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Fundamentals of Food Process Engineering

Romeo T. Toledo

Chapter 10

Refrigeration - all with Video Answers

Educators


Chapter Questions

07:56

Problem 1

An ammonia refrigeration unit is used to cool milk from $30^{\circ} \mathrm{C}$ to $1^{\circ} \mathrm{C}\left(86^{\circ} \mathrm{F}\right.$ to $\left.33.8^{\circ} \mathrm{F}\right)$ by direct expansion of refrigerant in the jacket of a shell and tube heat exchanger. The heat exchanger has a total outside heat transfer surface area of $14.58 \mathrm{~m}^2\left(157 \mathrm{ft}^2\right)$. To prevent freezing, the temperature of the refrigerant in the heat exchanger jacket is maintained at $-1^{\circ} \mathrm{C}\left(31.44^{\circ} \mathrm{F}\right)$.
(a) If the average overall heat transfer coefficient in the heat exchanger is $1136 \mathrm{~W} / \mathrm{m}^2 \cong \mathrm{K}$ ( $200 \mathrm{BTU} / \mathrm{h} \cong \mathrm{ft}^2 \cong{ }^{\circ} \mathrm{F}$ ) based on the outside area, calculate the rate at which milk with a specific heat of $3893 \mathrm{~J} / \mathrm{kg} \mathrm{K}\left(0.93 \mathrm{BTU} / \mathrm{lb} \cong{ }^{\circ} \mathrm{F}\right)$ can be processed in this unit.
(b) Determine the tons of refrigeration required for the refrigeration system.
(c) The high-pressure side of the refrigeration system is at $1.72 \mathrm{MPa}(250 \mathrm{psia})$. Calculate the horsepower of the compressor required for the refrigeration system assuming a volumetric efficiency of $60 \%$.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
04:20

Problem 2

A single-stage compressor in a freon 12 refrigeration system has a volumetric efficiency of $90 \%$ at a high side pressure of $150 \mathrm{psia}(1.03 \mathrm{MPa})$ and a low side pressure of $50 \mathrm{psia}(0.34$ $\mathrm{MPa}$ ). Calculate the volumetric efficiency of this unit if it is operated at the same high side pressure but the low side pressure is dropped to $10 \mathrm{psia}(68.9 \mathrm{kPa})$. Assume R12 is an ideal gas.

RZ
Rubeena Zulfiqar
Numerade Educator
03:49

Problem 3

Calculate the tons of refrigeration for a unit that will be installed in a cooler maintained at $0^{\circ} \mathrm{C}\left(32^{\circ} \mathrm{F}\right)$ given the following information on its construction and operation:
The cooler is inside a building.
Dimensions: $4 \times 4 \times 3.5 \mathrm{~m}(13.1 \times 13.3 \times 11.5 \mathrm{ft})$
Wall and ceiling construction:
$3.175 \mathrm{~mm}(1 / 8 \mathrm{in}$.) thick polyvinyl chloride sheet inside $(\mathrm{k}$ of $\mathrm{PVC}=0.173 \mathrm{~W} / \mathrm{m} \cong \mathrm{K}$ or $0.1 \mathrm{BTU} / \mathrm{h} \cong \mathrm{ft} \cong{ }^{\circ} \mathrm{F}$ )
$15.24 \mathrm{~cm}$ ( $6 \mathrm{in}$.) fiberglass insulation
$5.08 \mathrm{~cm}(2 \mathrm{in}$.) corkboard
$3.17 \mathrm{~mm}(1 / 2 \mathrm{in}$.) PVC outside
Floor construction:
$3.175 \mathrm{~mm}\left(1 / 8 \mathrm{in}\right.$.) thick floor tile $\left(\mathrm{k}=0.36 \mathrm{~W} /(\mathrm{m} \cong \mathrm{k})\right.$ or $0.208 \mathrm{BTU} /\left(\mathrm{h} \cong \mathrm{ft} \cong{ }^{\circ} \mathrm{F}\right)$
$10.16 \mathrm{~cm}$ (4 in.) concrete slab
$20.32 \mathrm{~cm}$ ( $8 \mathrm{in}$.) air space
Concrete surface facing the ground at a constant temperature of $15^{\circ} \mathrm{C}\left(59^{\circ} \mathrm{F}\right)$
Door:
$1 \mathrm{~m}$ wide $\times 2.43 \mathrm{~m}$ high $(3.28 \times 8 \mathrm{ft})$
Design for door openings that average four per hour at 1 minute per opening
Air infiltration rate:
$1 \mathrm{~m}^3 / \mathrm{h}\left(35.3 \mathrm{ft}^3 / \mathrm{h}\right)$ at atmospheric pressure and ambient temperature
Ambient conditions:
$32^{\circ} \mathrm{C}\left(89.6^{\circ} \mathrm{F}\right)$
Product cooling load:
Design for a capability to cool $900 \mathrm{~kg}$ of product $\left(\mathrm{C}_{\mathrm{P}}=0.76 \mathrm{BTU} / \mathrm{lb} \cong{ }^{\circ} \mathrm{F}\right.$ or $3181 \mathrm{~J} / \mathrm{kg}$ $\cong \mathrm{K})$ from $32^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}\left(89.6^{\circ} \mathrm{F}\right.$ to $\left.32^{\circ} \mathrm{F}\right)$ in 5 hours. The freezing point of the product is $-1.5^{\circ} \mathrm{C}\left(29.3^{\circ} \mathrm{F}\right)$.

Prashant Bana
Prashant Bana
Numerade Educator
05:00

Problem 4

The "stack effect" due to a difference in temperature between the inside and outside of a cooling room is often cited as the major reason for air infiltration. In this context, $\Delta \mathrm{P}$ is positive at the lowest section of a cooler and is negative at the highest section, with a zone, called the neutral zone, at approximately the center of the room where the $\Delta \mathrm{P}$ is zero. If the area of the openings at the lowest sections where $\Delta \mathrm{P}$ is positive equals the area of the openings in the highest sections where $\Delta \mathrm{P}$ is negative, air will enter at the top and escape at the openings in the bottom at the same volumetric rate of flow (assuming no pressure change inside the room). If the room allows air leakage at the rate of $2 \%$ of the room volume per minute at a $\Delta \mathrm{P}$ of $0.5 \mathrm{in}$. wg ( $124 \mathrm{~Pa})$, determine the rate of air infiltration that can be expected in a room that is $2 \mathrm{~m}(6.56 \mathrm{ft})$ high to the neutral zone if the interior of the room is at $-20^{\circ} \mathrm{C}$ $\left(-4^{\circ} \mathrm{F}\right)$ and ambient temperature is $30^{\circ} \mathrm{C}\left(86^{\circ} \mathrm{F}\right)$. The rate of gas flow through the cracks is proportional to the square root of $\Delta \mathrm{P}$. Assume air is an ideal gas. $\Delta \mathrm{P}$ due to a column of air of height $h$ at different temperatures $=g\left(\rho_1-\rho_2\right) h$, where $\rho_1$ and $\rho_2$ are the densities of the columns of air.

Niamat Khuda
Niamat Khuda
Numerade Educator
08:45

Problem 5

For a 1-ton refrigeration unit ( $80 \%$ volumetric efficiency) using refrigerant 12 at a high side pressure of $150 \mathrm{psia}(1.03 \mathrm{MPa})$ and a low side pressure of 45 psia $(0.31 \mathrm{Mpa})$, operating at an ambient temperature of $30^{\circ} \mathrm{C}\left(86^{\circ} \mathrm{F}\right)$, determine the effect of the following on refrigeration capacity and on $\mathrm{HP} /(\text { ton })_r$. Assume the same compressor displacement in each case.
(a) Reducing the evaporator temperature to $-30^{\circ} \mathrm{C}\left(-22^{\circ} \mathrm{F}\right)$. High side pressure remains at 150 psia (150 MPa).
(b) Increasing ambient temperature to $35^{\circ} \mathrm{C}\left(95^{\circ} \mathrm{F}\right)$. (Low side pressure remains at 45 psia [0.31 Mpa]). The high side pressure is to change such that $\Delta \mathrm{T}$ between the hot refrigerant gas and ambient air remains the same as in the original set of conditions.
(c) Air in the line such that the vapor phase of refrigerant always contain $10 \%$ air and $90 \%$ refrigerant by volume. Assume condensation temperature of hot refrigerant gas and temperature of cold refrigerant gas are the same as in the original set of conditions (partial pressure of refrigerant gas at the low and high side pressures are the same as in the original set of conditions, 45 and 150 psia or 0.31 and $1.03 \mathrm{MPa}$ ). Use $\mathrm{R}=1.987 \mathrm{BTU} /(\mathrm{lbmole}$ $\left.\cong{ }^{\circ} \mathrm{R}\right)$ or $8318 \mathrm{~J} /(\mathrm{kg} \cong \mathrm{K})$. The specific heat ratio $\mathrm{C}_{\mathrm{P}} / \mathrm{C}_{\mathrm{v}}$ for air is 1.4.
(d) Oil trapped in the vapor return line such that $\Delta \mathrm{P}$ across the constriction is 10 psi (68.9 $\mathrm{kPa}$ ). Assume evaporator temp. $=0^{\circ} \mathrm{C}$.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
06:45

Problem 6

Chopped onions are frozen in a continuous belt freezer at using $-50^{\circ} \mathrm{C}$ air at high velocity. When onions with a moisture content of $86 \%$ are loaded on the belt with a thickness of $2 \mathrm{~cm}$., it took 20 minutes for the temperature to drop from $10^{\circ} \mathrm{C}$ to $-20^{\circ} \mathrm{C}$. Onion juice is added and mixed with the chopped onions in a ratio 0.10 parts juice to 0.90 parts of the chopped onions. The freezing point of the chopped onions is $-0.5^{\circ} \mathrm{C}$. The juice contains $1.5 \%$ solids (all soluble). The juice has a freezing point of $-0.16^{\circ} \mathrm{C}$. Assuming that the rate of heat transfer is the same, calculate the time required to freeze the onions with the added juice.

Dr.  Satish  Ingale
Dr. Satish Ingale
Numerade Educator