Assignment 2 Solution

RYERSON UNIVERSITY Department of Aerospace Engineering AER  – ermodynamics Dr. J. V. Lassaline Assignment  Due: Oct. th,  Completed assignments can be le in the drop box opposite ENG .

Q 

A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of the following three processes in sequence: Process 1 – 2: Compression with U2 = U1 . Process 2 – 3: Constant-volume cooling to p3 = 140 kPa, V3 = 0.028 m3 . Process 3 – 1: Constant-pressure expansion with W31 = 10.5 kJ.

For the cycle, the net work is −8.3 kJ. ere is no change in kinetic or potential energy. .a)

Determine the volume at state 1, in m3 .

A

Use the information given about process 3–1 to determine the volume at state 1, given that state 3 is known.

W31 = p31 (V1 − V3 ) W31 10.5 kJ V1 = V3 + = 0.028 m3 + = 0.103 m3 p31 140 kPa

() ()

Note that the units of kJ/kPa yield m3 without further conversion. Also note that the order of the volume difference is important as V1 > V3 . In the general case for a constant pressure process the expansion or compression work is Wi j = p i j (Vj − Vi ).

.b)

Determine the work and heat transfer for process 1–2, each in kJ. A You are not told much about process 1–2, other than that the internal energy remains the same during the process. We do know that it must join the remaining two processes to form a thermodynamic cycle. An energy balance on process 1–2 yields

∆E12 = 0 = Q12 − W12 Q12 = W12

() ()

while the work done during process 2–3 is zero, the work done during process 3–1 is known, and the net cycle work is given. Calculating the net work of the cycle yields

W12 = Wc ycl e − W23 − W31 = −8.3 kJ − 0 − 10.5 kJ = −18.8 kJ

()

where the sign is consistent with the description of process 1–2 being a compression of the gas. Using this work with Eq.  yields Q12 = −18.8 kJ or heat rejection.

.c)

Is this a power or refrigeration cycle? Explain.

A

As the net cycle work is negative (work input) this must be a refrigeration (or heat pump) cycle. This can also be determined by sketching a p–V diagram and observing that the cycle is completed in the counter-clockwise direction.

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Q 

A cooking pot whose inner diameter is 20 cm is filled with water and covered with a heavy lid (4 kg) which provides a good seal at the pot rim. If the local atmospheric pressure is 101 kPa, determine the temperature at which the water will start boiling when heated.

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A

Consider a free-body diagram applied to the pot lid. The pressure of the vapour inside the pot can reach a maximum pressure p before lifting the lid. Thus the temperature at which the pot boils is Tsat for water at pressure p. Balancing the forces in the vertical direction, we have the weight of the lid acting down, atmospheric pressure acting over the top surface of the lid, and pressure p acting over the bottom surface of the lid. The vertical component of the force due to the constant pressure acting on either side is equal to the pressure multiplied by the surface area projected onto the horizontal plane. (The horizontal component of the force due to pressure will likely cancel due to symmetry of the lid, but it is irrelevant to this problem. A domed lid will behave the same as a flat lid with the same diameter.) For static equilibrium, summing the forces in the vertical direction (positive up) requires

π π ∑ F = −mg − p atm D2 + p D2 = 0 4 4

()

where D = 0.20 m is the inner diameter of the pot, p atm = 101 kPa is the atmospheric pressure, and p is the pressure of the water within the pot. Thus the pressure p is

p = p atm +

4 ⋅ 4 kg ⋅ 9.81 m/s2 4mg 3 = 101 × 10 Pa + = 102.2 kPa πD 2 π0.202 m2

()

which corresponds to Tsat = 100.2 °C using linear interpolation within Table A- of the textbook.

Q 

A rigid tank contains water vapour at 300 °C and an unknown pressure. When the tank is cooled to 180 °C, the vapour starts condensing. Estimate the initial pressure in the tank.

A

The word rigid implies that this is a constant volume process. The word cooling implies that there is heat rejection from the water to its surroundings. There is no other mention of work and it can be assumed that kinetic and potential energy are negligible. The initial state is at pressure p1 and temperature T1 = 300 °C. The final state is at T2 = 180 °C and is entirely saturated vapour. Thus the specific volume at 3 the final state is v2 = v g (180 °C) = 0.1941 m /kg as determined using Table A- of the textbook. Thus for a 3 constant volume process, the initial specific volume is v1 = v2 = 0.1941 m /kg. As the water is cooled down to a saturated vapour, the initial state must be a superheated vapour. This can be confirmed by examining the saturated vapour specific volume at 300 °C where v1 > v g (300 °C). Interpolating within Table A- of the textbook yields the following bracketing states.

p = 10 bar T ( °C) v ( m3/ kg) 280 0.2480 320 0.2678

p = 15 bar T ( °C) v ( m3/ kg) 280 0.1627 320 0.1765

Note that 300 °C does not appear explicitly in Table A- as needed, thus it is necessary to look for two subtables which would bracket the desired specific volume if we were to determine v at this temperature. Interpolating within each sub-table to determine the specific volume at 300 °C yields the new table

T = 300 °C v ( m3/ kg) p ( bar) 0.2579 10 bar 0.1696 15 bar Interpolating for pressure yields p1 = 13.6 bar.

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Q 

A rigid tank contains 5 lbm of a two-phase, liquid-vapour mixture of H2 O , initially at 260 °F with a quality of 60%. Heat transfer to the contents of the tank occurs until the temperature is 320 °F. Show the process on a p–v diagram. Determine the mass of vapour, in lbm, initially present in the tank and the final pressure, in lbf/in2 .

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A

Due to the rigid nature of the container, this is a constant volume process, which will appear as a vertical line on a p–v diagram (see below). As the initial state is given as having a quality of 0.6, the initial state must be a liquid-vapour mixture at 260 °F. Using Table A-E from the textbook, the pressure would be the saturation pressure at this temperature p1 = p sat (260 °F) = 35.42 lbf/in2 . From the definition of mixture quality we can determine that the initial mass of vapour is

mvapour,1 = χ1 m = 0.6 ⋅ 5 lbm = 3 lbm

()

To determine the final pressure at state , we can use the fact that v1 = v2 . The specific volumes of the 3 3 saturated liquid and vapour states at T1 = 260 °F are respectively v f = 0.01708 ft / lbm and v g = 11.77 ft / lbm. Thus the specific volume of the water is

v2 = v1 = 0.01708 + 0.6(11.77 − 0.01708) = 7.0688ft3/ lbm.

()

Examining Table A-E (or A-E) reveals the specific volume of saturated vapour water at 320 °F is 4.919 ft / lbm which is less than v2 . Thus state  must be in the superheated vapour region. To determine the pressure 3 at state  requires that we interpolate within Table A-E using v = 7.0688 ft / lbm. Searching the sub-tables yields the following four entries which bracket this value (the temperature 320 °F does not explicitly appear in these tables) 3

p = 60 lbf/in2 T( °F) v( ft3/ lbm) 300 7.26 350 7.82

p = 80 lbf/in2 T( °F) v( ft3/ lbm) 312.1 5.47 350 5.80

As this is superheated vapour we would expect that p2 < psat (320 °F). In this case the resulting pressure falls between 60 lbf/in2 and 80 lbf/in2 . At 80 lbf/in2 for a superheated vapour, the lower limit on temperature is 312.1 °F and thus the two subtables have slightly different temperature ranges. The above two tables can be reduced to the following table using linear interpolation by temperature

T = 320 °F v( ft3/ lbm) p( lbf/in2 ) 7.484 60 5.539 80 The final pressure p2 = 64.27 lbf/in2 can be determined by linear interpolation using specific volume. Note that your p–v diagram (see Fig. ) does not need to be to scale but it should indicate the position of the process end states relative to the vapour dome.

Q 

Piston 50 kg water 150 kPa, 20˚C

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A piston-cylinder device, as illustrated, contains 50 kg of water at 150 kPa and 20 °C. e cross-sectional area of the piston is 0.1 m2 . Assume that cylinder walls are frictionless. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0.2 m3 , the piston comes in contact with a linear spring whose spring constant is 100 kN/ m. More heat is transferred to the water until the piston rises an additional 20 cm while in contact with the spring. Determine (a) the final pressure  of 

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p 320˚F 260˚F 2

1

v

Figure : p–v diagram for Q 

and temperature, (b) the work done during this process. Include a p–V diagram.

A

Consider the water to be a closed system. The system undergoes a sequence of two expansion processes passing through three states. The intial state 1 begins at p1 = 150 kPa and T1 = 20 °C. Comparing T1 to Tsat (150 kPa) = 111.4 °C using Table A- reveals that this state is in the subcooled liquid phase region. Using the subcooled liquid approximation with Table A- yields v1 = v f (20 °C) = 1.0018 × 10−3 m3/ kg. The subcooled liquid is heated at constant pressure (the piston is free to move) until the specific volume at state 2 is

v2 =

V2 0.2 m3 = = 4 × 10−3 m3/ kg m 50 kg

()

For a constant pressure process the work is

W12 = m ∫

2 1

pdv = mp12 (v2 − v1 ) = 50 kg ⋅ 150 kPa ⋅ (4 × 10−3 − 1.0018 × 10−3 ) = 22.50 kJ

()

State 2 is in the saturated liquid-vapour phase region as indicated by the fact that v f (150 kPa) < v2 < v g (150 kPa). Process 2–3 involves the piston in contact with a linear spring, where the force exerted by the spring is equal to kx where k = 100 kN/ m and x is the distance by which the spring is compressed. The spring is compressed by a total of 0.20 m, which corresponds to a volume difference of V3 − V2 = (x3 − x2 )A = 0.20 m ⋅ 0.1 m2 = 0.02 m3 . The spring causes a linear increase in pressure as a function of x

p(x) = p2 +

kx A

()

Pressure during the process 2–3 can be described as a function of volume V using

V = V2 + Ax k(V − V2 ) p = p2 + A2 ©

 of 

() ()

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p (kPa)

350 150

3 1

2

V

Figure : p–V plot of processes in Q .

The final volume is V3 = 0.22 m3 , thus the final pressure is p3 = 350 kPa. The final specific volume is

v3 =

V3 0.22 m3 = = 4.4 × 10−3 m3/ kg m 50 kg

()

which falls between v f (350 kPa) = 1.0786 × 10−3 m / kg and v g (350 kPa) = 0.5243 m / kg, thus state 3 is within the vapour dome. The temperature at state 3 is T3 = Tsat (350 kPa) = 138.9 °C. The work of the expansion process 2–3 can be determined using 3

3

k(V − V2 ) kV2 k ) dV = (p2 − 2 ) (V3 − V2 ) + 2 (V32 − V22 ) 2 A A 2A V2 2 kN/ m 100 kN/ m ⋅ 0.2 m3 100 W23 = (150 kPa − ) ⋅ 0.02 m3 + ⋅ (0.222 − 0.22 ) = 5 kJ 0.12 m4 2 ⋅ 0.12 m4

W23 = ∫

3

pdV = ∫

V3

(p2 +

The total work done by the system is 27.5 kJ. A p–V diagram (not to scale) is illustrated in Fig. .

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() ()