Chapter 4
Energy Balance
Theory The overall energy balance equation for a system with one inlet (point 1) and one outlet (point 2) is: vml 2 v 2 dðmEÞ _ 1 H2 þ m2 þ z2 g m _ 2 þ q Ws ¼ þ z1 g m H1 þ dt 2 2 The overall energy balance equation for a system at steady state with more than two streams can be written as: X v2 _ ¼ q Ws H þ m þ zg m 2 where
H = enthalpy, J/kg vm = average velocity, m/s = correction coefficient (for a circular pipe = 1/2 for laminar flow, 1 for turbulent flow) z = relative height from a reference plane, m m = mass of the system, kg _ = mass flow rate, kg/s m q = heat transferred across the boundary to or from the system (positive if heat flows to the system), W Ws = shaft work done by or to the system (positive if work is done by the system), W E = total energy per unit mass of fluid in the system, J/kg t = time, s
In most of the cases, the overall energy balance ends up as an enthalpy balance because the terms of kinetic and potential energy are negligible compared to the enthalpy term, the system is assumed adiabatic (Q ¼ 0), and there is no shaft work (Ws ¼ 0). Then: X _ ¼0 mH
S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008
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4 Energy Balance
Review Questions Which of the following statements are true and which are false? 1. The energy in a system can be categorize as internal energy, potential energy, and kinetic energy. 2. A fluid stream carries internal energy, potential energy, and kinetic energy. 3. A fluid stream entering or exiting a control volume is doing PV work. 4. The internal energy and the PV work of a stream of fluid make up the enthalpy of the stream. 5. Heat and shaft work may be transferred through the control surface to or from the control volume. 6. Heat transferred from the control volume to the surroundings is considered positive by convention. 7. For an adiabatic process, the heat transferred to the system is zero. 8. Shaft work supplied to the system is considered positive by convention. 9. The shaft work supplied by a pump in a system is considered negative. 10. If energy is not accumulated in or depleted from the system, the system is at steady state.
Examples Example 4.1 1000 kg/h of milk is heated in a heat exchanger from 458C to 728C. Water is used as the heating medium. It enters the heat exchanger at 908C and leaves at 758C. Calculate the mass flow rate of the heating medium, if the heat losses to the environment are equal to 1 kW. The heat capacity of water is given equal to 4.2 kJ/kg8C and that of milk 3.9 kJ/kg8C. Solution Step 1 Draw the process diagram: q milk water
1000 kg/h 45°C 75°C
HEAT EXCHANGER
72°C 90°C
milk water
Examples
23
Step 2 State your assumptions: l
l l
l
The terms of kinetic and potential energy in the energy balance equation are negligible. A pump is not included in the system (Ws ¼ 0). The heat capacity of the liquid streams does not change significantly with temperature. The system is at steady state.
Step 3 Write the energy balance equation: _ w in Hw Rate of energy input ¼ m
in
_m þm
in
Hm in
_ w out Hw out þ m _ m out Hm out þ q Rate of energy output ¼ m (with subscript ‘‘w’’ for water and ‘‘m’’ for milk). At steady state rate of energy input ¼ rate of energy output or _ w in Hw in þ m _ m in Hm in ¼ m _ w out Hw out þ m _ m out Hm out þ q m Step 4 Calculate the known terms of eqn (4.1) i) The enthalpy of the water stream is: Input: Hw
in
Output: Hw
¼ cp T ¼ 4:2 90 ¼ 378 kJ=kg out
¼ cp T ¼ 4:2 75 ¼ 315 kJ=kg
ii) The enthalpy of the milk stream is: Input: Hm
in
Output: Hm
¼ cp T ¼ 3:9 45 ¼ 175:5 kJ=kg out
¼ cp T ¼ 3:9 72 ¼ 280:8 kJ=kg
Step 5 Substitute the above values in eqn (4.1), taking into account that: _ w in ¼ m _ w out ¼ m _ w and m _m m
in
_ m out ¼m
_ w 378 þ 1000 175:5 ¼ m _ w 315 þ 1000 280:8 þ 1 3600 m
(4:1)
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4 Energy Balance
Step 6 _w Solve for m _ w ¼ 1728:6 kg=h m Example 4.2 A dilute solution is subjected to flash distillation. The solution is heated in a heat exchanger and then flashes in a vacuum vessel. If heat at a rate of 270000 kJ/h is transferred to the solution in the heat exchanger, calculate: a) the temperature of the solution at the exit of the heat exchanger, and b) the amount of overhead vapor and residual liquid leaving the vacuum vessel. The following data are given: Flow rate and temperature of the solution at the inlet of the heat exchanger is 1000 kg/h and 508C, heat capacity of the solution is 3.8 kJ/kg8C, and absolute pressure in the vacuum vessel is 70.14 kPa. Solution Step 1 Draw the process diagram: mv, Hv
I II mFi TFi HFi
HEAT EXCHANGER
III mFo VACUUM TFo VESSEL HFo
q
mL, TL, HL
Step 2 State your assumptions: l
l l l
l
The terms of kinetic and potential energy in the energy balance equation are negligible. A pump is not included in the system (Ws ¼ 0). The heat losses to the environment are negligible. The heat capacities of the liquid streams do not change significantly with temperature and concentration. The system is at steady state.
Step 3 Write the energy balance equation in envelope II: _ Fi HFi þ q ¼ m _ Fo HFo m or
(4:2)
Examples
25
_ Fi cpF TFi þ q ¼ m _ Fo cpF TFo m
(4:3)
Substitute known values: 1000 3:8 50 þ 270000 ¼ 1000 3:8 TFo
(4:4)
Solve for TFo: TFo ¼ 121 o C Step 4 Write the mass and energy balance equations in envelope I: i) Overall mass balance: _ Fi ¼ m _Vþm _L m
(4:5)
_ Fi HFi þ q ¼ m _ V HV þ m _ L HL m
(4:6)
_ Fi cpF TFi þ q ¼ m _ V HV þ m _ L cpL TL m
(4:7)
ii) Energy balance:
or
Step 5 Calculate mv using equations (4.5), (4.6) and (4.7): i) From eqn (4.5): _L¼m _ Fi m _V m
(4:8)
ii) Substitute eqn (4.8) in (4.7): _ Fi cpF TFi þ q ¼ m _ V H V þ ðm _ Fi m _ V ÞcpL TL m
(4:9)
iii) Find the saturation temperature and the enthalpy of saturated vapor at 70.14 kPa from the steam tables: TL=908C V=2660 kJ/kg iv) Substitute numerical values in eqn (4.9): _ V 2660 þ ð1000 m _ V Þ 3:8 90 1000 3:8 50 þ 270000 ¼ m
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4 Energy Balance
_V v) Solve for m _ v ¼ 50:9 kg=h m Step 6 Alternatively, an energy balance in envelope III can be used instead of envelope I: i) Write the energy balance equation: _ Fo cpF TFo ¼ m _ V HV þ m _ L cpL TL m
(4:10)
ii) Combine eqns (4.5) and (4.10) and substitute numerical values: _ V Þ 3:8 90 _ V 2660 þ ð1000 m 1000 3:8 121 ¼ m _V iii) Solve for m _ v ¼ 50:9 kg=h m
Exercises Exercise 4.1 How much saturated steam with 120.8 kPa pressure is required to heat 1000 g/h of juice from 58C to 958C? Assume that the heat capacity of the juice is 4 kJ/ kg8C. Solution Step 1 Draw the process diagram: mji = 1000 kg/h juice 5°C ms steam 120.8kPa
mjo HEAT EXCHANGER
juice
95°C ms condensate 120.8kPa
Exercises
27
Step 2 Write the energy balance equation: _ ji Hji þ m _ s Hs ¼ ::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::: m or _ ji cpj Tji þ m _ s Hs ¼ ::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::: m Step 3 Substitute numerical values in the above equation. (Find the enthalpy of saturated steam and water [condensate] from steam tables): :::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::: þ :::::::::::::::::::::::: Step 4 _s Solve for m _ s ¼ ::::::::::::::::::::::::::::::::::::::kg=h m
Exercise 4.2 How much saturated steam with 120.8 kPa pressure is required to concentrate 1000 kg/h of juice from 12% to 20% solids at 958C? Assume that the heat capacity of juice is 4 kJ/kg8C. Solution Step 1 Draw the process diagram: water vapor
mv mji = 1000 kg/h juice o 95 C ms steam 120.8kPa
EVAPORATOR
mjo 95oC
juice
ms condensate 120.8kPa
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4 Energy Balance
Step 2 Write the overall mass balance equation on the juice side: _Vþm _ jo 1000 ¼ m Step 3 Write the solids mass balance equation: _ jo 0:12 1000 ¼ ::::::::::::::: m _ jo and m _V Solve for m _ jo ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=h m _ V ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=h m Step 4 i) Write the enthalpy balance equation: _ ji cpj Tji þ m _ s Hs ¼ ::::::::::::::::::::::: þ ::::::::::::::::::::::::: ¼ ::::::::::::::::::: m ii) From steam tables, find the enthalpy of water vapor at 958C, of saturated steam at 120.8 kPa, and of water (condensate) at 120.8 kPa. iii) Substitute numerical values in the above equation: :::::::::::::::::: þ ::::::::::::::::: ¼ :::::::::::::::::: þ ::::::::::::::::: þ ::::::::::::::::: _s iv) Solve for m _ s ¼ ::::::::::::::::::::::::::::::::::::::::kg=h m Exercise 4.3 2000 kg/h of milk is sterilized in a steam infusion sterilizer. The milk is heated to 1458C by introducing it into the steam infusion chamber H and then is cooled quickly by flashing in the flash vessel F. The vapor that flashes off in the vessel F is condensed in the condenser C by direct contact of the vapor with cooling water. To avoid dilution of the milk, the pressure in the vessel F must be such that the rate at which vapor flashes off in the vessel F is equal to the steam that is added in the vessel H. Calculate the cooling water flow rate in the condenser that will give the required pressure in the flash vessel. The following data are
Exercises
29
given: The temperature of the milk at the inlet of H is 408C, the temperature of the cooling water at the inlet of the condenser is 208C, the steam introduced into the chamber H is saturated at 475.8 kPa pressure, and the heat capacity of the milk is 3.8 kJ/kg8C at the inlet of the infusion chamber and 4 kJ/kg8C at the exit of the infusion chamber. Solution Step 1 Draw the process diagram:
milk mmi, Hmi I steam
cooling water mwi, Hwi III
II vapor mv, Hv
ms, Hs F
H
C
mms, Hms mmo, Hmo milk
mwo, Hwo
Step 2 State your assumptions: l
l l l
l
l
The terms of kinetic and potential energy in the energy balance equation are negligible. A pump is not included in the system (Ws ¼ 0). The heat losses to the environment are negligible. The water vapor pressure of the milk is equal to that of water at the same temperature. The water vapor pressure in the condenser is equal to the water vapor pressure in the flash vessel. The system is at steady state.
Step 3 Write the mass and energy balance equations in envelope I: i) Energy balance in envelope I: _ mi Hmi þ m _ s Hs ¼ m _ ms Hms m ii) Overall mass balance in envelope I: :::::::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::::
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4 Energy Balance
iii) Substitute numerical values and combine the last two equations: _ s ¼ ::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::: þ 2746:5 m _s iv) Solve for m _ s ¼ ::::::::::::::::::::::::::::::::::::::::::::::::kg=h m Step 4 i) Write the energy balance in envelope II: ::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: _s¼m _ v , in order to avoid ii) Substitute values taking into account that m dilution of the milk: ::::::::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::: or 1389158 7600 T ¼ 395:1 HV iii) Solve the last equation by trial and error to find the value of T that will give a value of HV in agreement with steam tables. T = . . .. . .. . .. . .. . .. . .. . ...8C. Step 5 Write the overall mass balance and energy balance in envelope III: i) Overall mass balance: :::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::::::: ii) Energy balance: :::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::::::: iii) Substitute numerical values in the last equation and solve for mwi. The temperature of the water at the exit of the condenser must be equal to :::::::::::::::::::: C, because the water vapor pressure in the condenser was assumed equal to that in the flash vessel F. _ wi ¼ :::::::::::::::::::::::::::::::::::::::kg=h m
Exercises
31
Exercise 4.4 Find the amount of saturated steam at 270.1 kPa required to heat 100 kg of cans from 508C to 1218C, if the heat capacity of the cans is 3.5 kJ/kg8C.
Exercise 4.5 One ice cube at 108C weighing 30g is added to a glass containing 200ml of water at 208C. Calculate the final water temperature when the ice cube melts completely. Assume that 3 kJ of heat are transferred from the glass to the water during the melting of the ice? Use the following values: the latent heat of fusion of the ice is 334 kJ/kg, the heat capacity of the ice is 1.93 kJ/kg8C, and the heat capacity of the water is 4.18 kJ/kg8C.
Exercise 4.6 For quick preparation of a cup of hot chocolate in a cafeteria, cocoa powder and sugar are added in a cup of water and the solution is heated by direct steam injection. If the initial temperature of all the ingredients is 158C, the final temperature is 958C, the mass of the solution is 150g initially, and the heat capacity of the solution is 3.8 kJ/kg8C, calculate how much saturated steam at 1108C will be used. State your assumptions.
Exercise 4.7 Calculate the maximum temperature to which a liquid food can be preheated by direct steam injection if the initial temperature and the initial solids concentration of the food are 208C and 33% respectively, and the final solids concentration must not be less than 30%. How much saturated steam at 121 kPa pressure will be used? Assume that the heat capacity of the food is 3.0 kJ/kg8C initially and 3.1 kJ/kg8C after the steam injection.
