energy balance

Energy Balances on Non-Reactive Process

Manolito E. Bambase JrAssistant Professor, Department of Chemical Engineering

CEAT, University of the Philippines, Los Banos, Laguna, Philippines

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hg

15.1 Energy Balance on a Closed System

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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

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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:

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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

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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.

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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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