Gas Compression and Expansion. How can you calculate the energy used or made available when the volume of a gas is changed?

Transcription

1 Gas Compression and Expansion How can you calculate the energy used or made available when the volume of a gas is changed? Gas Compression and Expansion page: 1 of 16 Contents Initial Problem Statement 2 Narrative 3-8 Solutions 9-14 Appendices 15-16

2 Gas Compression and Expansion Initial Problem Statement Many machines compress or expand a gas or fluid as part of their working design Examples include a simple bicycle pump, a refrigerator and an internal combustion engine To compress a gas you need to expend energy to reduce its volume When a gas is allowed to expand energy is released How can you calculate the energy used or made available when the volume of a gas is changed? Gas Compression and Expansion page: 2 of 16

3 Narrative 1 Isothermal change An isothermal change of volume of a gas is one where the temperature of the gas remains constant To achieve this, the change in volume must be slow and there must be good thermal contact with the surroundings In this case, for an ideal gas, the pressure, p, and volume,, are related through the equation p = k, This is Boyle s law where k is a constant The energy transfer, w, involved in changing the volume is given by w= p d Hint Activity 1 k The expression p = k can be re-written as p = Substitute this into w p = d and find the integral This is an indefinite integral; make sure you have a constant of integration Discussion 1 Hint How do you determine the constant of integration? Activity 2 Don t forget the minus sign One cylinder of an engine has a volume available to an air/fuel mix of 500 cm 3 when the piston is in its lowest position Convert this volume to m 3 olume available: 500 cm 3 Gas Compression and Expansion page: 3 of 16 Figure 1

4 The initial pressure at this volume is 101 kilopascals Find the constant, k, in p = k and the constant of integration in your expression for the energy transfer when the volume is changed Hint Don t forget the minus sign Discussion 2 What are the units of k? Activity 3 The gas is isothermally compressed to a volume of 50 cm 3 Figure 2 Find the energy transfer in joules Give your answer to 1 dp Hint Don t forget to convert the volume to m 3 Multimedia Compressed volume: 50 cm 3 Hint Don t forget the minus sign The activity Gas Compression and Expansion Interactive is available to demonstrate the change of volume that occurs as a pistol moves in a cylinder Gas Compression and Expansion page: 4 of 16

5 2 Isothermal change, definite integral The previous section used an indefinite integral to evaluate the energy change when a gas is compressed or expanded When changes in values are considered in problems involving integration, eg the change in energy when there is a change in volume, the result can be obtained by using limits to give a definite integral This avoids having to determine the constant of integration For the previous example where a volume of m 3 is compressed to a volume of m 3, the change in energy is written as a definite integral where k = 505 w = k 510 Discussion 3 1 d Since the variable in the integral is, the limits, and are values of Which of the limits, or is the larger? What does this tell you about what is happening? Activity 4 Find w using the definite integral above Give your answer to 1 dp A graph of how the pressure, p, varies with the volume,, in the isothermal system is shown below Pressure (Pa) 600E E E E E E+06 Final volume Initial volume 000E+00 00E+00 50E-05 10E-04 15E-04 20E-04 25E-04 30E-04 35E-04 40E-04 45E-04 50E-04 55E-04 60E-04 65E-04 Figure 3 olume (m 3 ) Discussion 4 What does the definite integral you have evaluated represent on the graph? Gas Compression and Expansion page: 5 of 16

6 Discussion 5 These calculations assume a slow change in volume so that the gas remains at the same temperature as its surroundings This is done by transferring energy between the gas and its surroundings Is this likely to be the case for an internal combustion engine? Gas Compression and Expansion page: 6 of 16

7 3 Adiabatic change This is a more challenging extension The previous sections have assumed a slow change in volume so that the gas remains at the same temperature as its surroundings This is done by transferring energy between the gas and its surroundings In many cases, however, this will not be the case For example, the compression of the air/fuel mixture in an internal combustion engine is very fast, typically taking only a few tens of milliseconds Under these circumstances energy does not have sufficient time to leave the system and instead heats the gas, changing its temperature In a diesel engine this effect is used to ignite the fuel by compression alone In the ideal case where no energy is lost the relationship between pressure and volumes is given for an ideal gas by p γ = k, where k is a constant (not equal to the constant used in the previous sections) The volume is raised to the power of γ which is a property of the gas being compressed For air γ may be taken as having a constant value of 14 The energy transfer involved in changing the volume, w, is still given by w= p d k The expression p γ = k can be re-written as p = γ, so that k w = γ d 1 = k d γ Activity 5 Find the above integral using γ = 14 Activity 6 One cylinder of an engine has a volume available to an air/fuel mix of 500 cm 3 when the piston is in its lowest position Convert this volume to m 3 The initial pressure at this volume is 101 kilopascals Find the constant, k, in p γ = k where γ = 14, and the constant of integration in your expression for the energy transfer when the volume is changed Hint Don t forget the minus sign Gas Compression and Expansion page: 7 of 16 Activity 7 The gas is adiabatically compressed to a volume of 50 cm 3 Find the energy transfer in joules Give your answer to 1 dp

8 Hint Don t forget to convert the volume to m 3 Activity 8 Introduce limits to the integral w= k 1 d and find the energy transfer using a γ definite integral Discussion 6 Do you think it is possible to compress a gas without a loss of energy? Gas Compression and Expansion page: 8 of 16

9 Solutions 1 Isothermal change Activity 1 solution For an isothermal system pressure and volume are related through p = k k where k is a constant This implies that p = The energy transfer involved in changing the volume, w, is given by w= p d Substituting for p, Finding the integral w = k d 1 = k d w= k 1 d = kln + c where c is a constant of integration Discussion 1 solution As k is a constant As 1 x dx = ln x + c To find the constant of integration you need to have a known value of w for a known value of For a change of volume of gas the most obvious volume to use is the starting volume As you haven t changed the volume of the gas yet the energy exchanged when the gas has this volume is zero (ie you haven t done anything to the gas yet!) Activity 2 solution For the specific case where the initial volume is 500 cm 3 you have an initial volume in m 3 of = m 3 The initial pressure is 101 kilopascals = pascals Using p = k, k = = 50 5 You have stated above that when is equal to the initial volume, the energy transfer, w, is zero, so that w= kln + c 0= 505 ln c Rearranging to make c the subject, c = 50 5ln 5 10 = Gas Compression and Expansion page: 9 of 16

10 ie k has the unit of energy, joules Discussion 2 solution The units of pressure are pascals = [N][m] -2 and the units of volume are [m] 3 The units of k in the expression p= k are therefore given by 2 3 [ ][ ] [ ] = [ ][ ]= N m m N m [] J Activity 3 solution Compressing the gas isothermally to a volume of 50 cm -3 = m 3 requires a transfer of energy equal to w= kln + c = 50 5ln = ie, the energy transfer is 1163 J (1 dp) Gas Compression and Expansion page: 10 of 16

11 2 Isothermal change, definite integral Evaluating, Discussion 3 solution The limits of the integration are or , of which, is the larger Notice how the lower limit gives the value of the variable at the starting point of the problem while the upper limit gives its value and the end-point of the problem As the lower limit is numerically larger than the upper limit, this tells you that the integral is evaluating an effect due to a reduction in volume Activity 4 solution The energy transfer is given by the definite integral w = d The energy transfer is 1163 J (1 dp) w= 50 5[ ln] = 50 5 ln5 10 ln 5 10 = ( ) Notice how using limits avoids having to calculate a constant of integration Pressure (Pa) 600E E E E E+06 Discussion 4 solution The absolute value of the result of the integration gives the area on the p- curve between the initial volume and the final volume 100E+06 Final volume 000E+00 00E+00 50E-05 10E-04 15E-04 20E-04 25E-04 30E-04 35E-04 40E-04 45E-04 50E-04 55E-04 60E-04 65E-04 Figure 4 Initial volume olume (m 3 ) Gas Compression and Expansion page: 11 of 16

12 Discussion 5 solution These calculations assume a slow change in volume so that the gas remains at the same temperature as its surroundings This is done by transferring energy between the gas and its surroundings In many practical cases, however, this will not be the case For example, the compression of the air/fuel mixture in an internal combustion engine is very fast, typically taking only a few tens of milliseconds Under these circumstances energy does not have sufficient time to leave the system and instead heats the gas, changing its temperature In a diesel engine this effect is used to ignite the fuel by compression alone Under these circumstances a different calculation must be made This is the topic of the next section Gas Compression and Expansion page: 12 of 16

13 3 Adiabatic change Activity 5 solution The energy transfer is given by the integral Re-write this in the form w= k 1 d γ w= k γ d and use the standard rules for integration (recall γ = 14) w= k γ d = k 14 d k = + c k = + c 04 Activity 6 solution n+ 1 n x x dx= + c n + 1 For the specific case where the initial volume is 500 cm 3 you have an initial volume in m 3 of = m 3 The initial pressure is 101 kilopascals = pascals Using p γ = k, where γ = 14, ( ) k = = You have stated above that when is equal to the initial volume, the energy transfer, w, is zero, so that k 0 = + c 04 Making c the subject and evaluating k c = ( ) = exactly = ( ) Gas Compression and Expansion page: 13 of 16

14 Activity 7 solution Compressing the gas adiabatically to a volume of 50 cm -3 = m 3 requires a transfer of energy equal to w = k + c 04 ( ) = 04 = ie, the energy transfer is 1909 J (1 dp) Activity 8 solution The calculation can be carried out using a definite integral The initial volume is m 3 and the final volume is m 3 This gives the energy transfer as the following definite integral w= k γ d = ie, the energy transfer is 1909 J (1 dp) 14 d = = ( 5 10 ) = Discussion 6 solution 04 ( ( ) ) U n+1 n x x dx= n + L 1 In real systems it is not possible for the change in volume to be adiabatic; some energy will be lost In these cases the change is said to be polytropic and the relationship between pressure and volume is given by where 1 < n < 14 p n = k, U L Gas Compression and Expansion page: 14 of 16

15 Appendix 1 using the interactives Gas Compression and Expansion Interactive The activity Gas Compression and Expansion Interactive is available to demonstrate the change of volume that occurs as a pistol moves in a cylinder Figure 5 The speed of the animation is controlled using the slider at the bottom of the screen In particular, if you stop the motion when the piston as at the very bottom of its motion and compare the volume available with that which available when the piston is at the very top of its motion, you will see that there is a volume ratio of 10:1 This is the compression ratio of the engine Figure 6 Figure 7 Gas Compression and Expansion page: 15 of 16

16 Appendix 2 mathematical coverage Use calculus to solve engineering problems Be able to interpret area Distinguish between definite and indefinite integrals and interpret a definite integral as an area Gas Compression and Expansion page: 16 of 16