]; r_h.s=2171 kJ/kg is the specific heat of water vaporization at p_h.s=0.35 MPa_[5]; 0.95 is a coefficient that takes into account 5% of heat loss.

Thermal load of the heating chamber:
Q=G_hot.steam*r_hot.steam
Q=4.89*2171*10³=10616.19*10³ Wt

3. 
   Calculation of the heat transfer coefficient.
   

31. The number of pipes of the heating chamber is determined by the (formula6.34):
n=355/(3.14*0.038*5*0.038)=595
The density of condensate passing outside the pipe is determined using the following formula:
Г=G_hot.steam/(n*d*
Г=4.89/(595*0.038*3.14)=0.068 kg/m*sec

The Re_пл criterion for condensate dynamics is determined by the formula(5.59);

Re_пл=4* Г/µ
Re_пл =(4*0.068)/(0.251*10^-3)=1083.6
Where µ=0.251*10³ P*sec is the dynamic viscosity coefficient of water at condensation temperature t=112.7˚C^.
Reduced film thickness:
δ_пр=[(0.251*10^-3)²/(948.5²*9.81)]^0.33=0.2*10^-4m
Using 𝑅𝑒_𝑝𝑙 > 400, we find 𝑁𝑢_𝑝𝑙:
Nu_pl=[Re_pl/((6.25*(Re_pl-400)/Pr^0.33)]
Nu_pl=[1083.6/((6.25*(1083.6-400)/1.56^0.33)=0.29
We determine the coefficient of heat transfer of condensed water to the pipe wall 𝛼₁ using the following formula:
α₁= Nu_пл* λ/ δ_пр
α₁=0.29*0.685/(0.2*10^-4)=9932.5Wt/m²*K.
3.2 Finding the heat transfer coefficient of the solution. Determination of the heat transfer coefficient 𝛼2 passing through the wall to the middle layer of the solution: we take the boiling temperature of the solution as constant. (Measurements of 77% solution. Boiling temperature of solution
𝑡=89.32⁰C)
Solution constants at boiling point_[9]:
p=1215.6 kg/m²; µ=1.24*10^-3 Pa*sec; λ=0.695 Wt/m*K; c=3475 J/kgK;
σ=12*10^-3 N/m³.
Properties of water vapor at P = 0.039 𝑀𝑃𝑎: 𝑡 = 75.4⁰C and heat of steam formation 𝑟 = 2297 𝑘𝑗/𝑘𝑔, 𝜌 =0.35 𝑘𝑔/𝑚³
Density of water vapor at P = 0.1 MPa (100 ⁰C): 𝜌 = 0.579 𝑘𝑔/𝑚³
α₂=(λ^1.3*p^0.5*p_{steam.mid.lay}^0.06*q^0.6)/(σ^0.5*r_{steam.mid.lay}^0.6*p_steam^0.66*c^0.3*µ^0.3)
α₂=(780*0.695^1.3*1215.6^0.5*0.35^0.06*q^0.6)/(0.12^0.5*2297000^0.6*0.579^0.66*3475^0.3*(1.24*10^-3)^0.3=4.09q^0.6
3.3 The heat transfer coefficient is found from the following formula:
K=1/((1/9932.5)+(0.002/16.4)+(1/5000)+(1/5800)+(1/4.09q^0.6))=1/(0.169*10^-3+0.24/q^0.6)
𝛿𝑡𝑟 = 0.002 𝑚-pipe wall thickness:
𝜆𝑡𝑟 = 16.4 𝑉𝑡/𝑚 ∙ 𝐾- heat transfer coefficient of X18H10T brand steel;
1/r_y1-5000 𝑉𝑡/𝑚² ∙ 𝐾- thermal conductivity of contamination from the mixture;
1/ry2=5800 𝑉𝑡/𝑚2 ∙ 𝐾- steam pollution thermal conductivity;
Comparative heat load:
_K=∆𝑡_{useful/0.169*10-3+0.24q0.6}
Useful temperature:
∆𝑡_{useful=0.169*10-3q+0.24q0.4
∆𝑡𝑓𝑜𝑦 = specific heat load at ∆𝑡useful=50.57 ℃ equal to 𝑞 = 56*103𝑉𝑡/𝑚
Heat transfer coefficient:
K=56*103/50.57=1107.3Wt/m2*K
Required heat exchange surface:}
F=10616.19*10³/(1107.3*50.57)=198.5m²
According to GOST 11987-81, we choose the nominal heat exchange surface: 𝐹𝑛 = 200 𝑚²
F𝑛 was significantly different from the previously calculated 𝐹_about. We determine K again.
The number of heating pipes is n = 335
n=200/(3.14*0.038*5)=335.57
Density of condensate outside the pipe:
Г=4.89/(335.57*3.14*0.038)=0.122 kg/m*sec
The thickness and criteria of the film for the condensate film The thickness does not change from the above;
^{𝛿пр=[(0.251*10-3)/(948.52*9.81)]0.33=0.2*10-4m}
Reynolds number:
Re_пл=4* Г/µ
Re_пл =(4*0.068)/(0.251*10^-3)=1083.6
Nusselt number:
Nu_pl=[108.6/((6.25*(1083.6-400)/1.56^0.33)=0.29
We determine the coefficient of heat transfer of condensed water to the pipe wall:
α₁=0.29*0.685/(0.2*10^-4)=9932.5Wt/m²*K.
3 The heat transfer coefficient is found from the following formula:
K=1/((1/9932.5)+(0.002/16.4)+(1/5000)+(1/5800)+(1/4.09q^0.6))=1/(0.169*10^-3+0.24/q^0.6)
Useful temperature:
_K=∆𝑡_{useful/0.169*10-3+0.24q0.6
∆𝑡𝑓𝑜𝑦 = specific heat load at ∆𝑡useful=55.68 ℃ equal to 𝑞=63*103𝑉𝑡/𝑚
K=56*103/5.57=1107.3Wt/m2*K
Required heat exchange surface:}
F=10616.19*10³/(1107.3*50.57)=189.5m²
According to GOST 11987-81, we choose the nominal heat exchange surface: 𝐹𝑛 = 200 𝑚²