500 kg/hr of orange juice is to be frozen in a heat exchanger. The solids conten

Question

500 kg/hr of orange juice is to be frozen in a heat exchanger. The solids content of the juice is 40%. If the initial temperature of the juice is 40 C and the final temperature is -20 C, determine the flow rate of Ethylene Glycol Solution that enters at -35 C and exits at -22C. The Ethylene Glycol solution has a freezing point of -37 C and a specific heat of 3.26 kJ/(kg K).

*This was all the information given, I looked up specific heat capacity of orange juice however & found above freezing: 0.93 Btu/lb*°F then below freezing: 0.42 Btu/lb*°F

Explanation / Answer

Now in heat exachanger:

We know that

Heat lost=Heat gain

So Heat lost by Orange juice =Heat gain by ethylene glycol

Heat lost by orange juice = m CpT

Cp,solid = 0.42 Btu/lb F=1758.45 J/kg K

Cp,liquid =0.93=3893.72 J/kg K

40% solids means =0.4*500=200 kg/hr

T1=40 C=40+273.15=313.15 K

T2=-20 C=-20+273.15=253.15 K

Remaining is liquid

So Heat lost =200*1758.45*(313.15-253.15)+300*3893.72*(313.15-253.15)=91188360 J/kg=91188.36 kJ/kg

Heat gain = 9188.36 kJ/kg

T1=-35 C=238.15 K

T2=-22 C =251.15 K

Q=mCpT

m= 9188.36/3.26*(251.15-238.15)=216.8 kg/hr

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