energy balance
DESCRIPTION
energy balance note.TRANSCRIPT
Energy Balances on Non-Reactive Process
Manolito E. Bambase JrAssistant Professor, Department of Chemical Engineering
CEAT, University of the Philippines, Los Banos, Laguna, Philippines
15
hg
15.1 Energy Balance on a Closed System
1
Closed SystemE = K + P + U
Q
W
DE = Q – W
DK + DP + DU = Q – W
D(mv2/2) + D(mgh) + DU = Q – W
m/2Dv2 + mgDh + DU = Q – W
15.2 Energy Balance on an Open System
2
Q W
Mass In
E1
Mass Out
E2
DE = Q – W
E2 – E1 = Q – W
If there are multiple inlets and outlets,
SE2 – SE1 = Q – W
The work appearing in the equation is the combined flow work and shaft work:
W = WF + WS
WF = flow work; work that is necessary to get mass into and out of the system
WS = shaft work; work produced or required beside getting mass into and out of the system.
Hence,
SE2 – SE1 = Q – (WF + WS)
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15.2 Energy Balance on an Open System
The net flow work is determined as
WF = (WF)2 – (WF)1
The flow work is usually expressed in terms of pressure and volume:
WF = (PV)2 – (PV)1
For multiple inlets and outlets,
WF = S(PV)2 – S(PV)1
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15.2 Energy Balance on an Open System
The energy balance becomes
SE2 – SE1 = Q – [(S(PV)2 – S(PV)1) + WS]
Since E = K + P + U, then
S(K + P + U)2 – S(K + P + U)1 = Q – [(S(PV)2 – S(PV)1) + WS]
Rearranging the terms,
S(K + P + U + PV)2 – S(K + P + U + PV)1 = Q – WS
S(K + P + H)2 – S(K + P + H)1 = Q – WS
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15.2 Energy Balance on an Open System
A cylinder with a movable piston contains 4.00 liters of a gas at
300C and 5.00 bar. The piston is slowly moved to compress the
gas to 8.00 bar.
(a) If the compression is carried out isothermally, and the work
done on the gas equals 7.65 L-bar, how much heat (in joules)
is transferred to or from (state which) the surroundings.
(b) Suppose instead that the process is adiabatic, what will
happen to the temperature of the gas.
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Example 15-1. Compression of an Ideal Gas in a Cylinder
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(a) Isothermal Compression
4.00 L
300C
5.00 bar
V2
300C
8.00 bar
1 2
Isothermal Compression
W = – 7.65 L-bar
This is a closed system and the energy balance is
DK + DP + DU = Q – W
Since the system is stationary, DK = DP = 0Since the process is isothermal, DT = 0 and DU = 0
Example 15-1. Compression of an Ideal Gas in a Cylinder
The energy balance is simplified to
0 = Q – W
Solving for Q:
8
8.314JQ W 7.65L bar 765J
0.08314L bar
(b) Adiabatic Compression
The energy balance reduces to DU = – W = – (–765 J) = 765 J
Since DU is positive, DT must also be positive. Hence, the temperature of the gas will increase.
Example 15-1. Compression of an Ideal Gas in a Cylinder
Example 15-2. Nitrogen Gas in a Piston-Fitted Cylinder
A piston-fitted cylinder with a 6-cm inner diameter contains 1.40
g of nitrogen. The mass of the piston is 4.50 kg, and a 20-kg
weigh rests on the piston. The gas temperature is 300C and the
pressure outside the cylinder is 1.00 atm.
(a) Calculate the pressure and volume of the gas inside the
cylinder if the piston-weight is at equilibrium.
(b) Suppose the 20-kg weigh is abruptly lifted and the piston rises
to a new equilibrium position in which the gas returns to
300C. Calculate the work done by the gas.
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(a) Pressure of the gas inside the cylinder
If the piston-weight is at equilibrium, then
FU = FD
Calculate FD:
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D ext piston piston 20 kgF P A W W
Example 15-2. Nitrogen Gas in a Piston-Fitted Cylinder
The cross-sectional area of the piston is
( )2
22 3 2piston
1mA r 3cm 2.83 10 m
100cm
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The force due to external pressure:
Example 15-2. Nitrogen Gas in a Piston-Fitted Cylinder
( )2
3 2ext ext piston
ext
101325N / mF P A 1.00atm 2.83 10 m
1.00atm
F 287 N
The force due to piston and 20-kg mass:
FP = (4.50 kg + 20.00 kg)(9.81 m/s2) = 240 N
The total downward force is
FD = (240 + 287) N = 527 N = FU
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Pgas = FU/Apiston = 527 N/(2.87 × 10-3 m2) = 1.86 × 105 N/m2
Assuming the gas is ideal:
Example 15-2. Nitrogen Gas in a Piston-Fitted Cylinder
( )( )nRT mRT
V 0.677LP MW P
(b) Work done by the gas when 20-kg mass is removed
Upon removal of the 20-kg mass, the downward force is reduced to:
FD = 287 N + (4.50 kg)(9.81 m/s2) = 331 N
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If the piston must attain a new equilibrium position, then
FD = FU = 331 N
The final pressure of the gas:
P2 = FU/Apiston = 331 N/(2.87 × 10-3 m2) = 1.16 × 105 N/m2
And the new volume of the gas is:
Example 15-2. Nitrogen Gas in a Piston-Fitted Cylinder
51
2 1 52
P 1.86 10V V 0.677L 1.08L
P 1.16 10
Change in volume: DV = V2 – V1 = (1.08 – 0.677)L = 0.403 L
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The change in volume can also be determined as:
DV = ApistonDx
Therefore, the displacement Dx can be computed as:
Dx = DV/Apiston = 0.142 m
Computing for work:
W = FDx = (331 N)(0.142 m) = +47 J
Since DK = DP = DU =0, this work must be accompanied by heat transfer to the gas equal to +47 J.
Example 15-2. Nitrogen Gas in a Piston-Fitted Cylinder
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Methane enters a 3-cm ID pipe at 300C and 10 bar with an
average velocity of 5.00 m/s and emerges at a point 200 m lower
than the inlet at 300C and 9 bar.
Calculate the DK and DP assuming the methane behaves as an
ideal gas.
Solution:
Mass flow must be the same at the inlet to attain steady-state
condition.
DK = m/2(v22 – v1
2) and DP = mg(h2 – h1)
Example 15-3. Methane Flowing in a Pipe
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Solving for mass flow:
Example 15-3. Methane Flowing in a Pipe
( )( )1
mRTV v A
MW P
Determine the mass flow:
Volumetric flow at the inlet = v1A
If methane behaves as an ideal gas:
( )( )( )1 1
1
v A MW Pm 0.0225kg / s
RT
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Solving for DP:
Example 15-3. Methane Flowing in a Pipe
( ) ( )2 1 2
kg mP mg h h 0.0225 9.81 200m
s s
JP 44.1 44W
s
D
D
Determine v2:
P1V1 = P2V2
P1(v1A) = P2(v2A)
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Solving for v2:
Example 15-3. Methane Flowing in a Pipe
12 1
2
P m 10barv v 5.00 5.555m / s
P s 9bar
Solving for DK:
( )2
2 2 2 22 1 2
m 1 kg mK v v 0.0225 5.555 5.00
2 2 s s
JK 0.0659 0.0659W
s
D
D
Example 15-4. Heating of Water in Boiler Tubes
A fuel oil is burned with air in a boiler furnace. The combustion
produces 813 kW of heat of which 65% is transferred as heat to
boiler that pass through the furnace. Water enters the boiler tubes
as a saturated liquid at 200C and leaves the tubes as saturated
steam at 20 bar absolute.
Calculate the mass flow rate (in kg/h) and volumetric flow rate
(in m3/h) at which the saturated steam is produced。
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Example 15-4. Heating of Water in Boiler Tubes
Boiler Tubes
Q = 0.65 (813 kW)
H2O (saturated liquid)
T = 200C
H2O (saturated steam)
Pabs = 20 bar
Using the energy balance for an open system,
S(K + P + H)2 – S(K + P + H)1 = Q – WS
Assuming DK = DP = W = 0, then
SH2 – SH1 = Q = 0.65 (813 kW) = 528 kW
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Example 15-4. Heating of Water in Boiler Tubes
In terms of specific enthalpy, Ĥ (kJ/kg)
m2Ĥ2 – m1Ĥ1 = 528 kW
where m = mass flow rate of water
If the process is under steady-state condition, then m1 = m2 = m
Hence,
m(Ĥ2 – Ĥ1) = 528 kW
Solving for m:
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2 1
528kWm
ˆ ˆH H
Example 15-4. Heating of Water in Boiler Tubes
From saturated steam table,
Ĥ1 = 83.9 kJ/kg and Ĥ2 = 2797.2 kJ/kg
And the mass flow rate is:
( )528kJ / s 3600s
m 701kg / h2797.2 83.9 kJ / kg 1h
Calculating for the volume flow rate:
3 3kg m mˆV mV 701 0.0995 69.7h kg h
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from steam table
Example 15-5. Mixing and Heating of Propane-Butane Mixtures
Three hundred L/h of 20 mole% C3H8-80% n-C4H10 gas mixture at 00C and 1.1 atm and 200 L/h of a 40 mole% C3H8-60% n-C4H10 gas mixture at 250C and 1.1 atm are mixed and heated to 2270C at constant pressure. Calculate the heat requirement of the process. Enthalpies of propane and n-butane are listed below. Assume ideal gas behaviour.
23
T (0C)
0
25
227
PropaneĤ (J/mol)
ButaneĤ (J/mol)
0
1772
20,685
0
2394
27,442
Example 15-5. Mixing and Heating of Propane-Butane Mixtures
Heating at
Constant
Pressure
n1 (mol/h)
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20% C3H8
80% C4H10
00C, 1.1 atm, 300 L/h
n2 (mol/h)
40% C3H8
60% C4H10
250C, 1.1 atm, 200 L/h
A (mol C3H8/h)
B (mol C4H10/h)
2270C
Mixture 3
Example 15-5. Mixing and Heating of Propane-Butane Mixtures
Simplified energy balance for open system:
Q = SHout – SHin
Q = (HP3 + HB3) – (HP2 + HB2 + HP1 + HB1)
Since mixture 1 is at 00C, then
Q = (HP3 + HB3) – (HP2 + HB2)
The total enthalpy of each component is determined as:
HP3 = AĤP3 HB3 = BĤB3
HP2 = 0.40n2ĤP2 HB2 = 0.60n2ĤB3
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Example 15-5. Mixing and Heating of Propane-Butane Mixtures
Find n1, n2, A, and B
n1 and n2 can be obtained from V1 and V2 using the ideal gas equation:
n1 = 14.7 mol/h ; n2 = 9.00 mol/h
A and B can be obtained using material balances for propane and butane:
Propane: A = 0.20n1 + 0.40n2 = 6.54 mol C3H8/h
Butane: B = 0.80n1 + 0.60n2 = 17.16 mol C4H10/h
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Example 15-5. Mixing and Heating of Propane-Butane Mixtures
Solving for the total enthalpies of the components:
HP3 = (6.54 mol/h)(20.865 kJ/mol) = 136.5 kJ/h
HB3 = (17.16 mol/h)(27.442 kJ/mol) = 470.9 kJ/h
HP2 = 0.40(9.00 mol/h)(1.772 kJ/mol) = 6.38 kJ/h
HB2 = 0.60(9.00 mol/h)(2.394 kJ/mol) = 12.93 kJ/h
Solving for the heat requirement of the process:
Q = (136.5 + 470.9 – 6.38 – 12.93) kJ/h
Q = 587 kJ/h
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