mae 91 su13: hw 3 solutions
DESCRIPTION
Homework 3 Solutions for MAE 91TRANSCRIPT
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
MAE 91 Summer 2013 Problem Set 3: Solutions
d e f
2
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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.
I Problem description
• Given:
Water m = 2kg
T1 = 20◦C
P1 = 300kPa
P2 = 3MPa
V2 = 0.1m3
• Find:
a. T2 =?
b. P − v diagram
c. 1W2 =?
• Figure
II. Analytical Solution
1. Conservation of mass: dmdt = 0
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
• State 1: From table B.1.1
v1 = vf@20◦C = 0.001002m3/kg
V1 = v1m = 0.001002× 2 = 0.002004m3
• State 2: From table B.1.2
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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.
I Problem description
• 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.
III. Analytical Solution
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
I Problem description
• 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
MAE 91 Summer 2013 Problem Set 3: Solutions
III. Graphical Solution
10
MAE 91 Summer 2013 Problem Set 3: Solutions
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.
I Problem description
• 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
II. Analytical and Numerical Solution
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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.
I Problem description
• 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.
II. Analytical Solution
1. Conservation of mass: dmdt = 0
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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 .
I Problem description
• 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.
III. Analytical Solution
1. Conservation of mass: System A+B
m1 = m2
mA1 +mB1 = mA2 +mB2
System A or B:
mA1 = mA2 = mA, mB1 = mB2 = mB
15
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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
I Problem description
• 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.
III. Analytical Solution
1. Conservation of mass: Tank A+B
17
MAE 91 Summer 2013 Problem Set 3: Solutions
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
MAE 91 Summer 2013 Problem Set 3: Solutions
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.
I Problem description
• 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.
II. Analytical and Numerical Solution
1. Conservation of mass and states:
m3 = m2 = m1 = m = 2kg
19
MAE 91 Summer 2013 Problem Set 3: Solutions
• 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
MAE 91 Summer 2013 Problem Set 3: Solutions
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.
I Problem description
• 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.
II. Analytical and Numerical Solution
1. Conservation of mass:
m2 = m1 = constant
mA2 +mB2 = mA1 +mB1
mA1 = VA
vA1, mB1 = VB
vB1
21
MAE 91 Summer 2013 Problem Set 3: Solutions
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