Lecture notes, H81ETD Booklet Section 2: Non -Flow Processes PDF

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STUDENT NAME: ________________________________ UNIVERSITY OF NOTTINGHAM DEPARTMENT OF CHEMICAL AND ENVIRONMENTAL ENGINEERING

MODULE TITLE:

H81ETD ENGINEERING THERMODYNAMICS: SECTION 2: Non-Flow Processes

MODULE CONVENOR: Dr. Vernon Collis Room: Coates A14

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 SECTION 2 11. Indicator diagrams: p / Pa

V / m3

Plots of TWO properties of State, such as p-V or T-V, provide useful graphical information on the processes that Systems undergo. These plots are called Indicator Diagrams. Specific variables:

p / Pa

V / m3 kg-1

PG. 2

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 12. Isothermal, isobaric, isometric, adiabatic and polytropic processes p / Pa Isobaric: constant pressure

Isothermal: constant temperature

Adiabatic: Zero Heat transfer Isochoric: constant volume V /m3 kg-1

Constant Volume (Isochoric or Isometric): Constant pressure (Isobaric): Constant Temperature (Isothermal): Zero Heat Flow (Adiabatic / Isocaloric): Isentropic index =



where  = cP / cV Multi-process (Polytropic): Polytropic index = n when: n = 0: isobaric process n = 1: isothermal process n = adiabatic process n = ∞: isochoric process

PG. 3

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Constant Volume (isochoric / isometric) p / Pa

Isochoric: constant volume

V /m3 kg-1 Ideal Gas Equation (IGE): p / T = constant Non-Flow Energy Equation (NFEE):

|rev

q12 = u2 – u1 Internal Energy (u) as a function cV: u = cV(T2 –T1)

|rev

Reversible Work Done:

∫

|rev

w12 = 0

PG. 4

|rev

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Constant pressure (isobaric) p / Pa

Isobaric: constant pressure

V /m3 kg-1 IGE: V / T = constant NFEE: q12 – w12 = u q12 = u + w12 q12 = (u2 – u1) + (p2V2 - p1V1) but: p = p1 = p2 q12 = (u2 +pV2) – (u1 + pV1) q12 = h2 – h1

Enthalpy (h) as a function of cP: h= cP(T2 – T1)

PG. 5

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Isobaric Work Done: p / Pa

Isobaric: constant pressure p

V1

V2

∫

|rev

w12 = Area under the pV curve w12 = p(V2 – V1) but pV =RgT w12 = Rg(T2 – T1)

PG. 6

V /m3 kg-1

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Constant Temperature (isothermal)

p / Pa

Isothermal: constant temperature

V /m3 kg-1

IGE: pV = constant

NFEE: u2 = u1 i.e. T1 = T2 q12 = w12

PG. 7

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012

Isothermal Work Done: p / Pa

Isothermal: constant temperature

V /m3 kg-1

∫

|rev ∫

w12 = RgT.ln(V2/V1) or w12 = RgT.ln(p1/p2)

PG. 8

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Adiabatic (isocaloric) (Prior knowledge: Maths revision for fractional indices) p / Pa

Adiabatic: Zero Heat transfer

V /m3 kg-1

IGE:   p1V1 = p2V2



- 1)

/(

- 1)

(V2/V1) = (T1/T2)1/( (p2/p1) = (T2/T1)

where “”, known as the isentropic index, is the ratio of principle specific heats:

NFEE: w12 = - (u2 – u1) w12 = - cV(T2 – T1)

PG. 9

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Adiabatic Work Done: p / Pa

Adiabatic: Zero Heat transfer

V /m3 kg-1

∫

|rev 

pV = c ∫ [

]

let c = p2V2 = p1V1

w12 =

w12 =Rg

PG. 10

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Polytropic process

p / Pa

Polytropic process

V /m3 kg-1 IGE:

n n p1V1 = p2V2

(V2/V1) = (T1/T2)1/(n

- 1)

(p2/p1) = (T2/T1) n/(n

- 1)

Polytropic index = n n = 0: isobaric process n = 1: isothermal process n = adiabatic process n = ∞: isochoric process NFEE: q12 = (u2 –u1) +

|rev

q12 = cV(T2-T1) + Rg

|rev

PG. 11

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012

Polytropic Work Done:

p / Pa

Polytropic process

V /m3 kg-1

∫

|rev

|rev

w12 =

|rev

w12 =Rg

PG. 12

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012

Polytropic processes: Example calculations: 1. A mass of 1.0 kg of helium initially at 20°C undergoes a reversible non-flow polytropic process of index 1.3 so that its volume is doubled. Find the change of internal energy, the work done and the heat transferred during the process. For helium, cP = 5.193 kJ kg-1 K-1 and the relative molecular mass is 4.003. Ans: U = -171.4 kJ, W12 = 380.8 kJ, Q12 = 209.4 kJ.

PG. 13

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 2. Oxygen, initially at 1 bar and 300 K, is compressed in a reversible non-flow adiabatic (isentropic) process to 1.5 bar and then expanded in a reversible non-flow polytropic process of index 1.26 so that the temperature returns to 300 K. Assuming the oxygen to be a perfect gas with a constant cP = 0.918 kJ kg-1 K-1, determine the final pressure, the overall volume ratio and the overall specific work done. Take the relative molecular mass of oxygen as 32. Sketch the complete process on a p-V diagram. Ans: p = 0.86 bar (8.6 X 104 N m-2), V3:V1 = 1.163, w123 = 12.46 kJ kg-1

PG. 14

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 3. A perfect gas initially at 10 bar, 298 K occupies a volume of 0.1 m3. It undergoes a cycle consisting of the following reversible non-flow processes: a) polytropic expansion, n = 1.3, while the specific enthalpy (h) is reduced by 177 kJ kg-1 (b) constant volume heating to 10 bar with heat transfer per unit mass of 570 kJ kg-1 (c) constant pressure compression to the initial state with the work done on the gas being 50 kJ. Find the value of cP and Rg for the gas, the work done during the polytropic expansion, the net work done and the heat transfer during the cycle. Ans: cP = 5.191 kJ kg-1 K-1, Rg = 2.078 kJ kg-1 K-1, W12 = 38.13 kJ, Wcycle -11.9 kJ, Qcycle = -11.9 kJ

PG. 15

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 13. Heat engines and efficiencies Reversible Cycles: p / Pa

Q H

Q C

V /m3 kg-1

∮

∮

|rev

W = QH - QC

PG. 16

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Carnot engine:

p / Pa WBY

S1

Q H

S2 TH

Q

Q=0

Q=0 S4

C

TC S3 WON V /m3 kg-1

S1 – S2: Isothermal expansion process S2 – S3: Adiabatic expansion process S3 – S4: Isothermal compression process S4 – S1: Adiabatic compression process

ALL processes are REVERSIBLE

TH

PG. 17

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Carnot engine

TH QH WOUT QC TC Conservation of Energy WOUT = QH - QC Carnot efficiency CARNOT = Wout / QH CARNOT = (QH - QC) / QH CARNOT = 1 – (QC / QH) this leads to the maximum efficiency of any engine:

CARNOT = 1 – (TC / TH)

Can you prove this? Irreversible processes: 

Joules experiment



Gas expanding into a vacuum



Hot and Cold Copper blocks

PG. 18

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 14. 2nd Law of Thermodynamics 1. Carnot stated that: “you cannot build an engine whose sole effect is to transfer HEAT from a cold reservoir to a hot reservoir”

TH

NOT ALLOWED

Q

TC 2. In 1850 Clausius clarified Carnot’s law by stating that: “it is impossible to construct a system operating in a cycle that transfers heat from a cooler body to a hotter body without work being done on the system by the surroundings.”

3. Another version of the 2nd law was formulated by Plank in 1927, where he states that: “it is impossible to construct a system operating in a cycle extracts heat from a reservoir and does an equivalent amount of work on the surroundings”

TH

QH

NOT ALLOWED WOUT = QH

PG. 19

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 15. Introduction to Entropy: We now introduce a new quantity called: ENTROPY (S / J K-1)

or

SPECIFIC ENTROPY (s / J kg-1 K-1)

This was a mathematically discovered quantity. Clausius Inequality: Recall Carnot efficiency: CARNOT = 1 – (QC / QH) and CARNOT = 1 – (TC / TH) this implies that: QC / QH = TC / TH rearranging gives QC / TC = QH / TH if we imagine summing this quantity around the Carnot cycle: (QH / TH) + 0 – (QC / TC) + 0 =0

∑ =0

(perfectly reversible cycle)

in integral form we write:

∮

|reversible

This is known as the Clausius statement for reversible cycles.

PG. 20

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Irreversible cycles: Given that the maximum efficiency of any heat engine is:

CARNOT = 1 – (QC / QH)

it follows that for an irreversible engine its efficiency will less than the reversible Carnot efficiency:

IRREVERSIBLE < CARNOT This implies that either less heat is flowing into the system (QH decreases) and/or more heat is flowing out of the system (QC increases) Recalling that (QH / TH) – (QC / TC) = 0 (Carnot cycle) then (QH / TH) – (QC / TC) < 0 (Irreversible cycle) therefore we may write:

∮

|irreversible

This is known as the Clausius statement for irreversible cycles

PG. 21

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012

Combining the two Clausius statements for reversible and irreversible cycles:

∮ i.e. the Clausius Inequality where Q depends on the path followed. We now use this expression to define a change in entropy (S) for the system:

S = S2 – S1 =

∫

|reversible

in differential form:

dS = dQ / T

where S/ J K-1 is found to be a property of the system

i.e. a “STATE VARIABLE” OR “PROPERTY”

PG. 22

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Principle of increase in Entropy Consider a cycle made up from TWO processes: process 1-2 is irreversible

1

2

process 2-1 is reversible

Recalling Clausius Inequality:

∮ we can express this cycle as: ∫

∫

where ∫ therefore ∫ giving ∫ showing that the entropy change ( S) is always greater than or equal to the entropy transfer ( Q/T) in any closed system PG. 23

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Order to disorder The 2nd Law can now be summarised by the following statement:

i.e. the change in entropy of the universe (system + surrounds) always increases. Definition of Entropy: General: We define the quantity entropy (S / J K-1) as: the ratio of HEAT transferred to the absolute temperature S=Q/T

|rev

|rev

dS = Q / T

Q = TdS

Specific entropy (s / J kg-1 K-1): ds = q / T

|rev

q = Tds Entropy is a State Variable.

PG. 24

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Let’s think about Carnot’s Law: Condition A

Condition B

TH

TH

Q

Q TC

TC

ALLOWED

NOT ALLOWED

Condition A CHANGE IN ENTROPY = HEATLOST + HEATGAINED TH TC

now TH > TC S > 0 Entropy has increased Condition

ALLOWED

Condition B CHANGE IN ENTROPY = HEATLOST + HEATGAINED TC TH

S < 0 Entropy has decreased Condition

NOT ALLOWED PG. 25

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Consider this: An Ideal Gas is contained within a thermally insulated rigid box. The box contains TWO compartments of equal volume. Initially one compartment contains gas at p=p1 and V = V1. The partition is now removed and the gas expands to fill the whole container.

p = p1 vacuum

State 1

V = V1

p = p1/2 V = 2V1

State 2

Questions: 1. If the initial temperature is T1, what is the final temperature of the gas? 2. Explain, in terms of entropy considerations, why we can move from State 1 to State 2, but we cannot move from State 2 to State 1.

PG. 26

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 Heat Engines: Example calculations 1. A heat engine cylinder contains 0.1 kg of air. The engine is assumed to operate on a cycle which consists of the following four reversible non-flow processes: (i) isothermal expansion at 1000 K during which the volume of air is doubled; (ii) adiabatic expansion until the temperature reaches 350 K and pressure 1 bar; (iii) isothermal compression of the air at 350 K; (iv) adiabatic compression until the air temperature reaches 1000 K and the cycle is completed. (a) Sketch the cycle on a pV diagram. (b) What is the maximum cylinder pressure? (c) Calculate the heat transferred and the work done during isothermal expansion. (d) How much work is done during adiabatic expansion and during adiabatic compression? (e) Show that during isothermal compression the volume of air is halved. (f) Determine the net heat transferred and the net work done per cycle. Treat the air as a perfect gas, taking Rg = 0.287 kJ kg-1 K-1, cV = 0.7175 kJ kg-1 K-1 and  = 1.4 Ans: (b) p = 78.84 bar (78.84 X 105 N m-2) (c) Q12 = +19.89 kJ; W12 = +19.89 kJ (d) W23 = 46.64 kJ, W41 = -46.64 kJ (f) Qcycle = + 12.93 kJ, Wcycle = +12.93 kJ

PG. 27

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 1/cont

PG. 28

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 2. A heat engine operates on a cycle consisting of the following reversible non-flow processes, assumed to take place in the engine cylinder. (a) Air, initially at 1 bar, 300 K is compressed adiabatically until the volume is reduced to 1/9th of the initial volume. (b) The air is then heated at constant volume. the heat supplied per unit mass of air being 1350 kJ kg-1 . (c) The air then expands adiabatically to its initial volume. (d) Finally, the air is cooled at constant volume to its initial state. Find the maximum cylinder temperature and pressure, the specific work done during compression and during expansion, and the net specific work done per cycle. How much heat per unit mass is rejected per cycle? Take  = 1.4 and cV =0.7175 kJ kg-1 K-1 Ans: T max = 2604 K, P max = 78.1 bar (78.1 X 105 Pa), Work done during compression = - 303 kJ kg-1, Work done during expansion = + 1092.5 kJ kg -1. Net specific work = - 789 kJ kg -1 . Heat rejected per cycle = - 561 kJ kg -1

PG. 29

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012 2/cont

PG. 30

H81ETD BOOKLET 2 Non Flow Processess Ver1 2012

Summary of ETD Section 2

11. Indicator diagrams: 12. Isothermal, isobaric, isometric, adiabatic and polytropic processes 13. Heat engines and efficiencies 14. 2nd Law of Thermodynamics 15. Entropy:

_________________________END OF SECTION 2__________________________

PG. 31...