Energy balance (leacture 6)

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LECTURE 6
Energy Balance
5
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 FORMS OF ENERGY: THE FIRST LAW OF THERMODYNAMICS
1. Kinetic Energy
2. Potential Energy
3. Internal Energy
5

Suppose a process system is closed, meaning that no mass is transferred
across its boundaries while the process is taking place. Energy may be
transferred between such a system and its surroundings in two ways:
1. As heat, or energy that flows as a result of temperature difference between
a system and its surroundings. The direction of flow is always from a higher
temperature to a lower one. Heat is defined as positive when it is
transferred to the system from the surroundings.
2. As work, or energy that flows in response to any driving force other than a
temperature difference, such as force, a torque, or a voltage.
5

- The principle that underlies all energy balances is the law of conservation of
energy, which states that energy can neither be created nor destroyed. This
law is also called as ‘first law of thermodynamics’.
- In its most general form, the first law states that the rate at which energy
(K.E.+P.E.+I.E.) is carried into a system by the input streams, plus the rate at
which it enters as heat, minus the rate at which it leaves as work, equals the
rate of accumulation of energy in the system (accumulation = input – output)
5

 Kinetic & Potential Energy
𝐸𝑘 = 1/2𝑚𝑣2 𝐸𝑝 = 𝑚𝑔𝑧
1. Water flows into a process unit through a 2-cm ID pipe at a rate of 2.00
𝑚3/h. Calculate 𝐸𝑘 for this stream in joules/sec.
2. Crude oil is pumped at a rate of 15 kg/s from a point 220 m below the
earth’s surface to a point 20 m above ground level. Calculate the
attendant rate of increase of potential energy.
5

 ENERGY BALANCES ON CLOSED SYSTEM
- A system is termed as open or close according to whether or not mass
crosses the system boundary during the period of time covered by the
energy balance.
- A batch process is by definition, closed, and semi batch and continuous
are open.
𝒂𝒄𝒄𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒐𝒏 = 𝒊𝒏𝒑𝒖𝒕 − 𝒐𝒖𝒕𝒑𝒖𝒕
5

𝑓𝑖𝑛𝑎𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 energy − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 𝑒𝑛𝑒𝑟𝑔𝑦
= 𝑛𝑒𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑛 − 𝑜𝑢𝑡
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑈𝑖 + 𝐸𝑘𝑖 + 𝐸𝑝𝑖
𝑓𝑖𝑛𝑎𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑈𝑓 + 𝐸𝑘𝑓 + 𝐸𝑝𝑓
𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = 𝑄 − 𝑊
Now the equation becomes,
𝑈𝑓 −𝑈𝑖 + 𝐸𝑘𝑓 − 𝐸𝑘𝑖 + 𝐸𝑝𝑓 − 𝐸𝑝𝑖 = 𝑄 − 𝑊
Or
∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 − 𝑊
This equation is the basic form of the first law of thermodynamics for a closed
system.

When we apply this equation to a given process, we should know the following
point:
1. The internal energy of the system depends almost entirely on the chemical
composition, state and temperature of the system materials. It is independent
of pressure for ideal gases and nearly independent of pressure for liquids and
solids. If no temperature changes, phase changes or chemical reactions
occur in a closed system and if pressure changes are less than a few
atmospheres, then ∆𝑼 ≈ 𝟎.
2. If a system is not accelerating, then ∆𝑬𝒌 = 𝟎. If a system is not rising or
falling then ∆𝑬𝒑 = 𝟎.
3. If a system and its surroundings are at the same temperature or the
system is perfectly insulated then 𝑸 = 𝟎. The process is then termed as
adiabatic.

Solution:
1.
∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 − 𝑊
∆𝐸𝑘 = 0 (𝑠𝑦𝑠𝑡𝑒𝑚 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦)
∆𝐸𝑝 = 0 (𝑛𝑜 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡)
𝑊 = 0 (𝑛𝑜 𝑚𝑜𝑣𝑖𝑛𝑔 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑖𝑒𝑠)
∆𝑈 = 𝑄
∆𝑈 = 2 𝑘 𝑐𝑎𝑙
∆𝑈 = 2 𝑘 𝑐𝑎𝑙 1000𝑐𝑎𝑙
1 𝑘 𝑐𝑎𝑙
1 𝐽
0.23901 𝑐𝑎𝑙
= 8370 𝐽

2.
∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 − 𝑊
∆𝐸𝑘 = 0 (𝑠𝑦𝑠𝑡𝑒𝑚 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦)
∆𝐸𝑝 = 0 (𝑎𝑠𝑠𝑢𝑚𝑒𝑑 𝑛𝑒𝑔𝑙𝑖𝑔𝑖𝑏𝑙𝑒)
∆𝑈 = 0
0 = Q − W
𝑊 = +100 𝐽
𝑠𝑜, 𝑄 = 100 𝐽
Thus an additional 100 J of heat is transferred to the gas as it expands and re-
equilibrates at 100˚C.

 ENERGY BALANCES ON OPEN SYSTEMS AT STEADY STATE
- An open system by definition has mass crossing its boundaries as the process
occurs.
- Work must be done on such system to push mass in and work is done on the
surroundings by mass that emerges.
- Both work term must be included in the energy balance.
5

 Flow work and Shaft work
- Net rate of work done by an open system on its surroundings may be written as:
𝑊 = 𝑊𝑠 + 𝑊𝑓𝑙
where,
𝑊𝑠 = shaft work, or rate of work done by the process fluid on a moving part within
the system.
𝑊𝑓𝑙 = flow work, or rate of work done by the fluid at the system outlet minus the
rate of work done on the fluid at the system inlet.

To derive an expression for 𝑊𝑓𝑙, we initially consider single-inlet-single-outlet
system shown here,
𝑊𝑖𝑛 = 𝑃𝑖𝑛 𝑉𝑖𝑛
(𝑁. 𝑚/𝑠) = (𝑁/𝑚2) (𝑚3/𝑠)
𝑊𝑜𝑢𝑡 = 𝑃𝑜𝑢𝑡 𝑉𝑜𝑢𝑡
𝑊𝑓𝑙 = 𝑃𝑜𝑢𝑡 𝑉𝑜𝑢𝑡 − 𝑃𝑖𝑛 𝑉
𝑖𝑛
Process Unit
𝑽𝒊𝒏(𝒎𝟑/𝒔)
𝑷𝒊𝒏(𝑵/𝒎𝟐)
𝑽 𝒐𝒖𝒕(𝒎𝟑/𝒔)
𝑷𝒐𝒖𝒕(𝑵/𝒎𝟐)
5

 Specific Properties & Enthalpy
- A specific property is an intensive quantity obtained by dividing an extensive
property (or its flow rate) by the total amount (or flow rate) of the process
material.
- Example:
i) volume of fluid is 200𝑐𝑚3 and mass of the fluid is 200g, then specific
volume of the fluid is 1𝒄𝒎𝟑/g.
ii) If the rate at which kinetic energy is transported by a stream is 300 J/min,
having mass flow rate 100 kg/min, then the specific kinetic energy of the
stream material is 3J/kg.

- This property can be denoted by a capital letter with a cap, example: 𝑉 will
denote specific volume, 𝑈 will denote specific internal energy and so on.
- If the temperature and pressure of a process material are such that the specific
internal energy of the material is 𝑈 (J/kg), then mass (kg) of this material has a
total internal energy:
U(J) = m(kg) 𝑼(J/kg)
Similarly, 𝑼 (J/s) = 𝒎 (kg/s) 𝑼(J/kg)
- A property that occurs in the energy balance equation for open systems is
specific enthalpy, defined as:
𝑯 = 𝑼 + 𝑷
𝑽
5

THE STEADY-STATE OPEN-SYSTEM ENERGY BALANCE
∆𝐻+ ∆𝐸 𝑘 + ∆𝐸 𝑝 = 𝑄− 𝑊
𝑠
We will use this equation as the starting point for most energy balance
calculation on open systems at steady state.
5

Example 7.4-2
Five hundred kilograms per hour of steam drives a turbine. The steam enters
the turbine at 44 atm and 450℃ at a linear velocity of 60 m/s and leaves at a
point 5m below the turbine inlet at atmospheric pressure and a velocity of 360
m/s. The turbine delivers the shaft work at a rate of 70 kW, and the heat loss
from the turbine is estimated to be 104 kcal/h. Calculate the specific enthalpy
change associated with the process.
Solution
500 kg/h
44 atm, 450℃
60 m/s
500 kg/h
1 atm,
360 m/s
5 m
𝑸 = −𝟏𝟎𝟒 𝒌𝒄𝒂𝒍/𝒉 𝑾𝒔 = 𝟕𝟎𝒌𝑾

Solution:
∆𝐻 = 𝑄 − 𝑊𝑠 − ∆𝐸 𝑘 − ∆𝐸 𝑝 → (
𝐴)
For units consistency,
𝑚 = 500
𝑘𝑔
ℎ
3600 = 0.139
𝑠 𝑘𝑔
ℎ 𝑠
1
2
∆𝐸 𝑘= 𝑚 2 1
𝑣2 − 𝑣2 =
1 𝑘𝑔
2 𝑠
0.139 3602 − 602
𝑚2
𝑠2
= 8757 𝑁 − 𝑚/𝑠
∆𝐸𝑘= 8.75 𝑥 103 𝑊 𝑜𝑟 8.75 𝑘𝑊
5

∆𝐸 = 𝑚 𝑔 𝑧 − 𝑧
𝑝 2 1
𝑘𝑔
= 0.139
9.81𝑚
𝑠2
0 − 5𝑚 = −6.82
N. m
s
∆𝐸𝑝 =
𝑠
𝑠
𝐽 1 𝑘𝑊
−6.82
1000 𝐽
= −6.82 𝑥 10−3 𝑘𝑊
𝑄 = −104 𝑘 𝑐𝑎𝑙 1 𝐽
0.239 𝑥 10−3 𝑘 𝑐𝑎𝑙
ℎ
𝑠
1 ℎ
3600 𝑠
1𝑘𝑊
1000 𝐽/𝑠
= −11.6𝑘𝑊
𝑊𝑠 = 70 𝑘𝑊
𝑁𝑜𝑤 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐴 ,
∆𝐻 = −11.6 𝑘𝑊 − 70 𝑘𝑊 − 8.75 𝑘𝑊 + 6.82 𝑥 10−3 𝑘𝑊
∆𝐻 = −90.3 kW

The question says we have to calculate specific enthalpy,
So,
∆𝐻 = 𝑚 𝐻2 − 𝐻1
𝑜𝑟
𝑚
∆𝐻
= ∆
𝐻
∆
𝐻 =
−90.3 𝑘𝑐𝑎𝑙
= −650
𝑘𝐽
0.139 𝑘𝑔/𝑠 𝑘𝑔
5

Example 7.6-1
Two streams of water are mixed to form the feed to a boiler. Process data are as
follows:
𝐹𝑒𝑒𝑑 𝑠𝑡𝑟𝑒𝑎𝑚 1
𝐹𝑒𝑒𝑑 𝑠𝑡𝑟𝑒𝑎𝑚 2
𝐵𝑜𝑖𝑙𝑒𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
120 𝑘𝑔/ min @ 30℃
175 𝑘𝑔/min @ 65℃
17 𝑏𝑎𝑟 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒
The exiting steam emerges from the boiler through a 6 cm ID pipe. Calculate the
required heat input to the boiler in kilojoules per minute if the emerging steam is
saturated at the boiler pressure. Neglect the kinetic energies of the liquid inlet
streams.
5

Energy balance (leacture 6)

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