Calculations for an Advanced Thermodynamics project: Refrigeration and Freezer Design

the project is Half done I need MORE calculations. I attached what I already have and those are the instructions:You have a new job in a consulting company which has been hired to design a walk-in freezer and refrigerator for a new food transit warehouse. Your predecessor at the company calculated that the freezer needs a 10 ton cooling capacity while the refrigerator needs 30 tons of cooling. After completing these calculations, he quit (Uh-oh) so your job is to compete the design and to write a report that your boss can present to the builders in fulfillment of the contract.It has been determined that the system will us Refrigerant 134a and that the freezer must be maintained between -10 and +5°F (-24 to -15°C) while the refrigerator must be kept between 37 and 45°F (3 to 7°C). Ambient conditions are expected to be as high as 100°F (38°C). Assume that refrigeration compressor efficiencies are 75%. Heat exchangers must maintain a 20°F (11°C) difference between the hot and cold sides at all locations.It is customary to use a separate vapor-compression refrigeration cycle for each of the warehouse sections, but the boss says that other options should be explored and compared, because the owners are interested in saving money on expenses. This includes both first cost and operating expenses. The largest operating expense is typically the electricity to run the compressors, so reducing compressor work input and increasing coefficient of performance is important here. In general compressor work is reduced by having more that one compression stage and cooling the gas between stages. One such configuration is to put a flash chamber at an intermediate pressure between the compressors. See Problem 10.024 in the homework. First cost is related to many things including how many compressor systems are needed. There are several ways to have both refrigeration and freezer systems using only one compressor. See Problem 10.027 and Reserve Problem 10.003. (Both have been included in the WileyPlus homework labeled Problem 10.024 and solutions are visible to the student.)In particular, you should find the power consumed in each scenario along with sufficient calculations and results to show which have the least consumption and the maximum coefficients of performance and also that each of them satisfies all of the constraints. You may use on-line tables for the refrigerant, but you need to be clear about the source of the data. You may work in either US or SI units as long as you find power consumption in kilowatts. You may assume that compressor inlets are at saturated vapor and that condenser outlets are saturated liquid.Also assume that the steam is generated by burning natural gas (methane) and that the products of combustion must be maintained at or above the maximum steam temperature to calculate the energy available.RequirementsThere are many aspects which go into this design. Your report must show that you have done enough calculations to convince me that you understand all of the aspects that need to be considered. This means showing trials that did not work and what changes you made to approach a solution that does work. Simply presenting a result is not sufficient. The thought process which goes into creating a design is of critical importance here.NOTE: This is an open-ended project. There is no correct answer, just looking for sufficient detail in the calculations that an engineer can be reasonably certain that (1) the calculations and approaches are correct, (2) there are enough results that trends can be discerned, and (3) you have sufficient discussion that the reader can be confident that your conclusions are reasonable.

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Multistage vapor pressure refrigeration is utilized when a few evaporators are required at
different temperatures, for example, in a general store or when the temperature of the evaporator
turns out to be low. Low evaporator temperature shows low evaporator weight and low refrigerant
thickness into the compressor. Two little compressors in arrangement have a littler dislodging and
ordinarily work more proficiently than one substantial compressor that covers the whole weight
territory from the evaporator to the condenser (Visek, Joppolo, Molinaroli & Olivani, 2014). The
same can be said for compressors with halfway or add up to infusion at a moderate weight. This is
particularly valid in refrigeration frameworks that utilization smelling salts on account of the vast
measure of superheating that happens amid the pressure procedure. So the decision for multistage
vapor pressure is self-evident. At the point when compressors are associated in an arrangement,
the vapor between stages ought to be cooled to bring the vapor to soaked conditions previously
continuing to the following phase of pressure.
Intercooling as a rule limits the dislodging of the compressors decreases the work
prerequisite and expands the COP of the cycle. On the off chance that the refrigerant temperature
between stages is above encompassing, a straightforward intercooler that expels warm from the
refrigerant can be utilized (Marques, Davies, Maidment, Evans, & Wood, 2014). If the temperature
is underneath surrounding, which is simply the typical case, the refrigerant must be utilized to cool
the vapor. This is refined with a glimmer intercooler. A typical suspicion is to work the intercooler
at about the geometric mean of the dissipating and consolidating weights. This working point gives
a similar weight proportion and almost level with volumetric efficiencies for the two compressors.
Floor construction
Thermal Resistance R (m2K/W)
5.3 – 7.0
Suspended Ceilings
7.9 – 8.8
8.8 – 10.6
Dry sand layer
Appropriate thickness = 300mm
Thermal properties of dry sand
Specific density = 1600kg/m3
Specific heat capacity = 794 J/kgK
Thermal conductivity = 0.35
Ventilation layer
Thermal properties of concrete
Specific density ? = 2310 kg/m3
Specific heat Cp = 880 J/kgK
Thermal conductivity k = 1.21 W/mK
Determining convection coefficient in the ventilation layer
A: Flow area of a ventilation duct
O: Wetted perimeter of the cross section of a ventilation duct
D: Hydraulic diameter of a ventilation duct
vair: air velocity in the ventilation ducts
Tair: average air temperature in the ventilation ducts
?: Specific density of the air
µ: Dynamic viscosity of the air
k: Thermal conductivity
Cp: Specific heat
L: Length of the duct
Reynolds number is calculated by
Re = ?vair D/ µ
Laminar forced convection
(Re < 2300) Fully developed flow ( L > 0.0575 ReD)
Convection coefficient in the ventilation layer
Area = 0.050 *0.14 = 0.007
O= 2*0.05 + 2*0.14 = 0.38
D= A/O *4 = 0.0737
Air load
When freezing starts, air fills the freezing room at ambient temperature
V air = (3*2*18)/2 + (1*2*18)/2 + 1*2*18 + 5 * 1 * 18 = 198 m3
Ambient temperature: tamb = 25°C
Ambient pressure: p = 1 atm = 101.325 kPa
Blast freezer operating temperature: tc = -40°C
Density of air 25°C: ? = 1.18435 kg/m3
Air mass: Mair = Vair x ? = 234.50 kg
Average relative humidity in Santander : f =77.5%
Saturation pressure at 25°C: pws = 3.1692 kPa
Partial pressure of water vapour at 25°C: pw = f x pws = 2.4561 kPa
The humidity ratio is given by
W = Mw/M da = 0.62198
Where Mw is the mass of water vapor and M da is the mass of dry air
W is therefore 0.015452
M air = M da + M w
M da = 1/ (1+w) * M air
M w = W/ (1+w) * M air
q= M da Cp air (t amb – t c ) + M w Cp w + (t amb – 0 ) +M w Lw + M w ? Cp ice dt
where by
Cpair: heat capacity of air = 1006.5 J/kgK
Cpw: heat capacity of water = 4220 J/kgK
Cpice: heat capacity of ice = 2062.3 + 6.0769 × t J/kgK
Lw : Latent heat of water congelation = 333.8 kJ/kg
teva: Evaporator temperature = -45°C
Total refrigerating load
Maximum Product load, which will be this one of tuna: 213.110 kW
Transmission load: 2.451 kW
Air load: 4.718 kW
Internal load: 44.532 kW
Defrost load: included in the safety factor
Safety factor: 10%
Qtotal = 291.291 kW
Conveyer belt
Lifting conveyer belt Tecnofish, with funnel (see figure 13.1).
Width: 800 mm
Material: P.V.C.
Length: 3.3 m
Chassis: stainless steel AISI-316
• Modular conveyer belt Tecnofish (see figure 13.2).
Width: 1000 mm
Material: stainless steel AISI-316
Length: 15.3 m
• Modular lifting conveyer belt Tecnofish
Material: P.V.C.
Width: 800 mm
Length: 4 m
Chassis: Stainless steel AISI-316
Glazing conveyer belt Tecnofish
Material: P.V.C.
Width: 800 mm
Length: 4 m
Chassis: Stainless steel AISI-316
Conveyer belt Tecnofish.
Material: P.V.C.
Width: 800 mm
Length: 4 m
Chassis: Stainless steel AISI-316
The refrigeration load and cooling time of the brine
Refrigeration load
Q trans = UA *change in temperature
Q trans is the heat gained
A is the area of the section
Change in T is the difference between the temperature outside and the temperature of the
refrigerated space.
Sensible heat gain through the free surface
Q u = s Sp ( T 4m-T 4eq ) + hS p ( Tm –Ta) + r dm/dt – as S p Is
Ta is the temperature of air
Tm is the temperature of brine
Teq is the equivalent temperature
Sp is the swimming pool`s free surface area
Is is the solar radiation intensity
As is the absorption coefficient
The heat convection coefficient is given by
h = 7+ 3.3?
? is the velocity of air
Marques, A. C., Davies, G. F., Maidment, G. G., Evans, J. A., & Wood, I. D. (2014). Novel
design and performance enhancement of domestic refrigerators with thermal
storage. Applied Thermal Engineering, 63(2), 511-519.
Visek, M., Joppolo, C. M., Molinaroli, L., & Olivani, A. (2014). Advanced sequential dual
evaporator domestic refrigerator/freezer: System energy optimization. International
Journal of Refrigeration, 43, 71-79.

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