mae 91 su13: hw 3 solutions

MAE 91 Summer 2013 Problem Set 3: Solutions

Problem 1

Illustrate the possible process in a P -v diagram for any of the �gures below:

I. Problem description

• Given:

Sketch of the initial state

• Find

Possible process, P -v diagram

II. Analytical Solution

a. Constant pressure process and constant volume process when V = Vstops (Minimum volume)

b. Linear relation between pressure and volume, considering that compression or expansion is done

against a linear spring and external pressure.

c. Constant pressure process and constant volume process when V = Vstops (Maximum volume).

d. Isochoric process until the pressure increases up to Pfloat.

e. Isochoric process.

f. Isochoric process until the pressure decreases up to Pfloat.

II. Graphical Solution

a b c

1


d e f

2


Problem 2

A 400L tank A, see �gure show, contains argon a gas at 250kPa, 30◦C. Cylinder B, having a frictionless

piston of such mass that a pressure of 150kPa will �oat it, is initially empty. The valve is opened and

argon a �ows into B and eventually reaches a uniform state of 150kPa, 30◦C throughout. What is the

work done by the argon?

I Problem description

• Known:

Argon

PA1 = 250kPa

TA1 = 30◦C

Constant pressure process

PFloat = 150kPa

VA = 0.4m3

P2 = 150kPa

T2 = 30◦C

• Unknown:

mA1 =?, m2 =?

• Find

1W2 =?

• Figure

II. Assumptions

The pressure drop in the valve is controlled in such a way that constant pressure is �lling tank B. Also,

if the process is considered to be quasi-static, then tank B is always at constant pressure. Argon is

treated as an ideal gas.

III. Analytical Solution

The mass of argon is conserved when it �ows from tank A to tank B. Ideal gas law allows us to �nd

the mass contained inside each tank.

3


1. Conservation of mass: dmdt = 0

m1 = m2

mA1 = mA2 +mB2

2. Ideal Gas law: PV = mRArT

mA1 = PA1VA

RArTA1

mA2 = PA2VA

RArTA2

mB2 = PB2VB

RArTB2

Conditions at state 2: Uniform state between tank A and B.

TA1 = TB2 = T2 and PA2 = PB2 = P2

Plugging each mass on conservation of mass:

PA1VA

RArTA1= P2VA

RArT2+ P2VB

RArT2(PA1VA

TA1− P2VA

T2

)= P2VB

T2

VB =(

PA1T2

P2TA1− 1

)VA

3. Work for a constant pressure process:

1W2 = 1W2B =´ VB2

VB1P dV = P2 (VB2 − VB1)

IV. Numerical solution

VB =(

PA1

P2− 1

)VA =

(250150 − 1

)0.4 = 0.2667m3

1W2 = P2 (VB2 − VB1) = 150 (0.2667− 0) = 40kJ

4


Problem 3

A piston cylinder contains 2kg of liquid water at 20◦C and 300kPa, as shown in the �gure. There is

a linear spring mounted on the piston such that when the water is heated the pressure reaches 3MPa

with a volume of 0.1m3.

a) Find the �nal temperature

b) Plot the process in a P − v diagram.

c) Find the work in the process.


• Given:

Water m = 2kg

T1 = 20◦C

P1 = 300kPa

P2 = 3MPa

V2 = 0.1m3

• Find:

a. T2 =?

b. P − v diagram

c. 1W2 =?

• Figure



m1 = m2 = m

• State 1: From table B.1.1 (Approximate from saturated liquid at given temperature)

v1 = V1

m

V1 = v1m

• State 2: From table B.1.2

v2 = V2

m

5


2. Work for a linear variation of pressure with volume: P = aV + b

1W2 =´ V2

V1P dV = Pavg (V2 − V1) =

12 (P1 + P2) (V2 − V1)

III. Numerical solution


v1 = vf@20◦C = 0.001002m3/kg

V1 = v1m = 0.001002× 2 = 0.002004m3


v2 = V2

m = 0.12 = 0.05m

3/kg

vg@P = 0.0668m3/kg, vf@P = 0.00122m

3/kg

Then vf < v2 < vg, so it is saturated:

T2 = 233.99◦C

1W2 = 12 (P1 + P2) (V2 − V1) =

12 (300 + 3000) (0.1− 0.002004) = 161.7kJ

6


Problem 4

Consider a piston cylinder with 0.5kg of R-134a as saturated vapor at −10◦C. It t is now compressed

to a pressure of 500kPa a in a polytropic process with n = 1.5. Find the �nal volume and temperature,

and determine the work done during the process.


• Given:

R-134a m = 0.5kg

T1 = −10◦C

x1 = 1

P2 = 500kPa

Polytropic process n = 1.5

• Find:

T2 =?, V2 =?, 1W2

Final volume of the balloon.

II. Assumption

Polytropic process with n = 1.5.


1. Conservation of mass: m1 = m2 = m

• State 1: From table B.5.1 (Saturated vapor at given temperature)

v1 = vg@T , P1 = Psat@T

V1 = v1m

2. Polytropic process:

P1Vn1 = P2V

n2 or P1v

n1 = P2v

n2

V2 =(

P1

P2

)1/n

V1

v2 = V2

m

• State 2: From table B.5.2 (Entry with pressure P2 and v2)

3. Work for a polytropic process P = CV n

1W2 =´ V2

V1P dV = P2V2−P1V1

1−n

7


IV. Numerical solution

• State 1: From table B.5.1 (Saturated vapor at given temperature)

v1 = vg@T = 0.09921m3/kg, P1 = Psat@T = 201.7kPa

V1 = v1m = 0.049605m3/kg

• State 2: From table B.5.2 (Entry with pressure P2 and v2)

V2 =(

P1

P2

)1/n

V1 = 0.02708m3

v2 = V2

m = 0.05416m3/kg

Then,

T2 = 79◦C

And work is,

1W2 = P2V2−P1V1

1−n = 500×0.02708−201.7×0.0496051−1.5 = −7.069kJ

8


Problem 5

Find the missing properties of (P , T , v, u, h and x) and indicate the states in a P -v and T -v diagram

for:

a. Water at 5000kPa, u = 1000kJ/kg (Table B.1 reference)

b. R-134a at 20◦C, u = 300kJ/kg

c. Nitrogen at 250K, 200kPa


• Given:

a. Water, P = 5000kPa, u = 1000kJ/kg

b. R-134a, T = 20◦C, u = 300kJ/kg

c. Nitrogen at T = 250K, P = 200kPa

• Find:

P , T , v, u, h and x

P -v and T -v diagrams

II. Analytical and Numerical Solution

a. Compressed liquid: Table B.1.4, at P = 5000kPa, interpolate between T = 220◦C and T = 240◦C.

T = 240−2201031.3−938.4 (1000− 1031.3) + 240 = 233.3◦C,

v = 0.001226−0.0011871031.3−938.4 (1000− 1031.3) + 0.001226 = 0.001213m

3/kg

h = 1037.5−944.41031.3−938.4 (1000− 1031.3) + 1037.5 = 1006.1kJ/kg

x = undefined

b. Two-phase liquid + vapor: Table B.5.1. P = Psat@T = 572.8kPa

uf@T = 227.03kJ/kg< u = 300kJ/kg <ug@T = 389.2kJ/kg

x =u−uf@T

ufg@T= 1000−227.03

162.16 = 0.45

v = vf@T + xvfg@T = 0.000817 + 0.45× 0.03524 = 0.01667m3/kg

h = hf@T + xhfg@T = 227.49 + 0.45× 182.35 = 309.55m3/kg

c. Superheated vapor: Table B.6.2. T = 250K > Tsat@P = 83.6K, P = 200kPa

v = 0.5 (0.35546 + 0.38535) = 0.3704m3/kg

u = 0.5 (177.23 + 192.14) = 184.69m3/kg

h = 0.5 (248.32 + 269.21) = 258.77m3/kg

x = undefined

9


III. Graphical Solution

10


Problem 6

A piston cylinder contains 1.5kg water at 200kPa, 150◦C. It is now heated in a process where pressure

is linearly related to volume to a state of 600kPa, 350◦C. Find the �nal volume, the work and the

heat transfer in the process.


• Given:

Water

m = 1.5kg

P1 = 200kPa

T1 = 150◦C

P2 = 600kPa

T2 = 350◦C

Linear process

P = aV + b

• Find:

V2, 1W2, 1Q2

• Sketch


1. Compute the states:

State 1: water is superheated at P1 and T1 given, table B.1.3.

v1 = 0.95964m3/kg , u1 = 2576.87kJ/kg

State 2: water is also superheated at P2 and T2 given, table B.1.3.

v2 = 0.47424m3/kg , u2 = 2881.12kJ/kg

2. Find the �nal volume:

V2 = mv2 = 0.7114m3

11


3. Find the work

1W2 = m´ v2v1

pdv = m2 (P1 + P2) (v2 − v1) =

1.52 (200 + 600) (0.47424− 0.9596) = −291.2kJ

4. Find the Heat: First Law: E2 − E1 = 1Q2 − 1W2

For a quasi-static process and no potential energy:

1Q2 = U2 − U1 + 1W2

1Q2 = m (u2 − u1) + 1W2 = 1.5 (2881.1− 2576.9)− 291.2 = 165.1kJ

12


Problem 7

A water-�lled reactor with volume of 1m3 is at 20MPa, 360◦C and placed inside a containment room

as shown in Figure. The room is well insulated and initially evacuated. Due to a failure, the reactor

ruptures and the water �lls the containment room. Find the minimum room volume so the �nal

pressure does not exceed 200kPa.


• Given:

V1 = 1m3

P1 = 20MPa

T1 = 360◦C

P2max ≤ 200kPa

• Find:

V2min

• Sketch

II. Assumption

The system is the reactor and the room. The walls of the room are rigid and adiabatic, so no work or

heat transfer are present.



m1 = m2

mR1 +mvacuum = m2

Expressing in terms of the speci�c volume,

mR1 = VR

v1, m2 = V2min

v2min

VR

v1= V2min

v2min

V2min = v2min

v1VR

13


2. State 2 is found from �rst law: E2 − E1 = 1Q2 − 1W2

Since 1Q2 = 0 and 1W2 = 0

The equilibrium condition of state 2 has no kinetic energy, but, the transition does.

E2 = E1 and then U2 = U1 or u2 = u1

II. Numerical Solution

1. State 1: Table B.1.4 at P1 = 20MPa, T1 = 360◦C

v1 = 0.001823m3/kg , u1 = 1702.8kJ/kg

2. State 2: from P2 = 200kPa, u1 = 1702.8kJ/kg, Table B.1.2. (Saturated)

uf@P = 504.8kJ/kg, ug@P = 2529.5kJ/kg, ufg@P = 2024.7kJ/kg

x2 =u2−uf

ufg= 1702.8−504.8

2024.7 = 0.59176

Then, the speci�c volume is:

vf@P = 0.001061m3/kg, vg@P = 0.88573m

3/kg, vfg@P = 0.88467m

3/kg

v2 = vf + x2vfg = 0.001061 + 0.59176× 0.88467 = 0.52457

3. The minimum volume is:

V2min = v2min

v1VR = 0.52457

0.0018231 = 287.7m3

14


Problem 8

An insulated cylinder is divided into two parts of 1m3 each by an initially locked piston, as shown

in Figure. Side A has air at 200kPa, 300K, and side B has air at 1.0MPa, 1000K. The piston is

now unlocked so it is free to move, and it conducts heat so the air comes to a uniform temperature

TA = TB . Find the mass in both A and B, and the �nal T and P .


• Given: Air

VA1 = VB1 = 1m3

PA1 = 200kPa

TA1 = 300K

PB1 = 1.0MPa

TB1 = 1000K

Rair = 0.287kJ/kgK

• Find:

mA, mB

T2, P2

• Sketch

II. Assumption

Air can be treated as an ideal gas. System A+B is isolated.


1. Conservation of mass: System A+B

m1 = m2

mA1 +mB1 = mA2 +mB2

System A or B:

mA1 = mA2 = mA, mB1 = mB2 = mB

15


2. Mass obtained from ideal gas law, PV = mRT

mA = PA1VA1

RairTA1

mB = PB1VB1

RairTB1

3. Balance of forces on the piston at equilibrium:

PAAp − PBAp = 0

PA2 = PB2 = P2

4. Second State is found from �rst law: E2 − E1 = 1Q2 − 1W2

For the system A+B, walls are adiabatic, 1Q2 = 0, and rigid, 1W2 = 0.

At equilibrium, condition of state 2 has no kinetic energy, and potential energy is neglected.

E2 = E1 and then U2 = U1

U1 = UA1 + UB1 = mAuA1 +mBuB1

For state 2, TA2 = TB2 = T2 at equilibrium, then u2A = u2B

U2 = UA2 + UB2 = mAuA2 +mBuB2 = u2 (mA +mB)

Solving for u2:

u2 = mAuA1+mBuB1

mA+mB

5. Final Pressure: State 2 as a uniform between A and B:

P2V2 = m2RairT2

P2 = (mA+mB)RairT2

VA+VB

IV. Numerical Solution

Mass on each part:

mA = PA1VA1

RairTA1= 200×1

0.287×300 = 2.323kg

mB = PB1VB1

RairTB1= 1000×1

0.287×1000 = 3.484kg

State 1: treated air as an ideal gas, internal energy is function of temperature only, found on Table

A.7.

TA1 = 300K → uA1 = 214.3kJ/kg

TB1 = 1000K → uB1 = 759.2kJ/kg

State 2:

u2 = mAuA1+mBuB1

mA+mB= 541.2kJ/kg→T2 = 736K

P2 = (mA+mB)RairT2

VA+VB= (2.323+3.484)0.287×736

1+1 = 613kPa

16


Problem 9

The cylinder volume below the constant loaded piston has two compartments A and B �lled with

water. A has 0.5kg at 200kPa, 150◦C and B has 400kPa with a quality of 50% and a volume of 0.1m3.

The valve is opened and heat is transferred so the water comes to a uniform state with a total volume

of 1.006m3.

a) Find the total mass of water and the total initial volume.

b) Find the work in the process

c) Find the process heat transfer


• Given:

PA1 = 200kPa

TA1 = 150◦C

PB1 = 400kPa

xA1 = 0.5

mPiston = 0.5kg

VB = 0.1m3

V2 = VA2 + VB2 = 1.006m3

• Find:

a. m2, V1

b. 1W2

c. 1Q2

• Sketch

II. Assumption

The process is assumed to be quasi-static, so the piston keeps the pressure constant in Tank A. The

valve is ideal and permits a quasi-static process.


1. Conservation of mass: Tank A+B

17


m2 = m1 = mA1 +mB1

State B1: Mass on tank B is obtained from table B.1.2.

mB = VB

vB1

State A1: volume on tank A is obtained from table B.1.3.

VA1 = mA1vA1

Initial volume is:

V1 = VA1 + VB

2. Balance of forces on the piston makes the process isobaric if quasi-static. Then, work is,

1W2 = 1W2B =´ V2

V1PdV = PA (VA2 − VA1)

3. The heat is found from First Law: E2 − E1 = 1Q2 − 1W2

For a quasi-static process and no potential energy:

1Q2 = U2 − U1 + 1W2

1Q2 = m2u2 − (mA1uA1 +mB1uB1) + 1W2

IV. Numerical Solution

State B1 (Saturation):

vB1 = vf@P + xB1vfg@P = 0.001084 + 0.5× 0.46138 = 0.2318m3/kg

uB1 = uf@P + xB1ufg@P = 604.3 + 0.5× 1949.3 = 1578.9kJ/kg

State A1 (Superheated):

vA1 = 0.95964m3/kg

uA1 = 2576.9kJ/kg

Then, the volume at A is,

VA1 = mA1vA1 = 0.5× 0.95964 = 0.47982m3

and the mass at B1 is,

mB = VB

vB1= 0.1

0.2318 = 0.4314kg

a. Total mass and initial volume are:

m2 = mA1 +mB1 = 0.5 + 0.4314 = 0.9314kg

V1 = VA1 + VB = 0.1 + 0.47982 = 0.57982m3

b. Work:

1W2 = PA (VA2 − VA1) = 200 (1.006− 0.57982) = 85.2kJ

c. Heat:

1Q2 = m2u2 − (mA1uA1 +mB1uB1) + 1W2

1Q2 = 0.9314× 2654− (0.5× 2576.9 + 0.4314× 1578.9) + 85.2 = 587.6kJ

18


Problem 10

Two kilograms of water is contained in a piston/cylinder with a massless piston loaded with a linear

spring and the outside atmosphere. Initially the spring force is zero and P1 = Po = 100kPa with a

volume of 0.2m3. If the piston just hits the upper stops the volume is 0.8m3 and T = 600◦C. Heat

is now added until the pressure reaches 1.2MPa. Find the �nal temperature, show the P�V diagram

and �nd the work done during the process.


• Given: water

m = 2kg

P1 = Po = 100kPa

V1 = 0.2m3

V2 = Vstops = 0.8m3

T2 = 600◦C

P3 = 1.2MPa

mPiston ≈ 0

• Find:

T3, P − V diagram and 1W2

• Sketch

II. Assumption

Pressure varies linearly with volume when the spring is compressed.


1. Conservation of mass and states:

m3 = m2 = m1 = m = 2kg

19


• State 1: Obtained from P1, initial mass and volume. Table B.1.2. shows this state is

saturated.

v1 = V1

m = 0.22 = 0.1m

3/kg

vf@P = 0.001043m3/kg, vg@P = 1.694m

3/kgx1 = 0.058

• State 2: from table B.1.3.

v2 = V2

m = 0.82 = 0.4m

3/kg

P2 = P@v2,T2 ≈ 1000kPa

• State 3: Since P3 > P3, the process is at constant volume from 2 to 3, following the path

123. If the pressure at the �nal state was lower, then the path would be 12'3', but is not

the case.

v3 = v2 = 0.4m3/kg and P3 = 1.2MPa, using table B.1.3,

T3 = 770◦C

2. Work is,

1W3 = 1W2 + 2W3 =´ V2

V1PdV +

´ V3

V2PdV = PAvg (V2 − V1) + 0

1W3 = PAvg (V2 − V1) + 0 = 12 (100 + 1000) (0.8− 0.1) = 330kJ

20


Problem 11

A piston/cylinder arrangement B is connected to a 1m3 tank A by a line and valve, as shown in �gure.

Initially both contain water, with A at 100kPa, saturated vapor and B at 400◦C, 300kPa, 1m3. The

valve is now opened and, the water in both A and B comes to a uniform state.

a. Find the initial mass in A and B.

b. If the process results in T = 200◦C, �nd the heat transfer and work.


• Given: Water

VA = 1m3

PA1 = 100kPa

xA1 = 1

PB1 = 300kPa

VB1 = 1m3

TB1 = 400◦C

T2 = 200◦C

• Find:

a. mA1, mB1

b. 1W2, 1Q2

• Sketch

II. Assumption

Quasi-static process, and constant pressure process on tank B. Valve is ideal.


1. Conservation of mass:

m2 = m1 = constant

mA2 +mB2 = mA1 +mB1

mA1 = VA

vA1, mB1 = VB

vB1

21


2. States:

• State 1A: saturated vapor, from Table B.1.2.

vA1 = vg@P = 1.694m3/kg

uA1 = ug@P = 2506.1kJ/kg

• State 1B: superheated vapor, from table B.1.3.

vB1 = 1.03151m3/kg

uB1 = 2965.5kJ/kg

• State 2: Since constant pressure process P2 = PB1. Then state is an uniform superheated

vapor, from table B.1.3.

v2 = 0.71629m3/kg

u2 = 2650.1kJ/kg

3. Initial and total mass:

mA1 = VA

vA1= 1

1.694 = 0.5903kg

mB1 = VB

vB1= 1

1.03151 = 0.9695kg

m2 = mA1 +mB1 = 1.5598kg

4. Final volume, tank B:

mA2 +mB2 = m2

VA

v2+ VB

v2= m2

VB2 = v2m2 − VA = 0.1172m3

5. Work: constant pressure process,

1W2 =´ V2

V1PdV = PB (VB2 − VB1) = 300 (0.1172− 1) = 264.82kJ

6. The heat is found from First Law: E2 − E1 = 1Q2 − 1W2

System is water only. The piston, which have potential energy, is not included into the system.

For a quasi-static process and no potential energy,

1Q2 = U2 − U1 + 1W2

1Q2 = m2u2 − (mA1uA1 +mB1uB1) + 1W2

1Q2 = −484.7kJ

22

mae 91 su13: hw 3 solutions

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