Keseimbangan Panas

1. Keseimbangan Panas

2. Heat balance calculations are treated in the same manner as material balances. • The amount of heatentering a system must equal the amount of heat leaving a system, or: • Heat in = heat out + accumulation • At a steady state, the accumulation term is zero and heat entering the system must equal what leaves the system. • Heat balance problems are facilitated by using diagrams that show process streams bringing heatand taking heat out of a system.

3. Calculate the amount of water that must be supplied to a heat exchanger that cools100 kg/h of tomato paste from 90◦C to 20◦C. The tomato paste contains 40% solids. The increasein water temperature should not exceed 10◦C while passing through the heat exchanger. There is nomixing of water and tomato paste in the heat exchanger.

4. T1 Inlet temperature = 20◦C, and T2 = exit water temperature = 20 + 10 = 30◦C. • Let 20◦C be the referencetemperature for enthalpy calculations. • The specific heat of water = 4187 J/(kgK), and • tomato paste, Cavg = 3349(0.6) + 837.36 = 2846.76 J/(kg K) • Heat content of entering tomato paste: • q1 = (100 kg) [2846.76 J/(kg K)](90 − 20) K= 19.927 MJ • Heat content of tomato paste leaving system: • q2 = 100kg[2846.76 J/(kg K)](20 − 20) K = 0 • Let W = kg water entering the system • q3 = Wkg[4187 J/(kg K)](20 − 20) K = 0 • q4 = heat content of water leaving the system • = Wkg [4187 J/(kg K)](30 − 20) K = 41, 870(W) J • The heat balance is • q1 + q3 = q2 + q4 • Because q2 and q3 = 0, q1 = q4, and: heat loss by the tomato paste = Heat gain by water

5. Type heat transfer • Liquids can be made to flow in either the same direction or in opposite directions in aheat exchanger. • Counter-current flow of fluids has a higher heat transferefficiency than co-current (or ‘parallel’) flow and is therefore widely used in heatexchangers. • However, the temperature difference varies atdifferent points in the heat exchanger and it is necessary to use a logarithmic mean temperature difference in calculations