ENERGY BALANCE

ENERGY BALANCE

Nur Istianah,ST.,MT.,M.Eng THP UB 2017

HEAT BALANCE OF PROCESS

INVOLVE PHASE CHANGES

Steam

(sat.vapor)

Steam (sat.liquid)

Guava juice (30°C)

Guava

concentrate (60°C)

Vapor

(60°C), 20 kPa Latent heat:

Steam Table

Phase changes

Example 1

Calculate the amount of water that must be supplied to a heat exchanger that cools 100 kg/h of tomato paste from 90◦C to 20◦C. The tomato paste contains 40% solids. The increase in water temperature should not exceed 10◦C while passing through the heat exchanger. Initial temperature of water is 20◦

C. There is no mixing of water and tomato paste in the heat exchanger.

q₄

Water

(T₂=T₁+10^oC), kg

water(T₁), W Kg q₃

q₂

tomato paste

20^oC tomato paste q₁

100 kg/h 90^oC 40% solids

Solution:

Q in = Q out T ref = 0 K

Q in

Q₁ = m C_p ΔT = 100 Kg * 2846,76 J/KgK * (90 –0 ⁰K) = 25621 kJ

Q₃ = m C_p ΔT = m* 4187 J/KgK *(20 – 0 K) = 83,740m kJ/Kg Q out

Q₂ = m C_p ΔT = 100 Kg * 2846,76 J/KgK * (20–0 K) = 5694 kJ

Q₄ = m C_p ΔT = m*4187 J/KgK *(30 –0 K) = 125,610m kJ/Kg By using energy balance :

Q₁ + Q₃ = Q₂ + Q₄

(25621- 5694) kJ = (125,610 - 83,740) m kJ/Kg m= 475,9 Kg

Excercise 2

Calculate the amount of steam at 121.1◦C (250◦F) that

must be added to 100 kg of a food product with a specific

heat of 3559 J/(kg ∼ =K) to heat the product from 4.44◦C

(40◦F) to 82.2◦C (180◦F) by direct steam injection.

Solution

The diagram of the system is shown in Fig. 5.5. Let x=kg of steam required. From steam table, the enthalpies of water and steam are as follows:

For steam at 121.1◦C or 250◦F, hgfrom Table A.3=1164.1 BTU/lb=2.70705 MJ/kg

For water at 82.2◦C or 180◦F, hf =148.00 BTU/lb=0.34417 MJ/kg=2.70705−0.34417=2.36288 MJ/kg

Total heat loss steam=x(2.36288) MJ

Heat gain by product=100kg[3559 J/(kgK)](82.2−4.44) K=27.67478 MJ x(2.36288)=27.67478

x=11.71 kg steam required

Excercise 3

Milk with the flow rate of 5000 kg/h was pasteurised using heat exchanger at 135 °C for 6 s. The initial temperature of milk is 15 °C and vapor pressure of HE is 313,18 kPa. Calculate steam (100% in quality) needed if the specific heat of milk 3,894 kJ/kg°C and Tref 0°C

Excercise

milk 15 ⁰C

condensat 135 ⁰C

Steam 313,1 KPa

product 135^oC

Jawab.

Q in = Q out T ref = 0 K

Q in

a. Qin milk = mCp(Tm-Tref) = (5000 kg/h)(3,894 kJ/kg°C)(15-0°) = 292.050,0 kJ/h

b. Qin uap = m.hg = m (2726,54 kJ/kg)

From steam table, hg at 135°C is 2726,54 kJ/kg

Q out:

a. Qout product = mCp(T-Tref)=5000 kg/h(3,894 kJ/kg°C)(135-0°) = 2.628.450,0 kJ/h

b. Qout condensat = m. hf = m (567,68 kJ/kg) From steam table, hf at 135°C is 576,68 kJ/kg Energy balance:

Qin milk + Qin steam = Q out product + Qout condensat 292.050,0 + m 2726,54=2.628.450,0 + m 567,68 So that, m= 1086,768 kg/h

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