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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.

Respuesta :

Answer:

rate of removal of heat is 100566 KJ/h

volume flow rate is 8626.35 m³/h

size of compressor is 23.83 kW

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

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