Last Week:Heat ExchangersRefrigeration

This Week:More on RefrigerationCombustion and SteamPasteurizationSteam Raising and Combustion

Refrigeration

Condenser

Evaporator

Compressor

Qout

Qin

Win

Refrigeration

Cond

Comp

Qout

Win

Fermenting Room

Lagering Cellar Cooler

Hop Storage Cooler

Flash Tank

EvaporatorSecondary Refrigerant

Storage Tank

Wort Cooler

Fermenting Vessels

Green Beer Chiller

Beer Chiller

Pasteurizer

Yeast Tanks

Air Conditioning

Primary RefrigerantsAmmonia, R-12, R-134aSaturation temp < Desired application temp

2 to 8C Maturation tanks0 to 1C Beer Chillers-15 to -20C CO2 liquefaction

Typically confined to small region of brewery

Secondary RefrigerantsWater with alcohol or salt solutionsMethanol/glycol, potassium carbonate, NaClLower freezing temperature of waterNon-toxic (heat exchange with product)Pumped long distances across brewery

Example 1

A maturation tank is maintained at 6C using a secondary refrigerant (glycol/water solution). The cylindrical tank has a diameter of 3 m and a length of 6 m. The air temperature in the room is 18C and the overall heat transfer coefficient between the maturation tank and surroundings is 12 W/m2K. Determine the rate of heat gain to the maturation tank.

The glycol water solution is supplied from a storage tank at -10C, it exits the maturation tank at 2C and its specific heat is 3.5 kJ/kg.K. Determine the mass flow rate of secondary refrigerant required.

Example 2A brewery refrigeration unit has to meet the following

cooling duties simultaneusly.

1. Cool 800 hL of wort from 35 to 8C in two hours

2. Maintain two cold rooms at 0C – 40 kW ea.

3. Lager chiller cooling 50 m3/hr of product to 0C – 500 kW

4. Beer chiller cooling 50 m3/hr of product to 5C – 250 kW

5. Air conditioning, hop stores and yeast tanks – 100 kW

If the primary circuit uses R-134a and the secondary circuit uses 22.5% sodium chloride, estimate, stating all assumptions that you make, the maximum flow rates of R-134a and brine and the refrigerant compressor power.

Specific heat of brine – 3.7 kJ/kg.K. Min temp diff in evap and condenser, 20CCooling water temp to condenser, 15C

Wort BoilingImportance

• Flavor development• Trub formation• Wort stabilization• Wort concentration

Time and temperature – color, flavor, sterilization, etc.

Turbulence – trub formation and volatile removal

Rolling boil required.

Temperature above boiling (C)

Hea

t tr

ansf

er c

oef.

Interface Evaporation

(forced convection) <2C

Film Boiling >25C

Bubbles(nucleate boiling)

2C < T < 25C

Wort BoilingIn wort boiling it is important to maintain a temperature

difference below the critical difference between the wort and heating element surface (25C) If the wort is boiling at 105C, calculate the maximum operational steam pressure you would recommend for an indirect steam heated wort boiler. The wall of the steam heating element is 1.0 mm thick and has a thermal conductivity of 15 W/m.K. The condensing steam’s heat transfer coefficient is 12,000 W/m2.K and the maximum heat flux is 160,000 W/m2.

0.35 MPa 139.0C0.40 MPa 143.5C0.45 MPa 148.0C0.50 MPa 152.0C

0.55 MPa 155.5C0.60 MPa 159.0C0.65 MPa 162.0C

CombustionFuel + Oxidizer Heat + Products

Oxidizer: Air (79% N2, 21% O2 by Volume)

Fuels: Typically hydrocarbonsMethane CH4

Ethane C2H6 GasesPropane C3H8 Natural Gas = 95% CH4

Butane C4H10

C6 – C18 LiquidsGasoline (Average C8)Fuel Oil No. 1 (Kerosene)Fuel Oil No. 2 (Diesel)

Fuel Oil No. 3-6 (Heating Oils)

CombustionTo Balance Stoichiometric Combustion Reaction:

1. Balance Carbon (CO2 in products)

2. Balance Hydrogen (H2O in products)

3. Balance Oxygen (O2 in reactants)

4. Balance Nitrogen (N2 in products)

Example: (a) Determine the theoretical quantity of air required for combustion of natural gas. Give results in kg of air per kg of natural

gas. Assume that natural gas is 100% CH4.

(b) Determine the mass of CO2 emitted per kg of natural gas burned.

CombustionActual combustion process Excess air

Complete combustion (reduce CO, UHC)

Reduce flame temperature (reduce NOx)

Example: Determine the composition of CH4 combustion products with 25% excess air.

CombustionFlue gas analysis – Work backwards to find %

excess air.

Example: Determine the excess air used for CH4 combustion when the O2 concentration in the products is 5.5% volume. (Note, for ideal gas mixtures, volume fraction = mole fraction).

Calorific Value of Fuels (= Heating Value)

Solids, Liquid: MJ/kg

Gases: MJ/m3

LHV = H2O vapor in products, HHV = liquid

SteamHigh latent heat, cheap, non-toxic, available

Combustion/Steam ProblemA 5 m3 wort kettle is heated from 70C to 95C with

steam at 3 bar (gauge) in an external heating jacket. The steam enters as saturated vapor and it exits as saturated liquid. Natural gas (LHV = 40 MJ/kg).

a. Calculate the total mass of steam required for the heating process.

b. What mass of fuel is required and what will the fuel cost be if natural gas can be purchased for $1.00/Therm (1 Therm = 100,000 BTU)

PasteurizationMicroorganisms growing in beer

• Wild yeast strains• Lactic acid bacteria

No – Homogeneous population of microbesN – Remaining number of microbest – time in minutesD – Decimal reduction time at temperature T

Time (min) Number of microbes per Liter

0 10,000

2 1,000

4 100

8 1

10 0.1

D

t

oN

N

10

min 260 D

PasteurizationTypically choose D value of most resistant organism1.0 P.U. = “one minute of heating at 60C”

An average Z value of 6.94C is used

Z

T

T DD60

6010

TLogDLogD

TZ

60

60

tPU TTotal

60394.1

Flash PasteurizationT

ime

(min

)0.

1

1

1

0

100

50 60 70Temperature (C)

Over Pasteurization

Under Pasteurization

Minimum Safe Pasteurization

5.6 min

PasteurizationFor the data given below, calculate the total number of

pasteurization units (PU). Assume a Z value of 6.94C.

What type of pasteurizer is this?

Minute Mean Temp (C)

PU’s

21 49.7

22 53.0

23 55.9

24 58.3

25 60.2

26 61.5

27 62.25

28 62.65

Minute Mean Temp (C)

PU’s

29-34 62.8

35 62.6

36 61.2

37 58.6

38 56

39 53.7

40 51.75

41 50

Total

Flash Pasteurization

Beer in = 0C

Pasteurizer60-70C30 sec - 2 min

90-96%regeneration

Flash PasteurizationP

ress

ure

(Bar

)

Tem

pera

ture

(C

)

Time (sec)

Pressure in Pasteurizer

CO2 equilibrium pressure

Temperature in Pasteurizer

Flash PasteurizationTypical Conditions:

Beer inlet: 3COutlet from regenerative heating: 66CHolding tube: 70COutlet from regenerative cooling: 8COutlet from cooling section: 3CHolding Time: 30 sec

Advantages

Little space required

Relatively inexpensive equipment and operation

Short time at “intermediate” temperatures where chemical changes occur without pasteurization

Tunnel PasteurizationPasteurized after bottled or canned

Bottles or cans move slowly down conveyer system

Hot water sprays heat beer to pasteurization temperature

Cool water sprays cool beer after pasteurization is complete

Pressure builds in headspace- Volume of headspace- CO2 concentration in beer

Bottles could break (Typical 1 in 500)

CO2 could leak if bottles are not sealed well

Tunnel PasteurizationP

ress

ure

(Bar

)

Tem

pera

ture

(C

)

Time (min)

Spray water temperature

Product Temperature

Tunnel Pasteurization

Simpler system than flash pasteurization

Slow process (may take up to 40 minutes)

Energy intensive process

Beer near outside of can/bottle over pasteurized

Mechanical failure, other stoppage could cause over pasteurization, effecting beer flavor