Download Brewing Week 4

So Far:
Conservation of Mass and Energy
Pressure Drop in Pipes
Flow Measurement Instruments
Flow Control (Valves)
Types of Pumps and Pump Sizing
This Week:
Energy Balance
Heat Transfer
Conservation of Energy
dE


Ein  Eout 
dt system
For steady flow systems
Ein  Eout  ESystem
dEsystem
dt
0
Energy = Heat (Q), Work (W), mass (h)
No Phase Change, E = m c ΔT
Phase Change, E = m hfg where
hfg = enthalpy of vaporization or fusion
A 2 m3 water tank is filled with 1.25 m3 of
hot water at 80C and 0.75 m3 of cold
water at 10C. Assume that the specific
heat of water is 4.2 kJ/kg.K.
a) Determine the temperature in the tank
after it has been filled.
b) How much heat must be added to the
tank to bring its temperature to 65C?
c) If a 30 kW electric heater is used, how
long will the heating process take?
500 kg of grain (25C) is mixed with hot
(80C) and cold (10C) water for mashing.
The water to grain ratio (by weight) is 3:1
and the specific heat capacities of the
water and grain are 4.2 and 1.7 kJ/kg.K,
respectively.
a) If the desired “mash in” temperature is
38C, how much hot and cold water
should be added?
(Continued) A three step mashing process,
with 20 minute-long rests at 50, 62 and 72C,
is desired. The mash should be heated
quickly, but not too quickly between rests;
with an optimal rate of 1C per minute.
Neglect heat losses to the surroundings.
b) Plot the mash temperature vs. time.
c) Determine the heating power required, in
kW.
d) Determine the total heat required for the
mashing process, in kJ.
Two types of heat sources are available for
mashing, electric resistance heaters and
steam. The steam enters a heating jacket
around the mash as dry, saturated steam at
300 kPa and it exits the system as wet,
saturated steam at the same pressure
(enthalpy of vaporization = 2150 kJ/kg).
(e) What is steam flow rate required, in kg/s?
(f) If steam is used, what is the total mass of
steam required, in kg?
At the location of our brewery, electricity
costs $0.14/kW-hr and the steam can be
generated for $0.03 per kg.
(g) What is the mashing cost when electric
resistance heaters are used?
(h) What is the cost with steam?
Energy Balance Example
The power goes out at your brewery due to
an overheated transformer, shutting down
your fermentation cooling mechanism.
Consider a 25 m3 cylindroconical vessel
that is full with a product at 10oC, specific
heat of 3.8 kJ/kg.K, and density of 1025
kg/m3. Assuming that the sum of heat
gains from the surroundings and
conversion from fermentation is 7 kW,
determine the temperature after 8 hours.
How would the 7 kW load change over
time?
Heat Transfer Equipment
Mash Tun – External heating jacket
Kettle – External jackets/panels, internal coils,
internal or external calandria
Wort cooler – Plate heat exchanger
Fermenter – Internal or external coils or panels
Pasteurizers – Plate heat exchangers, Tunnel
Refrigeration equipment – Shell and tube heat
exchangers, evaporative condensers
Steam and hot water equipment – Shell and tube
Heat Transfer Equipment
Steam
in
Wort
Steam
out
Mash Tun – External heating jacket
Heat Transfer Equipment
Mash Tun – External heating jacket
Heat Transfer Equipment
Steam
Wort kettle – Internal calandria
Heat Transfer Equipment
Steam
Wort kettle – External calandria
Heat Transfer Equipment
Wort kettle – Internal calandria
Heat Transfer Equipment
Plate Heat Exchanger
Heat Transfer Equipment
Plate Heat Exchanger
Heat Transfer Equipment
Shell and tube heat exchanger
Heat Transfer
Transfer of energy from a high temperature
to low temperature
Conservation of Energy
Ein – Eout = Esystem
Qin = m(u2 – u1) = mc(T2-T1)
Qin
Wort
Heat Transfer
Rate of Ein – Rate of Eout = Rate of E Accumulation
 Q out  m (hin  hout )  0
 c p Tin  Tout 
Q out  m
Qout
min
Wort
Calculate the rate of heat transfer required to
cool 100 L/min of wort from 85 to 25C. The wort
has a density of 975 kg/m3 and specific heat of
4.0 kJ/kg.K.
Heat Transfer
Rate of Ein – Rate of Eout = Rate of E Accumulation
H2O
Wort


Q in, H 2O  m H 2O c p , H 2O TH 2O,in  TH 2O,out  0
 Q out, wort  m wort c p , wort Twort,in  Twort,out   0


 wort c p,wort Twort,in  Twort,out   m
 H 2Oc p, H 2O TH 2O,in  TH 2O,out  0
m
Heat Transfer
Rate of Ein – Rate of Eout = Rate of E Accumulation
H2O
Wort
Wort is being cooled with chilled water in a heat
exchanger. The wort enters at 85C with a flow
rate of 100 L/min and it exits the heat exchanger
at 25C. The chilled water enters at 5C with a
flow rate of 175 L/min. The specific heat of the
wort and water are 4.0 and 4.2 kJ/kg.K Determine
the exit temperature of the chilled water.
Conduction
Transfer of microscopic kinetic energy from one
molecule to another
1-D Heat Transfer, Fourier Equation:
T

Q  kA
x
Q  UAT
1 x1 x 2 x 3
    ...
U k1 k 2 k 3
A 0.5 m2, 1.75 cm thick stainless steel plate (k =
50 W/m.K) has surface temperatures of 22.5 and
20C. Calculate the rate of heat transfer through
the plate.
Conduction
Same equations apply for multi-layer systems
1-D Heat Transfer, Fourier Equation:
Q  UAT
1 x1 x 2 x 3
    ...
U k1 k 2 k 3
How would the rate of heat transfer change if a
2.5 cm thick layer of insulation (k = 0.05 W/m.K)
were added to the “low” temperature side of the
plate?

Convection
Transfer of heat due to a moving fluid
Natural convection – buoyant forces drive flow
Temperature
Forced convection – mechanical forces drive flow
Fluid
Wall
Tfluid
Twall

Q convection  hA T fluid  Twall

Heat Transfer
Overall Heat Transfer Coefficient
1
U conduction
Q  U o AT
1
1
x


U convection h
k
For “thin walled” heat exchangers, Ai = Ao
1
1
x
1



Uo houtside kw hinside
Convection
A tube-in-tube heat exchanger carries hot wort at
85C in the inner tube and chilled water at 5C in
the outer tube. The tube wall thickness is 4 mm
and its thermal conductivity is 100 W/m.K. The
wort film coefficient is 750 W/m2.K and the chilled
water film coefficient is 3000 W/m2K. Determine
the overall heat transfer coefficient and the rate of
heat transfer per meter of heat exchanger length.
The diameter of the pipe is 4.0 cm.
Convection
Condensation
Constant temperature process
Occurs when a saturated comes in contact
with a surface with temperature below Tsat
for the vapor
Film coefficients: 5,000-20,000 W/m2.K
Boiling
Constant temperature process
Some surface roughness promotes boiling
Bubbles rise – significant natural convection
Fraction of surface “wetted” effects Q
Fig 9, page 114 in Kunze.
T1
T
T2
Length
Temperature
Temperature
Log Mean Temperature Difference
Parallel Flow
Counter Flow
T1
T
T2
Length
Log Mean Temperature Difference
T1  T2
Tm 
T1
ln
T2
A tube-in-tube, counterflow heat exchanger carries hot
wort at 85C in the inner tube and chilled water at 5C in
the outer tube. The tube wall thickness is 4 mm and its
thermal conductivity is 100 W/m.K. The wort film
coefficient is 750 W/m2.K and the chilled water film
coefficient is 3000 W/m2K. Determine the overall heat
transfer coefficient and the rate of heat transfer per meter
of heat exchanger length.
Calculate the LMTD.
Fouling
Layers of dirt, particles, biological growth, etc.
effect resistance to heat transfer
1
U o, dirty
1

 Ro  Ri
Uo
We cannot predict fouling factors analytically
Allow for fouling factors when sizing heat transfer
equipment
Historical information from similar applications
Little fouling in water side, more on product
Typical values for film coefficient, p. 122
Heat Exchanger Sizing
Beer, dispensed at a rate of 0.03 kg/s, is chilled in an ice
bath from 18C to 8C. The beer flows through a
stainless steel cooling coil with a 10 mm o.d., 9 mm i.d.,
and thermal conductivity of 100 W/m.K. The specific
heat of the beer is 4.2 kJ/kg.K and the film heat transfer
coefficients on the product and coolant sides are 5000
W/m2.K and 800 W/m2.K, respectively. The fouling
factors on the product and coolant sides are 0.0008 and
0.00001 m2K/W. Assume that the heat exchanger is thin
walled.
a. Determine the heat transfer rate
b. Determine the LMTD
c. Determine the overall heat transfer coefficient
d. Determine the outside area required
e. Determine the length of tube required
Radiation
Vibrating atoms within substance give off photons
Energy Radiated  T
4
Emissivity of common substances
Polished aluminum: 0.04
Stainless steel:
0.60
Brick:
0.93
Water:
0.95
Snow:
1.00
Radiation between surface and surroundings:

4
4

Q   surf Asurf Tsurf  Tsurr

Radiation
Sometimes, we’ll make an analogy to convection

Q
rad  hrad Asurf Tsurf  Tsurr 
A 3 cm diameter, 15 m long pipe carries hot wort
at 85C. The pipe has 1.0 cm thick insulation,
which has thermal conductivity of 0.08 W/m.K.
The insulation exterior surface temperature is
35C and its emissivity is 0.85. The temperature of
the surroundings is 20C. Determine the rate of
heat loss by radiation.
Heat Losses
Total Heat Loss = Convection + Radiation
Preventing heat loss, insulation
Air – low thermal conductivity
Air, good
Water – relatively high thermal conductivity
Water, bad
Vessels/pipes above ambient temperature –
open pore structure to allow water vapor out
Vessels/pipes below ambient temperature closed pore structure to avoid condensation
Conduction
Hollow cylinders (pipes)
r1
Q  UAm T
r2




r2  r1 

Am  2L
 r2 
 ln

r1 
