Question

You as the engineer of the food processing plant "chick" are tasked with designing a new line to process chicken for sale in supermarkets. The new line is planned for a production of 120 chickens per hour working 8 hours per day, from Monday to Friday. On average each chicken has a mass of 1.2 kg. The chicken enters the process at 20 ° C and is submerged in a hot water bath at 90 ° C for two minutes, which raises the average chicken temperature to 40 ° C. Additionally, the hot water bath loses 10% of heat to the environment. The hot water bath consists of 4 gallons of water for each chicken and should be kept at 90 ° C at all times. The heat that maintains the hot water is supplied by a boiler that produces saturated steam at 150 kPa that comes out as saturated liquid at the same pressure. After the process the chicken is subjected to a bath of cold water to lower the temperature from 40 ° C to 15 ° C. Calculate the steam flow necessary to keep the process within the desired parameters. Determine and select the boiler (or boilers or any equipment that performs the function) necessary to meet the steam requirement of the plant. Include as many details of the selected equipment as brand, capacity, etc. Calculate the flow of cold water necessary to cool the product to the desired temperature if the water enters the process at 10 ° C and should not flow out at more than 14 ° C. Determine and select the "chiller" (or the "chillers" or any equipment that performs the function) necessary to satisfy the need of the plant. Include as many details of the selected equipment as brand, capacity, etc. Cp of chickens 3.32 kg/(kj K) Using the second law of thermodynamics

Answer 1

2 minutes ime cold wader Hot wader bath bah㈠ chicken cut chick en at 40 chicken cold wede o to loose 1o of supplieoae hecu to in Col en o f chill e Steam in Steam out liquid apour Given ass of chicken a 12kg Hed absosbed by chicken cu hot bath (m CPAT)( chicken = 1.2 x832x (40-20) - 74.6& kJ 'This hect 15 equal to head given by hol…cd er 77.an mCp(AT ho 1outex

71,68 = (4x3.39) x 4 1.87x (90-7) 1.403 90-T hot moder trop. afteγ processinq one chick en ed to Keep hot vwl er bath cunp ct But this heat is go% of he supplied by steam →(10./. head suppl e-d) lied to ervimmnm Hence total hea supplicd by sieai 79 68=- x Heap supplied by steam so O Heat supplied by Steauna: 88-.53 kJ he Mears it ive seuaed liqda ladens hel ont Latenl hec steam = nb× 22 2 6.22 cuble n & minutes

Hence, st eam flow al e IRO sec He lost by chick en f chicken cold toath q.6 low heat lost by cuiot en in cold Head gain by cold wcd es bath in two minutes mass flow nale of woder sec Hecu sequsement fos hot buth is very less So instead of boiles, any hecde is pseffexcble becouse boiles has huge no of equu pemen ts,