Question

A refrigeration system is to cool bread loaves with an average mass of 350 g from 30 to –10°C at a rate of 1200 loaves per hour with refrigerated air at –30°C. Taking the average specific and latent heats of bread to be 2.93 kJ/kg·°C and 109.3 kJ/kg, respectively, determine (a) the rate of heat removal from the breads, in kJ/h (b) the required volume flow rate of air, in m3/h, if the temperature rise of air is not to exceed 8°C (c) the size of the compressor of the refrigeration system, in kW, for a COP of 1.2 for the refrigeration system.

Answer 1

The rate of heat removal from the breads is calculated based on sensible and latent heats, and the volume flow rate of air is determined using the heat removal rate and the properties of air. The size of the compressor is calculated using the specified coefficient of performance.

Step-by-step explanation:

Rate of Heat Removal from the Breads

To calculate the rate of heat removal from the breads (part a), we'll first need to determine the total energy required to cool a single loaf of bread from 30 to -10°C. This includes both the sensible heat to cool the bread from 30°C to 0°C and the latent heat of freezing, as well as the sensible heat from 0 to -10°C. The sensible heat (Q_sensible) is given by Q_sensible = m × c × ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change. The latent heat (Q_latent) is Q_latent = m × L, where L is the latent heat of the bread.
Combing these and considering the rate of 1200 loaves per hour, the total heat removal rate can be calculated and given in kJ/h.

Volume Flow Rate of Air

To determine the required volume flow rate of air (part b), we use the principle of conservation of energy. The specific heat of air and the maximum temperature rise of 8°C determine how much air is required to remove a certain amount of heat per hour. Using the heat removal rate from part a and the properties of air, we can compute the volume flow rate in m³/h.

Size of the Compressor

For part c, we calculate the size of the compressor using the coefficient of performance (COP) of 1.2. The power input to the compressor (P) can be found using the relationship COP = Q_cooling / P, where Q_cooling is the rate of heat removal. By rearranging the equation, we get P = Q_cooling / COP, which allows us to calculate the size of the compressor in kW.

Answer 2

rate of removal of heat is 100566 KJ/h

volume flow rate is 8626.35 m³/h

size of compressor is 23.83 kW

Step-by-step explanation:

Given data

mass m = 350 g = 0.37 kg

temperature T1 = –10°C

temperature T2 = 30°C

number of cool bread loaves = 1200

average specific heat = 2.93 kJ/kg·°C

latent heats = 109.3 kJ/kg

solution

we will find here first mass rate of bread that is = number of cool bread × mass

mass rate bread = 1200 × 0.37 = 444 kg/h = 0.123 kg/s

so removal of heat = sum of heat of bread and heat of freeze

so removal of heat = mass of bread C(T2-T1) + mass of bread ×latent heat

removal of heat = 444 (2.93)(30 -(-10)) + (444 × 109.3)

so rate of removal of heat = 100566 KJ/h

and

we know that specific heat of air = 1.005 kJ/kg from property of ideal gas table

so volume flow rate is mass of air / density of air

and density of air = P/RT

density = 101.325 / 0.287(-30+273)

density = 1.45 kg/m³

so mass flow rate = mass of air ×C×(Δt)

100566 = mass of air (1.005) 8

mass of air = 12508.208 kg/h

so volume flow rate will be = mass of air / density

volume flow rate = 12508.208 / 1.45 = 8626.35 m³/h

and

power capacity of compressor = heat removal / size of compressor

so

size of compressor = 100566 / 1.2 = 83805 kJ/h = 23.83 kW

so size of compressor is 23.83 kW