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1. (a)With simple sketch, compare and discuss the differences among freezing of water, solution

and food. [0.5]

(b) Sketch a typical freezing curve for food, and describe the three periods during freezing.

[0.5]

2. A meat ball can be assumed as a sphere. If the meat ball has a diameter of 120 mm and is

to be cooled under the following conditions:

Specific heat c = 3800 Jkg-1K

-1

Thermal conductivity k = 0.47 Wm-1K

-1

Density = 1050 kgm-3

Initial product temperature Ti = 35oC

Cooling air temperature Tm = -2

oC

Heat transfer coefficient h = 11.0 Wm-2K

-1

Desired final centre temperature Tc = 5oC

Using the graphical method with Figures 1-3 to calculate the cooling time and the mean

temperature for the meat ball.

General form of the cooling time model is

j

f Y

ln

2.303

[1.0]

3. (a) A 150 kg beef carcass is frozen to -20oC, if the mass fraction of water and protein are

58.21% and 17.48% respectively, and the initial freezing point is -1.7oC, what is the mass

of frozen and unfrozen water at -20oC? The mass fraction of ice can be calculated by

ln 1

0.8765 1

1.105

t t

x

x

f

wo

ice

[0.5]

(b) If the initial temperature of the carcass is 10oC, how much heat must be removed

during this freezing process? The enthalpy of unfrozen and frozen foods can be calculated

respectively by

( )(4.19 2.30 0.628 )

3

f f s s H H t t x x (for unfrozen foods)

t t

x x L t

H t t x

r

wo b f

r s

0

( )

( ) 1.55 1.26 (for frozen foods)

where tr = reference temperature (zero enthalpy) = -40oC; Hf = enthalpy of food at initial

freezing temperature, kJ/kg; L0 = latent heat of fusion of water = 333.6 kJ/kg. (Hint: The

equation for frozen foods can be used to calculate Hf). [0.5]

4. Mechanical vapour compression cycle shown in Figure 4 and absorption cycle shown in

Figure 5 are two main cycles used in food refrigeration, compare and discuss the

differences among these two cycles and describe their operating principles.

[1.0]

0.1 1.0 10 100

0.1

1.0

10

100

Bi

f/L2

Slab

Cylinder

Sphere

Figure 1. Plot of f/L

2

against Biot number.

10-1 100

101

102

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

Bi

jc

Sphere

Cylinder

Slab

Figure 2. Plot of jc against Biot number.

10-1 100

101

102

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Bi

jav

Slab

Cylinder

Sphere

Figure 3. Plot of jav against Biot number.

Figure 4. Mechanical vapour compression cycle.

Figure 5. Absorption refrigeration cycle.

1. Figure 1 shows the schematic of the mechanical vapour compression cycle,

(1) Explain its operating principle; [0.5]

(2) If a standard vapour compression cycle operates with a condensing temperature of

35oC and an evaporating temperature of -10oC, draw the standard vapour compression

cycle on the pressure-enthalthy diagram in Figure 2. [0.5]

2. Determine the thermal conductivity and density of lean pork shoulder meat which is at a

temperature of -40oC. Use both the parallel and perpendicular thermal conductivity

models. The composition of the meat is xwo = 0.7263, xf = 0.0714, xp = 0.1955, xa =

0.0102, and other components can be omitted. The initial freezing point of the meat is –

2.2oC.

The following equations may be used in the calculation:

t

t

x x x

f

ice wo b 1

i

v

i

k x k

w = 9.9718×102 + 3.1439×10-3

t – 3.7574×10-3

t

2

ice = 9.1689×102

– 1.3071×10-1

t

p = 1.3299×103

– 5.1840×10-1

t

f = 9.2559×102

– 4.1757×10-1

t

c = 1.5991×103

– 3.1046×10-1

t

fb = 1.3115×103

– 3.6589×10-1

t

a = 2.4238×103

– 2.8063×10-1

t

kw = 5.7109×10-1 + 1.7625×10-3

t – 6.7036×10-6

t

2

kice = 2.2196 – 6.2489×10-3

t + 1.0154×10-4

t

2

kp = 1.7881×10-1 + 1.1958×10-3

t – 2.7178×10-6

t

2

kf = 1.8071×10-1

– 2.7604×10-3

t – 1.7749×10-7

t

2

kc = 2.0141×10-1 + 1.3874×10-3

t – 4.3312×10-6

t

2

kfb = 1.8331×10-1 + 1.2497×10-3

t – 3.1683×10-6

t

2

ka = 3.2962×10-1 + 1.4011×10-3

t – 2.9069×10-6

t

2

3. An infinite slab beef with thickness of 0.04 m is to be frozen in an air blast freezer. The

initial temperature of the beef is 10oC, freezing air temperature is -30oC and surface heat

transfer coefficient is 40 W/(m2K). Use Cleland and Earle method to calculate the time

required for the thermal centre of the beef to reach a temperature of -15

oC.

Some thermal properties of the beef are listed below:

Property At -40oC At -10oC At -1.7oC At 10oC

, kg/m3 s = 1018 s = 1018 l = 1075 l = 1075

H, kJ/kg – Hs = 83.4 Hl = 274.2 –

c, kJ/kgK cs = 2.11 – – cl = 3.52

k, W/mK ks = 1.66 – – –

The following equations may be used:

ref m

c m

f m Freeze s s T T

T T

k

Ste

k

RD

h

PD

T T E

H

ln 1.65 1

( )

2

1 0

P = 0.5 [1.026 + 0.5808 Pk + Ste (0.2296 Pk + 0.1050)]

R = 0.125 [1.202 + Ste (3.41 Pk + 0.7336)]

The dimensionless numbers are defined as:

k

hD Bi

H

T T

Pk C

i f

l

H

T T

Ste C

f m

s

4. Vacuum cooling is an established precooling method to rapidly remove the field heat of

leafy vegetables such as lettuce after harvest.

(1) Explain its cooling principle; [0.5]

(2) In recent years, some advanced applications of vacuum cooling have been researched,

identify these advanced applications and discuss their advantages and disadvantages.

[0.5]

Figure 1. Mechanical vapour compression cycle.

Figure 2. Pressure-enthalphy diagram for refrigerant 134a.

1. With sketches, compare the differences during freezing of water, solution and food.

2. Figure 1 shows the schematic of the mechanical vapour compression cycle,

(a) Explain its operating principle; [0.5]

(b) If a standard vapour compression cycle operates with a condensing temperature of

35oC and an evaporating temperature of -10oC, draw the standard vapour compression

cycle on the pressure-enthalthy diagram in Figure 2. [0.5]

3. (a) A 150 kg beef carcass is frozen to -20oC, if the mass fraction of water and protein are

58.21% and 17.48% respectively, and the initial freezing point is -1.7oC, what is the mass

of frozen and unfrozen water at -20oC? The mass fraction of ice can be calculated by

ln 1

0.8765 1

1.105

t t

x

x

f

wo

ice

[0.4]

(b) If the initial temperature of the carcass is 10oC, how much heat must be removed

during this freezing process? The enthalpy of unfrozen and frozen foods can be calculated

respectively by

( )(4.19 2.30 0.628 )

3

f f s s H H t t x x (for unfrozen foods)

t t

x x L t

H t t x

r

wo b f

r s

0

( )

( ) 1.55 1.26 (for frozen foods)

where tr = reference temperature (zero enthalpy) = -40oC; Hf = enthalpy of food at initial

freezing temperature, kJ/kg; L0 = latent heat of fusion of water = 333.6 kJ/kg. (Hint: The

equation for frozen foods can be used to calculate Hf). [0.6]

4. An infinite beef slab with thickness of 40 mm is to be frozen in an air blast freezer. The

initial temperature of the beef is 10oC, freezing air temperature is -30oC and surface heat

transfer coefficient is 40 Wm-2K

-1

.

(a) Explain the physical meaning of the dimensionless numbers: Bi, Pk and Ste; [0.3]

(b) Calculate the time required for the thermal centre of the beef to reach a temperature of

-10oC. [0.7]

The following data and equations may be used in the calculation:

Densities of frozen and unfrozen beef are 1018 kg m-3

and 1075 kg m-3

respectively;

Enthalpies at –10oC and at the initial freezing point –1.7oC are respectively 83.4 kJ kg-1

and 274.2 kJ kg-1

;

Specific heats of frozen and unfrozen beef are 2.11 kJ kg-1 K

-1

and 3.52 kJ kg-1 K

-1

respectively;

Thermal conductivity of frozen beef is 1.66 W m-1 K

-1

.

f m s

k

RD

h

PD

T T

H

2

10

Bi = hD / k; Pk = Cl (Ti –Tf) / H; Ste = Cs (Tf – Tm) / H

P = 0.5072 + 0.2018 Pk + Ste (0.3224 Pk + 0.0105/Bi + 0.0681)

R = 0.1684 + Ste (0.2740 Pk – 0.0135)

Figure 1. Mechanical vapour compression cycle.

Figure 2. Pressure-enthalphy diagram for refrigerant 134a.

1. Figure 1 shows the schematics of a mechanical vapour compression cycle and absorption cycle,

which are two common cycles for food refrigeration, describe their operating principles, and

compare and discuss their differences.

2. Blast freezers are the most common freezing system used in the food industry. There are a wide

range of blast freezer systems available. List these various types of blast freezers and discuss their

operating principles and applications.

3. A meat ball can be assumed to be a sphere. If the meat ball has a diameter of 120 mm and is to be

cooled under the following conditions:

Specific heat c = 3800 Jkg-1K-1

Thermal conductivity k = 0.47 W m-1K-1

Density = 1050 kg m-3

Initial product temperature Ti = 35oC

Cooling air temperature Tm = -2

oC

Heat transfer coefficient h = 11.0 W m-2K-1

Desired final centre temperature Tc = 5oC

Using the graphical method with Figures 2-4 to calculate the cooling time and the mean temperature

for the meat ball.

General form of the cooling time model is

4. The freezing time can be considered as the summation of precooling, phase change and subcooling

times. Based on this concept, the following equation was developed to estimate the freezing time:

4

1

/ 2

2

2

1

1 s

Freeze

Bi

T

H

T

H

E h

D

Bis = Biot number for fully frozen food = hD/ks; EFreeze = equivalent heat transfer dimensionality;

H1 and T1 are volumetric enthalpy change and temperature difference for precooling period,

and H2 and T2 are those for combined freezing-subcooling period:

H1 = Hi – Hfm = Cl(Ti –Tfm)

H2 = Hfm – Hc = [Lf + Cs(Tfm –Tc)]

T1 = (Ti+Tfm)/2 – Tm

T2 = Tfm – Tm

Tfm is “mean freezing temperature”: Tfm = 1.8 + 0.263 Tc + 0.105 Tm

If a meat product is frozen in air at Tm = -40oC from its initial temperature of 35oC to a final

temperature of -18oC, calculate the freezing time (the meat product can be assumed to be a sphere

with a radius L = 0.06 m). Some data are as follows:

heat transfer coefficient h = 11.0 W m-2K-1

; latent heat, Hs = 209000 J kg-1

;

specific heat cl = 3800 J kg-1K-1

; cs = 1900 J kg-1K-1

;

thermal conductivity kl = 0.47 Wm-1K-1

; ks = 1.35 W m-1K-1

;

density l = 1050 kgm-3

; s = 970 kg m-3

.

j

f Y

ln

2.303

Figure 1. The schematics of mechanical vapour compression cycle and absorption cycle.

0.1 1.0 10 100

0.1

1.0

10

100

Bi

f/L2

Slab

Cylinder

Sphere

Figure 2. Plot of f/L

2

against Biot number.

10-1 100

101

102

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

Bi

jc

Sphere

Cylinder

Slab

Figure 3. Plot of jc against Biot number.

10-1 100

101

102

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Bi

jav

Slab

Cylinder

Sphere

Figure 4. Plot of jav against Biot number.

Question?