Design Optimization

m.Minimize(Ctot)

(:title Heat Integration Optimization:)

milk_pasteurization_heat_pump.png

A heat pump is a type of heating and cooling system that uses an external heat source or sink to provide cooling in summer and heating in winter. It can also be used to provide heat for industrial processes such as milk pasteurization. The system works by transferring heat from the source to the milk. The heat exchanger acts as a buffer, to heat the incoming milk and cool the outgoing milk. This type of heat integration is highly efficient and helps to reduce energy costs.

In the pasteurization of milk, the temperature is raised to 73°C, held for 20 sec., and then cooled. The milk arrives at a temperature of 7°C and is delivered from the pasteurizing process for packaging at a temperature of 4°C.

A heat pump with a regenerative heat exchanger is a type of heating and cooling system that uses an external heat source or sink to provide cooling in summer and heating in winter. The system works by transferring heat from one area to another. The heat exchanger acts as a buffer, allowing the heat pump to operate in both heating or cooling mode without having to switch directions. The regenerative heat exchanger is a type of heat exchanger that stores energy from the environment, such as the heat from the sun, and uses it to heat or cool the desired space. This type of system is highly efficient and helps to reduce energy costs.

1. Inlet temperature (degC)

1. Outlet temperature (degC)

1. Pasteurization Temperature (degC), target temperature

1. Variables
2. compressor work (kW)

1. heat exchanger areas (m^2)

t1 = m.Var(value = ti) # Temperature 1 (degC) t2 = m.Var(value = ti) # Temperature 2 (degC) te = m.Var(value = ti) # Evaporator Temperature (degC) tc = m.Var(value = ti) # Condenser Temperature (degC) Ctot = m.Var(value = 100000) # Total Cost

1. exchanger), size of the compressor, and temperatures t1, t2, te and tc
2. that result in the minimum total cost

1. areas of the heat exchangers (evaporator, condensers, and regenerative
2. exchanger), size of the compressor, and temperatures t1, t2, te and tc
3. that result in the minimum total cost

Solution Help

See GEKKO documentation and additional example problems.

(:source lang=python:) from gekko import GEKKO

1. Initialize Gekko model

m = GEKKO()

1. Optimal heat pump and regenerative exchanger that minimizes the total present
2. worth of costs (capital cost and operating cost). Specifically, determine the
3. areas of the heat exchangers (evaporator, condensers, and regenerative exchanger),
4. size of the compressor, and temperatures t1, t2, te and tc that result in the
5. minimum total costOptimize
6. Parameters

U = 0.6 Ureg = 0.5 mdotm = 4 cpm = 3.75 cpw = 1.00 intrate = 0.09 nper = 6

1. Inlet temperature (degC)

ti = 7

1. Outlet temperature (degC)

to = 4

1. Pasteurization Temperature (degC), target temperature

tp = 73

1. Variables
2. compressor work (kW)

W = m.Var()

1. heat exchanger areas (m^2)

Ae = m.Var(value = 10, lb = 1.0, ub = 1000) Areg = m.Var(value = 10, lb = 1.0, ub = 1000) Afc = m.Var(value = 10, lb = 1.0, ub = 1000) Aac = m.Var(value = 10, lb = 1.0, ub = 1000)

t1 = m.Var(value = ti) #Temperature 1 (degC) t2 = m.Var(value = ti) #Temperature 2 (degC) te = m.Var(value = ti) #Evaporator Temperature (degC) tc = m.Var(value = ti) #Condenser Temperature (degC) Ctot = m.Var(value = 100000) #Total Cost

1. evaporator

Qe = m.Var(value = 1) dapp1e = m.Var(value = 1) dapp2e = m.Var(value = 1)

1. regenerative heat exchanger

Qreg = m.Var(value = 1) dapp1reg = m.Var(value = 1) dapp2reg = m.Var(value = 1)

1. fore condenser

Qfc = m.Var(value = 1) dapp1fc = m.Var(value = 1) dapp2fc = m.Var(value = 1)

1. after condenser

Qac = m.Var(value = 1) dapp1ac = m.Var(value = 1) dapp2ac = m.Var(value = 1)

1. temps of approach

dappe = m.Var(lb = 0.1) # >= 10 dappc = m.Var(lb = 0.1) # >= 10 dappreg1 = m.Var(lb = 0.1) # >= 10 dappreg2 = m.Var(lb = 0.1) # >= 10

1. ratios

ratio_e = m.Var(value = .05, lb = 0.001, ub = 0.999) ratio_reg = m.Var(value = .05, lb = 0.000, ub = 1.0) ratio_fc = m.Var(value = .05, lb = 0.001, ub = 0.999) ratio_ac = m.Var(value = .05, lb = 0.001, ub = 0.999)

1. Log-mean temperature differences

delte = m.Var(value = 1) deltreg = m.Var(value = 1) deltfc = m.Var(value = 1) deltac = m.Var(value = 1)

1. Coefficient Of Performance

COP = m.Var(value = 1)

1. Intermediates
2. Cost elec year at 4 hours per day

Celyear= m.Intermediate(W * 4.0 * 365.0 * 0.07)

1. Cost total over time
2. Multiplier is about 4.5 (instead of 6.0) because of the
3. interest rate on unspent capital

mult = m.Intermediate((((1.0+intrate)**nper)-1.0) / (intrate*(1.0+intrate)**nper)) Celtot = m.Intermediate(Celyear * mult)

1. Cost equipment

Cequip = m.Intermediate(Ae*200.0 + Afc*200.0 + Aac*200.0 + Areg*200 + W*240.0)

1. temperatures in Kelvin

TeK = m.Intermediate(te + 273) TcK = m.Intermediate(tc + 273)

1. Equations

m.Equations([

    # Total cost
    Ctot == Celtot + Cequip,

    # overall energy balance on regen heat exchanger
    # mdotm and cpm cancel from each term
    t1 + t2 == tp + ti,

    # log-mean temperature difference
    #   lmtd(thi,tco,tho,tci)

    # evaporator 
    Qe == mdotm * cpm *(t2 - to),
    dapp1e == to - te,
    dapp2e == t2 - te,

    # regenerative heat exchanger 
    Qreg == mdotm * cpm * (tp-t2),
    dapp1reg == t2 - ti,
    dapp2reg == tp - t1,

    # fore condenser 
    Qfc == mdotm * cpm * (tp - t1),
    dapp1fc == tc - tp,
    dapp2fc == tc - t1,

    # after condenser 
    Qac == W + mdotm * cpm * (ti - to),
    dapp1ac == tc - 35.0,
    dapp2ac == tc - 30.0,

    # approach temperatures 
    dappe == to - te,         # Temp app evap
    dappc == tc - tp,         # Temp app fore cond
    dappreg1 == t2 - ti,      # Temp app reg1
    dappreg2 == tp - t1,      # Temp app reg2

    # ratios
    ratio_e   * dapp2e   == dapp1e, 
    ratio_reg * dapp2reg == dapp1reg,
    ratio_fc  * dapp2fc  == dapp1fc,
    ratio_ac  * dapp2ac  == dapp1ac,

    # Log-mean Temperature Difference
    delte   * m.log(ratio_e)   == (dapp1e - dapp2e),
    #deltreg * m.log(ratio_reg) = (dapp1reg - dapp2reg)
    deltfc  * m.log(ratio_fc)  == (dapp1fc - dapp2fc),
    deltac  * m.log(ratio_ac)  == (dapp1ac - dapp2ac),

    # Average temperature difference
    #delte    = (dapp1e   + dapp2e)   / 2
    deltreg  == (dapp1reg + dapp2reg) / 2,
    #deltfc   = (dapp1fc  + dapp2fc)  / 2
    #deltac   = (dapp1ac  + dapp2ac)  / 2

    # coefficient of performance
    # 75% of Carnot Efficiency
    COP * (TcK-TeK) == 0.75 * TeK,

    # Work to compressor 
    W * COP == Qe,

    # heat transferred for each exchanger
    Ae   * U    * delte   == Qe,
    Areg * Ureg * deltreg == Qreg,
    Afc  * U    * deltfc  == Qfc,
    Aac  * U    * deltac  == Qac
    ])

1. objective function

m.Obj(Ctot)

1. minimize ctot

m.options.SOLVER = 1 m.options.IMODE = 3

m.solve()

print('Minimum cost: ' + str(Ctot[0])) (:sourceend:)

This assignment can be completed in groups of two. Additional guidelines on individual, collaborative, and group assignments are provided under the Expectations link.

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(:title Heat Pump Optimization Problem:) (:keywords heat pump, nonlinear, optimization, engineering optimization, two-bar optimization, engineering design, interior point, active set, differential, algebraic, modeling language, university course:) (:description Engineering design and optimization of a heat pump system for pasteurization of milk. Optimization principles are used to design the system.:)

In the pasteurization of milk the temperature is raised to 73°C, held for 20 sec., and then cooled. The milk arrives at a temperature of 7°C and is delivered from the pasteurizing process for packaging at a temperature of 4°C.

We will consider using a heat pump with a regenerative heat exchanger to do this. One possible cycle is shown in the figure below. The incoming milk is preheated in a regenerative heat exchanger and then heated further in the fore-condenser of the heat pump. As it exits the fore-condenser, the temperature of the milk is 73°C. Thereafter the milk is cooled as it flows through the other side of the regenerative heat exchanger and then through the evaporator of the heat pump.

Full Heat Pump Assignment

Find the optimal heat pump and regenerative exchanger that minimizes the total present worth of costs (capital cost and operating cost). Specifically, determine the areas of the heat exchangers (evaporator, condensers, and regenerative exchanger), size of the compressor, and temperatures t₁, t₂, t_e and t_c that result in the minimum total cost. Run the problem with no constraints on temperatures of approach (Case 1); then rerun with constraints that all temperatures of approach (evaporator, condensers, regenerative heat exchanger) are at least 10 °C (Case 2).

This problem is based on a problem from W. Stoecker, Design of Thermal Systems, 3rd ed.

This assignment can be completed in groups of two. Additional guidelines on individual, collaborative, and group assignments are provided under the Expectations link.