Energy Balance Equations for open Systems

1. Chemical Process
Calculations II
Lecture 3
Energy balance for open system

2. Lecture Outcomes
• Student will be able to explain
– Energy balance for open unsteady state system
– Explain Flow work
– Explain Energy balance for open steady state system
– Apply Energy balance at heat exchanger.
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3. Energy Balance For Open Unsteady State System
• Some system are these open?
– Internal combustion engines
– Air compressors
– Water pump
– Steam engine
– Boiler Turbine
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4. For open & unsteady state system
• The accumulation term (∆E) in the energy balance Equation can be
non-zero
1. The mass in the system changes
2. The energy per unit mass in the system changes
And both 1 & 2 occur.
General Energy balance Equation:
∆𝐸 = 𝐸𝑡1 − 𝐸𝑡2 = 𝑄 + 𝑊 − ∆ 𝐻 + 𝐾. 𝐸 + 𝑃. 𝐸
Here change in ∆ represents:
First one change inside the system with respect to time
Second shows energy with mass out – energy with mass in.
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5. For Open & Unsteady State System
• Components of general equation of energy are:
1. Accumulation in the system from T1 to T2 :
∆𝐸 = 𝑚 𝑡2 𝑈 + 𝐾. 𝐸 + 𝑃. 𝐸 − 𝑚 𝑡1 𝑈 + 𝐾. 𝐸 + 𝑃. 𝐸
2. Energy transfer in with mass flow = 𝑚1 𝑈 + 𝐾. 𝐸 + 𝑃. 𝐸
3. Energy transfer out with mass flow = 𝑚2 𝑈 + 𝐾. 𝐸 + 𝑃. 𝐸
4. Net energy Transfer (in and out of system) by heat = Q
5. Net energy Transfer (in and out of system) by shaft, mechanical or
electrical work = W
6. Net energy Transfer by work to introduce or remove mass from T1 to T2 =
𝑝1 𝑉1 𝑚1 − 𝑝2 𝑉2 𝑚2
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6. Homework.
• For open system
– ∆𝑬 𝒊𝒏𝒔𝒊𝒅𝒆 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 = ∆𝑬 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒆𝒙𝒄𝒉𝒂𝒏𝒈𝒆 𝒐𝒓 𝒇𝒍𝒐𝒘 + ∆𝑬 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝑬𝒏𝒆𝒓𝒈𝒚 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒇𝒓𝒐𝒎 𝒃𝒐𝒖𝒏𝒅𝒓𝒚.
– Also ∆𝑬 𝒊𝒏𝒔𝒊𝒅𝒆 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 = ∆𝑬 𝒊𝒏 − ∆𝑬 𝒐𝒖𝒕
Using the energy balance component and above Equation Prove that
∆𝑬 = 𝑬 𝒕𝟏 − 𝑬 𝒕𝟐 = 𝑸 + 𝑾 − ∆ 𝑯 + 𝑲. 𝑬 + 𝑷. 𝑬
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7. Flow work
• Flow work occurs at areas of a system’s boundary across which there is material flowing.
• The flow work is associated with the force and displacement required to push the material
into the system (input streams), or with the force and displacement required to push
material out of the system (output streams).
• A fluid cannot flow unless it creates space for itself when it enters or exits a system.
• The forces that do the necessary pushing are exerted by particles of the flowing material
inside the system on the particles of the material outside the system, and are evaluated at
the stream inlets and outlets where the transfer of material into/out of the system takes
place.
• These forces, expressed per area, are the pressure P in the stream.
• The flow work W fl,j, performed by the system on stream j in order to transfer a volume of
material Vj into or out of the system, is straightforwardly calculated from the pressure Pj in
the stream
Wfl,j = PjVj
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8. Crystallizer Filling a fixed volume tank with water Batch distillation
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9. Problem (to learn when ∆U= ∆H for a open unsteady system)
• A rigid well insulated tank is connected to two
system.
• One valve goes to a steam line that has a steam at
1000 kPa and 600K, and the other to a vacuum
pump.
• Both valves are closed initially.
• Then the valve to the steam line is opened, the tank
is evacuated, and the valve closed.
• Next to the valve to the steam line is opened so
that the steam flows in the evacuated tank very
slowly until the pressure in the tank equals the
pressure in the steam line.
• Calculate the final temperature of the steam in
the tank?
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10. Solution
1. Define the system type:
as open- unsteady state.
2. Mention your Assumption (related to energy balance depending on the system)
∆𝑬 = 𝑬 𝒕𝟏 − 𝑬 𝒕𝟐 = 𝑸 + 𝑾 − ∆ 𝑯 + 𝑲. 𝑬 + 𝑷. 𝑬
 No change occurs within the system for the P.E and K.E , ∆E= ∆U
 No work is done or by the system on the work thus W= 0
 NO heat is transferred in or out of the system, thus Q= 0
 The kinetic and potential energy of the stream entering or leaving are zero.
 NO stream exist the system, hence Hout = 0
 Initially no mass exist in the stream thus Ut,in = 0
So the general Equation reduces to
𝒎 𝒕𝟐 𝑼 𝒕𝟐 − 𝟎 = − 𝑯 𝒐𝒖𝒕 − 𝑯𝒊𝒏
3. Take the require properties data from steam tables.
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11. For open & unsteady state system
• Steady state means that all the state variables and the mass within the
system remains the same.
• Therefore ∆E = 0
• Now the general energy balance for open system becomes
𝑸 + 𝑾 = ∆ 𝑯 + 𝑲. 𝑬 + 𝑷. 𝑬
• In most of the cases K.E and P.E are negligible.
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12. Boiler to generate steam Evaporator that concentrates a solute Plate extraction Column.
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13. Energy Balance for a
Heat exchanger.
∆𝑯 = 𝟎
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14. Problem-2 (to learn when ∆H= 0 for a open steady system)
• Milk (essentially water) is heated from 15
⁰C to 25 ⁰C by hot water that goes from 70
⁰C to 35⁰C as shown in figure.
• What assumption can you make for
simplifying the problem?
• What is rate of water flow in kg/min per
kg/min of milk.
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15. Solution
1. Define the system type:
– milk + water is picked as system.
– open- steady state.
2. Mention your Assumption (related to energy balance depending on the system)
𝑸 + 𝑾 = ∆ 𝑯 + 𝑲. 𝑬 + 𝑷. 𝑬
 For steady system ∆E= 0
 No work is done or by the system on the work thus W= 0
 NO heat is transferred in or out of the system, thus Q= 0 (system is insulated )
 The kinetic and potential energy of the stream entering or leaving are zero.
So the general Equation reduces to
∆𝑯 = 𝟎
3. Take the require properties data from steam tables.
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16. Properties given.
4. Select the Basis for system.
– Basis = 1 min (and 1 kg of milk)
5. Calculate the desired value.
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