ENGR Thermodynamics

(1)

HW #1

Problem : 1 - Mass, Force, Density and Acceleration - 4 pts 6-Apr-11

Read : Given : n 0.5 lbmole g 30.5 ft/s2 V 0.145 ft3 _g c 32.174 lbm-ft/lbf-s 2 MW 18.016 lbm/lbmole Find : a.) W ??? lbf b.)  ??? lbm/ft 3 Diagram : None.

Assumptions: 1 - Gravitational acceleration is uniform. 2 - Density of the water is uniform. Solution :

Part a.)

Eqn 1 Eqn 2

We can solve Eqn 2 for the weight of the water, W.

Eqn 3

Eqn 4

Plugging values into Eqn 1 yields : m 9.008 lbm

Now, we can plug values into Eqn 3 to evaluate the weight of the water.

W 8.539 lbf

Part b.) This is a straightforward application of the definition of density.

Eqn 5  62.12 lbm/ft 3 Verify : Answers : a.) b.) W 8.54 lbf  62.12 lbm/ft 3

The next step in determining the weight of the water is to convert from the given number of moles to mass using:

None of the assumptions made in the solution of this problem can be verified based on the given information. Since we determined m in part (a) and the volume of the water was given, all we need to do is plug values into Eqn 5. A closed system consists of 0.5 lbmole of liquid water and occupies a volume of 0.145 ft3. Determine the weight of the system, in lbf, and

the average density, in lbm/ft 3

, at a location where the acceleration of gravity is g = 30.5 ft/s2.

The key ideas here are the meaning of lbmole, the use of gc in Newton's 2nd Law of Motion and the definition of density.

The key here is the ralationship between mass and force, in this case weight, given by Newton's 2nd Law of Motion.

m

 

n MW

c

g F



m a

g W

_c



m g

c

g

W

m

g



m

V

 

Dr. Baratuci - ChemE 260

hw1-sp11.xlsm, WB-1

4/11/2011

(2)

Baratuci HW #1

Problem : 1.7E - Mass, Weight and Acceleration - 3 pts 6-Apr-11

Read :

Given : WEarth 210lbf gmoon 5.47 ft/s

2 gearth 32.10ft/s 2 gc 32.174ft-lbm/lbf-s 2 Find : Wmoon ??? N Diagram : None. Assumptions: None. Solution : Eqn 1 Eqn 2 Eqn 3 Eqn 4 Eqn 5

This leaves us with : Eqn 6

Now, we can solve Eqn 6 for the unknown variable Wmoon.

Eqn 7

Plugging values into Eqn 7 yields : _W

moon 35.79 lbf

Verify : None.

Answers : Wmoon 35.8lbf

ENGR 224 - Thermodynamics

A man weighs 210 lbf at a location where g = 32.10 ft/s 2

. Determine his weight on the moon where g = 5.47 ft/s2.

The keys to this problem are to understand that mass does not depend on location, but weight does and how to apply Newton's 2nd Law of Motion.

The key here is the ralationship between mass and force, in this case weight, given by Newton's 2nd Law of Motion.

The key is that gc and the mass of the object are constants. So, if we take the ratio of Eqn 3 to

Eqn 4, these terms will cancel !

We can now apply Eqn 2 to the object when it is on the moon and again to the object when it is on the surface of the Earth. c

g F



m a

g W

_c



m a

c Earth Earth

g W



m g

c moon moon

g W



m g

c moon moon c Earth Earth

g W

m g

g W



m g

moon moon Earth Earth

W

g

W



g

moon moon Earth Earth

g

W

g



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HW #1

Problem : 1.11E - Specific Heat: Unit Conversions - 4 pts 6-Apr-11

Read :

Given : CP 1.005kJ/kg-oC

Find : a.) CP ??? kJ/kg-K c.) CP ??? kcal/kg-oC

b.) CP ??? J/g-oC d.) CP ??? Btu/lbm -o F Assumptions: None. Solution : Part a.) 1 oC = 1 K Eqn 1 Eqn 2 CP 1.005 kJ/kg-K Part b.) 1 kJ = 1000 J Eqn 3 1 kg = 1000 g Eqn 4 Eqn 5 CP 1.005 J/g-oC

The only unit we need to convert in this part is the temperature unit. We need to get from oC to Kelvins. This is an easy one since a change in T of 1oCelsius is the same as a change in T of 1oKelvin. So, here is the conversion factor.

In this part, we need to convert units of both energy and mass. We need to get from kJ to J and from kg to g. Here are the conversion factors.

The constant-pressure specific heat of air at 25oC is 1.005kJ/kg-oC. Express this value in kJ/kg-K, J/g-oC, kcal/kg-oC, and Btu/lbm

-o

F.

This problem is an exercise in unit conversions. The key to the problem is to recognize that the temperature unit in the heat capacity corresponds to a change in temperatureof a magnitude of one with the given units.

P

kJ

1 C

C

1.005 kg

C 1 K



=

⋅

⋅ 

P

kJ

1 kg

1000 J

C

1.005 kg

C 1000 g

1 kJ

=

⋅

⋅ 

(4)

Part c.) 4.184 kJ = 1 kcal Eqn 6 Eqn 7 CP 0.2402 kcal/kg-oC Part d.) 1 kJ = 0.94782 Btu Eqn 8 1 kg = 2.2046 lbm Eqn 9 1 oC = 1.8 oF Eqn 10 Eqn 11 CP 0.2400 Btu/lb_m-oF Verify : None.

Answers : a.) CP ??? kJ/kg-K c.) CP ??? kcal/kg-oC

b.) CP ??? J/g-oC d.) CP ??? Btu/lbm

-o F In this part, we only need to convert units energy. We need to get from kJ to kcal. Here is the conversion factor.

In this part, we need to converat all three units from SI to AE. Here are the conversion factors.

P

kJ

1kcal

C

1.005 kg

C 4.184 kJ

=

⋅

⋅ 

P m

kJ

1 kg

0.94782 Btu

1 C

C

1.005 kg

C 2.2046 lb

1 kJ

1.8 F



=

⋅

⋅ 



(5)

HW #1

Problem : 1.37E - Temperature Conversions - 2 pts 6-Apr-11

Read : This problem provides practice converting temperature units.

Given : T(oC) = 150 Ratio = 1.8o F / oC Tfreezing 32oF Rzero F = 459.67oR Find : T(oF) = ??? T(oR) = ??? Assumptions: None. Solution :

Part a.) The key here is the relationship between the Celsius and Fahrenheit temperature scales.

Eqn 1 Plugging values into Eqn 1 yields :

T = 302 o

F

Part b.) There are seveal ways to solve this part of the problem.

One is to use the relationship between the Fahrenheit and Rankine temperature scales.

Eqn 2

Plugging values from part (a) into Eqn 2yields : _{T =} _761.67 o

R Verify : None.

Answers : a.) T = 302 o

F b.) T = 762 o

R What is the temperature of heated air at 150oC in oF and in oR?

 

o

 

o

T

F



T

C



1.8 

32  

o o o o o

F

T

F

150 C 1.8

32 F

C







 

o o o o o

R

T

R

T

F

1 459.67 R

F







(6)

Baratuci HW #1

Problem : 1.41E - Temperature Change - 3 pts 6-Apr-11

Read : Given : T 10o F Find : a.) T ??? o C c.) T ??? o R b.) T ??? K Assumptions: None. Solution :

Part a.) A change in temperature of 1 oC is equal to a change in temperature of 1.8oF.

Ratio = 1.8o

F / oC

Therefore : Eqn 1

T 5.56 o

C Part b.) A change in temperature of 1 oC is exactly the same as a change in temperature of 1 K.

RatioKC= 1 K / oC

Therefore : Eqn 2

Therefore it is easy to write the answer to part (b). T 5.56 K

Part c.)

RatioRK = 1.8oR / K

Eqn 3

Plugging values into Eqn 3 yields : T 10 o

R

Eqn 4

Plugging values into Eqn 3 yields : T 10 o

R Verify : None. Answers : a.) T 5.6 o C c) T 10 o R b.) _T 5.6 K

The temperature of water changes by10oF during a process. Express this temperature change in Celsius, Kelvin and Rankine units.

ENGR 224 - Thermodynamics

It is very important to understand that a change in temperature is not the same as a value of the temperature. It seems obvious, but it is common point of confusion.

One way to do this part of the problem is to use the result from part (b) and the converion factor for Kelvins and degrees Rankine.

Of course, it would be even easier to recognize that a change in temperature of 1 oF is exactly the same as a change in temperature of 1 oR.

 

o

 

o

R

T

R

T K

1.8 K



 



10 F

T

1.8 F / C



D =



T

5.56 C 1 K / C

D =

 ⋅



 

o o o o

F

T

F

T

R

1 R



 



(7)

HW #1

Problem : 1.78 - Differential, Multi-Fluid Manometer - 6 pts 6-Apr-11

Hg 13590 kg/m3

Read : This problem requires the careful application of the Manometer Equation. The key is that none of the fluid in the manometer tubes is moving.

Given : h1 0.60 m FW 1000 kg/m3

h2 0.10 m SW 1035 kg/m3

h3 0.70 m Hg 13590 kg/m3

h4 0.40 m

Find : PSW-PFW ??? kPa

Assumptions: 1 - The acceleration of gravity is uniform. 2 - The density of the mercury is uniform. 3 - The density of the water is uniform. 4

-5 - None of the fluids inside the tubing are moving. Solution :

Eqn 1

Eqn 2

Eqn 3

Eqn 4 Because the density of air is three orders of magnitude smaller than the density of liquid water or mercury :

Let's use the Manometer Equation to work our way around from the seawater pipe to the fresh water pipe.

Because the elevation at points 3 and 4 is the same, the manometer fluid is not moving and points 3 and 4 are connected by a tube filled with mercury only: P4 = P3.

The density of the air is negligible. Therefore, P1 = P2.

Fresh water and seawater flowing in parallel, horizontal pipelines are connected to each other by a double u-tube manometer, as shown in the figure. Determine the pressure difference between the two pipelines in kPa. Assume the density of seawater to be 1035 kg/m3. Can the air column be ignored in this analysis ?

By working from one pipe to the other we can determine the difference in the pressure between the two pipes.

SW 1 SW 4 c

g

P

h

g



 

1 4 2 3 h4= h2= = h3 = h1 2 1

P



P

4 FW FW 1 c

g

P

h

g



 

3 2 Hg 2 c

g

P

h

g



 

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, 1.78

4/11/2011

(8)

Combining Eqns 1 & 2 yields :

Eqn 5

Solving Eqn 3 for P2 yields :

Eqn 6

Eqn 7

Eqn 8

Solving Eqn 7 for P2 - PFW yields :

Eqn 9

Now, we can plug values into Eqn 9 to complete the solution of this problem.

PSW-PFW -3383 N/m 2

PSW-PFW -3.383 kPa

Verify :

Answers : PSW-PFW -3.38 kPa

Could you have guessed just by looking at the fluid levels and fluid densities where the pressure was higher, in the fresh or salt water pipe ? I could have. Think about it. Look at the diagram and the equations.

None of the assumptions made in the solution of this problem can be verified based on the given information.

2 3 Hg 2 4 Hg 2 c c

g

P

h

P

h

g



 



 

SW 2 SW 4 c

g

P

h

g



 

SW 4 Hg 2 SW 4 c c

g

P

h

g









 



 





SW FW FW 1 Hg 2 SW 4 c c c

g

P

h

g





_



_









 



 



 















SW FW FW 1 Hg 2 SW 4 c c c

g

P

h

g



 

 

 

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, 1.78

4/11/2011

(9)

HW #1

Problem : WB-2 - Absolute and Gauge Pressures - 5 pts 6-Apr-11

Read : The focus of this problem is the meaning of gauge pressure.

In this case, the pressure gauge indicates PA - PB, where PA and PB are absolute pressures.

Given : PA,gauge 1.4 bar gauge Patm 101 kPa

Hg 13.59 g/cm3 g 9.81 m/s2

h 20 cm gc 1 kg-m/N-s2

Find : PA ??? kPa PB ??? kPa

Assumptions: 1 - The acceleration of gravity is uniform. 2 - The density of the mercury is uniform. 3

-4 - The mercury in the manometer is not moving. Solution :

Because the manometer tube is open at the end, P1 = Patm. P1 101 kPa The general form of the Manotmeter Equation is :

Eqn 1

Applying Eqn 1 to our manometer gives us :

Eqn 2

Because the elevation at points 2 and 3 is the same, the manometer fluid is not moving and points 2 and 3 are connected by a tube filled with mercury only: P3 = P2.

Let's use the Manometer Equation to work our way around from point 1 on the diagram all the way to the inside of tank A.

The key is that a pressure guage indicates the difference between the pressure inside the pipe or vessel to which it is attached and the pressure outside of the pressure gauge.

The density of the air is negligible. Therefore, the pressure inside each tank is uniform.

Tank A lies inside of Tank B, as shown in the figure. Pressure gauge A is located inside Tank B and reads 1.4 bar. Both tanks contain air. The manometer connected to Tank B contains mercury ( = 13.59 g/cm3). The manometer reading is h = 20 cm, atmospheric pressure is 101 kPa and g = 9.81 m/s2. Determine the absolute pressures inside Tank A and Tank B in kPa.

1

3

2





down up fluid up down

c

g

P

z

g



 



2 1 Hg c

g

P

h

g



 

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, WB-2

4/11/2011

(10)

Now, because the density of the air in Tank B is negligible, PB = P3 = P2 ! Therefore :

Eqn 3

Plugging values into Eqn 3 yields :

Eqn 4

PB 127.66 kPa

The key to determining PA from PB is the meaning of gauge pressure.

Usually, we use : Eqn 5

Eqn 6 Plugging values into Eqn 4 yields :

PA 267.66 kPa

Verify :

Answers : PA 267.7 kPa PB 127.7 kPa

But in this problem the pressure on the outside of the gauge is not atmospheric, but PB instead. Also, the Pabs that we are

looking for here is PA. As a result, the form of Eqn 5 that we need to use here is :

B atm Hg c

g

P

h

g



 

6 3 2 B 3 3 2 3 2

g

1kg

10 cm

9.81m / s

1m

kPa

P

101kPa

13.59 20 cm

1000 g

100 cm

cm

1m

1kg m / N s

10 N / m









_



_







abs gauge atm

P



P



P

A A ,gauge B

P



P



P

A

100 kPa

P

1.4 bar

127.66 kPa

1bar







Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, WB-2

4/11/2011

(11)

HW #1

Problem : 3.26 - Steam Table Fundamentals - 4 pts 6-Apr-11

Complete the following table for water.

T (oC) P (kPa) H (kJ/kg) x (kg vap/kg) 200 0.7 140 1800 950 0 80 500 800 3162.2 Read :

It also tests your understanding and ability to use quality, x.

Given : Two pieces of data for each part, (a) through (e). Find : Complete the table.

Assumptions: The system is in an equilibrium state. Solution :

Part a.)

Here is the relevant data :

Sat. Liq Sat. Vap

200 120.21 504.71 2706.3

T = Tsat 120.21oC

Eqn 1

Plugging values into Eqn 1 yields : H 2045.82 kJ/kg

Part b.)

140 361.53 589.16 2733.5

Since :

P= Psat 361.53 kPa

Eqn 2

Plugging values into Eqn 2 yields : x 0.5647 kg vap/kg

the system contains a saturated mixture and the pressure must be equal to the saturation pressure.

This problem is designed to test how well you understand how to use tables of thermodynamic properties.

Temp. (oC)

Pressure (kPa)

H (kJ/kg) The key equation that relates the specific enthalpy of the system to the specific enthalpy of the saturated liquid and of the saturated vapor is:

The first step here is to use the given T to obtain data from the Saturation Temperature table.

This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases

present in the system at equilibrium.

Pressure (kPa)

Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent.

Since we are given the quality, x = 0.7, we know that vapor and liquid are both present in the system at equilibrium.

Because both vapor and liquid are present at equailibrium, the temperature must be equal to the saturation temperature at P = 200 kPa. We can find this temperature in the Saturation Pressure Table of the steam tables.

Because both saturated liquid and saturated vapor are present in the system at equilibrium, we must use the quality to evaluate the overall average specific enthalpy, as follows.

H (kJ/kg) Temp.

(oC) Phase Description

Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:





sat sat sat

mix vap liq

ˆ

H



x H



1 

x H

sat sat liq vap

ˆ

H



H



H

sat liq

sat vap sat liq

ˆ

H

x

ˆ

H







Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, 3.26

4/11/2011

(12)

Part c.)

950 177.66 752.74 2775.2

T = Tsat 177.66oC H = Hsat liq 752.74 kJ/kg

Part d.)

Tsat(500 kPa) = 151.83oC

80 47.414 335.02 (saturated liquid)

80 500 ??? (subcooled liquid)

80 5000 338.96 (subcooled liquid)

Now, we can determine H from this table by interpolation. _H _{335.38 kJ/kg}

Part e.)

H 3162.2 kJ/kg 800 170.41 720.87 2768.3 Since : 800 300 3056.9 800 350 3162.2 T 350o C 800 400 3267.7 Verify : Answers : _{T (}o C) P (kPa) H (kJ/kg) x (kg vap/kg) 120.21 200 2045.82 0.7 140 361.53 1800 0.565 177.66 950 752.74 0 80 500 335.38 N/A

350 800 3162.2 N/A Superheated Vapor

Pressure (kPa) Temp. (oC) H (kJ/kg) Pressure (kPa) Phase Description Sat'd Mixture (VLE)

Subcooled Liquid Sat'd Mixture (VLE) Sat'd Liquid

Since T = 80oC is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is

undefined ! So, we need to look at the Subcooled Liquid tables.

H (kJ/kg) Pressure (kPa) Temp. (oC) Temp. (oC) Pressure (kPa) H (kJ/kg) Temp. (oC)

Here things get a bit difficult because the first pressure table in the Subcooled Liquid tables is associated with 5 MPa. Our given P of 500 kPa is less than this value, so it seems we do not have 2 values to interpolate between to determine H.

This is not true. It just means that we must interpolate between the saturated liquid state at T =80oC and the subcooled liquid state at 80oC and 5 MPa. Here is the relevant data from the Steam Tables, arranged in a convenient way.

The first step here is to use the given P to obtain data from the Saturation Pressure table.

H (kJ/kg)

the system contains a superheated vapor, quality is undefined and we must use the Superheated Vapor Table to determine T.

We are fortunate that one of the Superheated Steam tables is associated with a P of 800 kPa. Because the quality is zero, we immediately know the system contains a saturated liquid. This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given).

So, we can look up 950 kPa in the Saturation Pressure table.

Here we are given both the T and P. Let's use the given P, and the Saturation Pressure table, to determine the Tsat associated

with P. Then, we can determine the phases present by comparing the given T to Tsat(P).

sat vap

ˆ

H



H

(13)

HW #1

Problem : 3.29E - R-134a Table Fundamentals - 4 pts 6-Apr-11

Complete the following table for R-134a. T (oF) P (psia) H (Btu/lbm) x (lbm vap/lbm) 80 78 15 0.6 10 70 180 129.46 110 1.0 Read :

It also tests your understanding and ability to use quality, x.

Given : Two pieces of data for each part, (a) through (e). Find : Complete the table.

Assumptions: The system is in an equilibrium state. Solution :

Part a.) P 80 psia H 78 Btu/lbm

80 65.89 33.394 112.20

Since :

T = Tsat 65.89oF

Eqn 1

Plugging values into Eqn 1 yields : x 0.5660 kg vap/kg

Part b.) T 15 oF x 0.6 lbm vap/lbm

This tells us that T = Tsat = 15 o

F (given) and P = Psat.

So, we can look up 15oF in the Saturation Temperature table. Here is the relevant data :

15 29.759 16.889 105.27

P = Psat 29.76 psia

Eqn 2

Plugging values into Eqn 2 yields : H 69.92 kJ/kg

Pressure (psia)

Temp. (oF)

Because both saturated liquid and saturated vapor are present in the system at equilibrium, we must use the quality to evaluate the overall average specific internal energy, as follows.

The key equation that relates the specific internal energy of the system to the specific internal energy of the saturated liquid and of the saturated vapor is:

This problem is designed to test how well you understand how to use tables of thermodynamic properties.

Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent.

the system contains a saturated mixture and the temperature must be equal to the saturation temperature.

H (Btu/lbm)

Phase Description

Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:

Temp. (oF)

H (Btu/lbm)

Because the quality lies between 0 and 1, we immediately know the system contains a saturated mixture. sat sat liq vap

ˆ

H



H



H

sat liq

ˆ

H

x

ˆ

H











sat sat sat

mix vap liq

ˆ

H



x H



1 

x H

(14)

Part c.) T 10 oF P 70 psia

Tsat(70 psia) = 58.30oF

10 26.646 15.259 (saturated liquid)

10 70 ??? (subcooled liquid)

10 100 15.169 (subcooled liquid) Now, we can determine H from this table by interpolation.

H 15.21 Btu/lbm

Part d.) P 180 psia H 129.46 Btu/lbm

180 117.69 51.497 117.96

Since :

T 160 o

F

Part e.) T 110 oF x 1.0 lbm vap/lbm

Because the quality is 1.0, we immediately know the system contains a saturated vapor. This tells us that T = Tsat = 100

o_F_(given),_{H = H}

sat vap and P = Psat .

So, we can look up 110oF in the Saturation Temperature table.

110 161.07 48.698 117.23

P = Psat 161.07 psia H = Hsat vap 117.23 Btu/lbm

Verify : Answers : T (oF) P (psia) H (Btu/lbm) x (lbm vap/lbm) 65.89 80 78 0.566 15 29.76 69.92 0.6 10 70 15.21 N/A 160 180 224 N/A 110 161.07 117.23 1.0 Pressure (psia) Temp. (oF) H (Btu/lbm)

We are amazingly fortunate that one of the Superheated R-134a tables is associated with a P of 180 psia and H = 129.46 Btu/lbm at

T = 160oF !

the system contains a superheated vapor, quality is undefined and we must use the Superheated Vapor Table to determine T.

Temp. (oF) Pressure (psia) H (Btu/lbm) Subcooled Liquid Superheated Vapor Saturated Vapor Phase Description Sat'd Mixture (VLE) Sat'd Mixture (VLE) Temp. (oF) Pressure (psia) H (Btu/lbm)

Here we are given both the T and P. Let's use the given P, and the Saturation Pressure table, to determine the Tsat associated with

P. Then, we can determine the phases present by comparing the given T to Tsat(P).

Since T = 10o_F_{is less than}_T

sat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined

! So, we need to look at the Subcooled Liquid tables.

Here things get a bit difficult because (1) CB did not give you a subcooled liquid table for R-134a and (2) the first pressure table in the LT Subcooled Liquid tables is associated with 100 psia. Our given P of 70 psia is less than this value, so it seems we do not have 2 values to interpolate between to determine H.

This is not true. It just means that we must interpolate between the saturated liquid state at T =10oF and the subcooled liquid state at 10oF and 100 psia. Here is the relevant data from the R-134a Tables, arranged in a convenient way.

sat vap

ˆ

H



H

(15)

HW #1

Problem : 3.37E - Expansion of R-134a in a Spring-Loaded P&C Device - 5 pts 6-Apr-11

Read : Given : m 0.2 lbm k 37 lbf/in T1 30 oF D 12 in x1 0.8 kg vap/kg V2 / V1 1.4 Find : T2 ??? oF Assumptions: Solution : Eqn 1 We can determine the initial specific volume because we know both T1 and x1.

Eqn 2 30 40.813 0.01234 1.153

V1 0.925 ft3/lbm V1 0.185 ft3

Next, we can determine V2 because we know it is 40% larger than V1.

V2 1.295 ft3/lbm V2 0.259 ft3

Eqn 3

Where : Eqn 4 Eqn 5

Plugging values into Eqns 4, 5 & 3 yields : Across 0.7854 ft

2

1 ft 12 in 113.1 in2

1 ft2 = 144 in2 h1 2.83 in

h2 3.96 in

Be very careful with ft and in units when using Eqn 3 ! P2 41.18 psia We know the mass, so if we can determine the specific volume, we

can determine the total volume using:

A spring-loaded piston-and-cylinder device is initially filled with 0.2 lbm of a vapor-liquid mixture of R-134a at -30oF and a quality of 80%. The spring constant in the spring force relation F = k x is 37 lbf/in, and the piston diameter is 12 in. The R-134a undergoes a process that increases its volume by 40%. Calculate the final temperature and enthalpy of the R-134a.

The key to this problem is that the mass of R-134a inside the cylinder remains constant.

The initial state of the R-134a is fixed because we know both T1 and x1. This will allow us to determine the specific volume and

then the total volume in the initial state.

We can then determine the total volume and specific volume in the final state.

Once we know the total volume in the final state, we can use the spring constant to determine P2.

Finally, knowing P2 and V2, we will be able to determine all the other intensive properties for state 2, including T2 and H2 that we

need to answer the question.

- Both the initial and final states are equilibrium states. Lets begin by determining the total volume of the R-134a in the initial state.

Here is the relevant data from the Saturation

Temperature table for R-134a. _Temp.

(oF)

V (ft3/lbm)

Now, we need to determine P2. We can accomplish this using the linear spring equations given in the problem statement.

As the R-134a expands, the pressure increases because the spring exerts additional force opposed to the expansion. The key to the relationship between the force exerted by the spring and the pressure inside the cylinder is the cross-sectional area of the piston, Across.

We don't how much force the spring exerts on the piston in the initial state, but that is not a problem. What we need to know is how much additional force the spring exerts on the pistons as the R-134a expands.

1

ˆ

1

V

=

m V

⋅





sat sat sat

mix vap liq

ˆ

V



x V



1 

x V

spring 2 1 2 1 1 cross cross

F

k (h

h )

P

A

⋅

-=

+

=

+

2 cross

A

D

4 p

=

cross

V

h

A

=

(16)

40 29.01 0.01232 1.1760

45 34.86 0.01242 1.0497

Here is the double interpolation table for

T

:

T(

o

F )

40

41.18

50

60 1.2768

1.244280

1.0019

76.7385

1.295263

80 1.3389

1.305197

1.0540

T

2

76.74

o

F

Here is the double interpolation table for

H

:

T(

o

F )

40

41.18

50

60

113.79

113.71

113.11 76.7385

117.33

117.26

116.73

80

118.02

117.95

117.43 H

2

117.26 Btu/lb

m Verify : None. Answers :

T

2

76.7

o

_F

H

2

117.3 Btu/lb

m Notes:

The correct answers for this problem using T1= 30

o

F are: T2 79.08

o F

H2 117.72 Btu/lbm

As you can see, the errors due to interpolation are significant, but not disasterous. If we had used T1 = -30oF, the errors would have been disasterous !

Here is the relevant data from the Saturation Temperature table for R-134a.

Temp. (oF)

V (ft3/lbm)

Now, we know the values of two intensive variable for the final state: V2 and P2, so we can determine the values T2 and H2 by

interpolation on the R-134a tables.

The reason I changed T1 from -30

o

F to +30oF was to reduce the error that is an unavoidable problem in all interpolations, but especially in double interpolations.

Since V2 is greater than Vsat vap even at the lower P of 10 psia, we can conclude that the R-134a is a superheated vapor in the

final state. So, we need to look at the Superheated table for R-134a.

Because there is no table for 11.29 psia in the Superheated R-134a table and our value of V2, 4.965 ft 3

/lbm, does not appear in

both the 10 psia and 15 psia tables, double interpolation is required to solve this problem.

Pressure (psia)

This double interpolation is a little different

because

V

2

is known and

T

2

is unknown. I

chose to interpolate on pressure first and

then to interpolate on volume. I don't think

you can do this problem in using a different

interpolation order.

Pressure (psia)

I chose to interpolate on pressure first and

then to interpolate on temperature. If you do

the interpolations in the opposite order, you

will get a slightly different answer. Either

method is satisfactory.

(17)

HW #1

Problem : 3.53 - Isobaric Expansion of Water - 6 pts 6-Apr-11

a.) What is the initialtemperature of the water in the cylinder in oC? b.) What is the total mass of the water in the cylinder in kg ? c.) Calculate the final volume in m3.

d.)

Read :

Given : V1,sat liq 0.005 m3 P1 = P2 600 kPa

V1,sat vap 0.9 m3 T2 200 o C Find : a.) T1 ??? o C b.) m ??? kg c.) V2 ??? m3

d.) Construct a fully labeled PV diagram of this process.

Assumptions : 1 - The system is closed. No mass enters or leaves the cylinder. 2 - The intial and final states are equiibrium states.

Diagram : Solution : Part a.) Tsat(600kPa) = 158.83oC T1 158.83 oC Part b.) Eqn 1 Eqn 2 m1,sat liq 4.541 kg

Sat. Liq Sat. Vap m1,sat vap 2.852 kg

600 158.83 0.001101 0.31560 x1 0.386 kg vap/kg

The total mass is the sum of the masses of the sat'd liquid and sat'd vapor : m 7.393 kg Because the process is isobaric, state 2 is a saturated vapor at 300 kPa and we can lookup all of its properties in the saturation pressure table.

In the initial state, liquid and vapor exist in equilibrium at 600 kPa. Therefore the initial temperature must be equal to the saturation temperature at 600 kPa. We can look this value up in the saturated steam pressure table.

We know the volume of the saturated liquid and the volume of the saturated vapor in the cylinder, so we can determine the mass of each from the specific volume using :

A piston-and-cylinder device contains 0.005 m3 of liquid water and 0.9 m3 of water vapor in equilibrium at 600 kPa. Heat is added to the water at constant pressure until the temperature reaches 200oC.

Show the process on a completely labeled PV Diagram. Be sure to include the two-phase envelope.

The keys to this problem are that the process is isobaric and that the cylinder is a closed system. The mass in the system is the same in the initial and final states.

We can determine the mass of water in the system in the initial state using the given T, P and V.

We can look up the specific volumes of the saturated liquid and saturated vapor in the saturated steam pressure table and then use Eqns 1 & 2 to determine the mass o f each phase in the initial state.

Pressure (kPa) Temp. (oC) V (m3/kg) sat vap sat vap sat vap V m ˆ V 

Isobaric

Heating

1 P

1

, V

1

, T

1

P

2

= P

1

V

2

> V

1

T

2

> T

1

2

sat liq sat liq sat liq V m ˆ V 

Dr. Baratuci - ENGR

hw1-sp11.xlsm, 3.53

4/11/2011

(18)

Part c.) Eqn 3 V2 0.35212 m3/kg V2 2.603 m3 Part d.) Verify : Answers : a.) T1 158.83 oC b.) m 7.39 kg c.) V2 2.60 m3 d.) See above.

We can determine V2 if we can first determine the final specific volume. Since we know both T2 and P2, we can look up the final

specic volume in the steam tables. Because T2 > Tsat(600 kPa), we will need to look in the superheatd steam tables.

2 sat vap

ˆ

2

V



m



V

(19)

HW #1

Problem : 3.66 - R-134a Table Fundamentals - 4 pts 6-Apr-11

A 0.5 m3 vessel contains 10 kg of R-134a at -20oC. Determine… a.) The pressure in kPa

b.) The total internal energy in kJ

c.) The volume occupied by the liquid phase in m3

Read : Parts (a) and (b) are just a review of how to use thermodynamic data tables. The key is to use the given volume and mass to calculate the overal specific volume.

Given : Vtotal 0.5 m3 T -20 oC

mtotal 10 kg

Find : a.) P ??? kPa

b.) Utotal ??? kJ

c.) Vliq ??? m3

Assumptions : - The system exists in an equilibrium state. Solution :

Part a.) We know T, but we need to know the values of 2 intensive variables. So, lets use V and m to calculate the overall specific volume.

Eqn 1 V 0.05 m3

/kg

Next, we neeed to use the given T to obtain data from the Saturation Temperature table.

Sat. Liq Sat. Vap Sat. Liq Sat. Vap

-20 132.82 7.3620E-04 0.14729 25.39 218.84 Since : P= Psat 132.82 kPa Part b.) Eqn 2 Eqn 3 We can determine the quality from the specific volume of the mixture using :

Eqn 4

Now, we have everything we need to know to complete part (b). Plugging values into Eqns 4, 3 & 2, in that order yields :

x 0.3361 kg vap/kg

U 90.42 kJ/kg Utotal 904.2 kJ

Part (c) requires you to USE the liquid specific volume and vapor specific volume to determine how much of the mass is in the liquid and vapor phases.

Pressure (kPa) V (m3/kg) Temp. (oC) U (kJ/kg)

the system contains a saturated mixture and the pressure must be equal to the saturation pressure.

We need to know the specific internal energy of the system or vapor-liquid mixture so we can calculat the total internal energy in the system. The key equation is:

We could determine the specific internal energy using the following equation, if we only knew the quality !

total total

V

ˆ

V

m



sat sat liq vap

ˆ

V



V



V

total total

ˆ

U



m



U





sat sat sat

mix vap liq

ˆ

U



x U



1 

x U

sat liq

ˆ

V

x

ˆ

V







Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, 3.66

4/11/2011

(20)

Part c.)

Eqn 5 So, we need to determine the mass of liquid in the system.

This is not difficult because we know the quality ! The equation we need to use is :

Eqn 6 Now, we can plug values into Eqns 6 & 5, in that order, to complete the solution of this problem.

msat liq 6.639 kg Vsat liq 4.8873E-03 m

3

4.887 L Verify :

Answers : a.) P 132.8 kPa

b.) Utotal 904 kJ

c.) Vliq 4.887 L

We know the specific volume of the saturated liquid in the system. So, we could determine the volume occupied by the saturated liquid using :

sat liq sat liq

ˆ

sat liq

V



m



V





sat liq total

m



1 

x m

(21)

HW #1

Problem : 3.83 - Inflating an Automobile Tire - 6 pts 6-Apr-11

Read : Given : V 0.025 m3 T2 50 o C T1 25 o C Patm 100 kPa

P1,gauge 210 kPa R 8.314 J/mol-K

P1,abs 310 kPa MW 29 g/mol

Find : a.) P2 - P1 ??? kPa b.) m2-3 ??? g

Vamb ??? L

Assumptions : - The air in the tire is an ideal gas throughout both processes: heating and pressure relief. Solution :

Part a.)

Eqn 1 Eqn 2

No air leaves the tire during the heating process, so : n1 = n2 = 3.126 moles

Eqn 3 P2 336.0 kPa

The rise in pressure for the heating process is: P2- P1 26.0 kPa

Part b.)

Eqn 4 m1 90.67 g

n3 2.885 moles

Next, we apply Eqn 4 to determine the mass of air in the tire in state 3. m3 83.65 g

Eqn 5 m2-3 7.01 g

Now, we can solve Eqn 1 for V to determine the volume that the air removed from the tire would occupy at 100 kPaand 25oC.

Eqn 6 Vamb 6.00 L

Now, we know n2, V, T2, so we can apply the Ideal Gas EOS to state 2 (at the end of the heating process) to determine P2.

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25oC, the pressure gauge reads 210 kPa. If the volume of the tire is 0.025 m3 , determine the pressure rise in the tire when the air temperature rises to 50oC. Also determine the amount of air that must be bled off to restore pressure to its original value at this temperature (in kg and in L at Patm and

25oC). Assume atmospheric pressure is 100 kPa. Assume that the air in the tire behaves as an ideal gas, but then check the validity of this assumption

The keys to this problem are that the air is to be treated as an ideal gas and that the volume of the air in the tire remains constant throughout both processes: heating and pressure relief.

The first thing we need to do is determine the number of moles of air initially inside the tire. We can use the Ideal Gas EOS directly because we were given the values of P1, V and T1 in the problem statement. Be sure to use the absolute pressure !

Lets start by determining the mass of air initially inside the tire. We know the number of moles and the MW of air, so this should not be too difficult.

Next, lets determine how many moles of air must be in the tire at 50oC when the pressure is 210 kPa (gauge). This is the final state, state 3. We can do this by applying Eqn 2 to state 3.

Now, we can determine the mass of air that must be bled off from the tire and the volume this air would occupy at ambient T and P. P Vn R T n P V R T  2 2 2 n R T P V 

m

=

**n * MW**

2 3 2 3

m

_-

m

D

=

-

2 3



amb amb amb n n R T V P   

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, 3.83

4/11/2011

(22)

Verify :

Eqn 7 Plugging values into Eqn 7 for states 1 yields : V1 8.0 L/mol

Note that the molar volume in state 2 is the same as the molar volume in state !

Answers : a.) P2 - P1 26 kPa b.) m2-3 7.0 g

Vamb 0.006 m3

We need to verify the validity of the Ideal Gas EOS by determining the molar volume of the air in the tire in both the initial and final states.

Use the fact that the molar volume is V/nand rearrange Eqn 1 to obtain :

The molar volume is too small. It is not accurate to treat the air in the tire as an ideal gas. The error in doing so may exceed 1%.

If we consider air to be a diatomic gas, then the criterion for ideal gas behavior is :

V 20 L / mole R T V P  

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, 3.83

4/11/2011

(23)

HW #1

Problem : WB-3 - An Application of Equations of State - 8 pts 6-Apr-11

A 0.016773 m3 tank contains 1 kg of R-134a at 110oC. Determine the pressure of the refrigerant using: a.) The Ideal Gas EOS

b.) The Compressibility Factor EOS c.) The R-134a Tables

d.) The van der Waal EOS e.) The Soave-Redlich-Kwong EOS Read : Not much to say here.

Given : m 1kg Tc 374.25K

T 110oC Pc 4.067E+06Pa

383.15K  0.32684

V 0.016773m3 MW 102.03g / mol

R 8.314J/mol-K

Part a.) Ideal Gas EOS : Eqn 1

Solve Eqn 1 for pressure : Eqn 2

We must determine the molar volume before we can use Eqn 2 to answer the question.

Use the definition of molar volume: Eqn 3

Where : Eqn 4

Plugging values into Eqn 4 yields : n 9.80 mol H2O

Now, plug values back into Eqns 3 & 2. V 1.711E-03 m3/mol

Be careful with the units.

P 1.86E+06 Pa P 1861 kPa

Part b.) Compressibility EOS :

Eqn 5 Eqn 6

Plugging values into Eqn 6 yields : TR 1.0238

Definition of the ideal reduced molar volume : Eqn 7

Plugging values into Eqn 7 yields : VR

ideal

2.2369 Read the Generalized Compressibility Chart for PR = 0 to 1 :

PR 0.39 Z 0.865

We can use the definition of PR to calculate P :

Eqn 8 Eqn 9

Plugging values into Eqn 9 yields : _P _{1586 kPa}

Or, we can use Z and its definition to determine P : Eqn 10

Plugging values into Eqn 10 yields : _P _{1610 kPa}

Given TR and the ideal reduced molar volume, use the compressibility charts to evaluate either PR or

the compressibility, Z P V R T R T P V  _ V V n   m n MW  P V Z R T   ideal R c c V V R T / P    R c P P P  R c PP P IG R T P Z Z V V     R c T T T 

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, WB-3

4/11/2011

(24)

Part c.) The R-134a Tables provide the best available estimate of the pressure in the tank.

Eqn 11

Using the MW of R-134a yields : v 1.677E-05 m3/g

v 1.677E-02 m3/kg

The Superheated R-134a Table gives us : At P = 1.6 MPa

v = 0.016773 m3/kg

Fortunately, no interpolation is required : P 1600 kPa

Part d.) van der Waal EOS :

Eqn 12

Eqn 13 Eqn 14

a 1.0043 Pa-mol2/m6 b 9.56E-05 m3/mol

Now, we can plug the constants a and b into Eqn 12 to determine the pressure.

P 1629 kPa

Part e.) Soave-Redlich-Kwong EOS : Eqn 15

Eqn 16 Eqn 17

Eqn 18 Eqn 19

Eqn 20 Where :  0.327

Now, plug values into Eqns 15 - 20 :

TR 1.0238 a 1.01762 Pa-mol 2 /m6 m 0.9756 b 6.629E-05 m3/mol  0.97707 P 1610 kPa Answers : P (kPa)

a.) The Ideal Gas EOS 1861

b.) The Compressibility Factor EOS 1586

c.) The R-134a Tables 1600

d.) The van der Waal EOS 1629

e.) The Soave-Redlich-Kwong EOS 1610

We can determine the values of a and b, which are constants that depend only on the chemical species in the system, from the following equations.

We can determine the values of a, b and , which are constants that depend only on the chemical species in the system, from the following equations.

In either case, we begin by converting the molar volume into a specific volume :

Because T > Tc, the properties of the R-134a in the tank must be obtained from the superheated vapor table, even though the

water is actually a supercritical fluid in this system.

At this point we can make use of the fact that we have a pretty good idea of what the actual pressure is in the tank (from parts a-d) or we can scan the superheated vapor tables to determine which two pressures bracket our known value of the specific volume.

2 R T a P V b V      2 2 c c 27 R T a 64 P  c c R T b 8 P  V ˆ V MW  





R T a P V b V V b         2 2 c c R T a 0.42748 P  c c R T b 0.08664 P 





2 R 1 m 1 T    _   _ 2 m0.485081.55171 0.1561 R c T T T 

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, WB-3

4/11/2011

(25)

HW #1

Problem : WB-4 - Relative Humidity and Fogged Glasses - 5 pts 6-Apr-11

Read :

Given : Tin 20 oC

Tout 10 oC

hr 55%

Find : Can the humidity gauge be correct if the glasses do not fog ? Assumptions : 1- The humid air behaves as an ideal gas.

2- The glasses are at a temperature of 10oCwhen they enter the house. Solution :

Relative humidity is defined by : Eqn 1

Eqn 2 P*w = Psat(20

o

C) 2.3392 kPa

Pw 1.2866 kPa

Eqn 3 Now, let's consider what happens when the humid indoor air is cooled down to its dew point.

Eqn 4

P*w (Tdew) 1.2866 kPa

10 1.2281

Tdew 1.2866

15 1.7057

Tdew 10.61oC

On entering a dwelling maintained at 20oC from the outdoors where the temperature is 10oC, a person's eye-glasses are observed not to become fogged. A humidity gauge indicates the relative humidity in the dwelling is 55%. Can this reading be correct? Provide supporting calculations.

The key to this problem is to assume that the glasses are at 10oCwhen they come in from the outdoors and to recognize that if the dew point of the indoor air is GREATER THAN 10oC, then when the humid air hits the cold glasses, water will condense onto the glasses and they will "fog". Since the glasses did not actually fog, we know that the dew point temperature of the humid air in the house is LESS THAN 10oC. So, we can calculate Tdew of indoor air at hr =55% and compare to 10

o C to see if the 55% relative humidity is possibly correct in light of the fact that the glasses did not fog.

The water in the humid air in the house will condense on the glasses if 10oC is less than or equal to the dew point temperature of the air in the house.

So, we can determine whether the humidity gauge is correct by comparing the dew point of the humid air in the house to 10oC. If Tdew > 10

o

C, then the glasses SHOULD have fogged and we can conclude that the humidity gauge is not correct. If Tdew <

10oC then we cannot determine whether the humidity gauge is correct

We know the relative humidity of the indoor air and we can look up the vapor pressure of water at the indoor T of 20oC, so we can solve Eqn 1 for the partial pressure of water in the indoor air and evaluate it.

The key to solving this problem is that the mole fraction of water in the humid indoor air remains constant as it approaches the surface of the cold glasses. The total pressure in the room also remains constant. So, if the indoor air is an ideal gas, then the partial pressure of water in the indoor air also remains constant because :

Temp. (oC)

Pressure (kPa) At the dew point temperature, the relative humidity is 100%.

Since the partial pressure of water in the humid indoor air is constant, we can use Eqn 4 to evaluate the vapor pressure of water at the dew point temperature.

Now, we can go back to the saturation temperature tables and interpolate to detemine the T at which the vapor pressure is 1.2866 kPa. This T is the dew point !

 

w r * w P h P T 





* w w dew P P T

 

* w r w P h P T w w total P y P

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, WB-4

4/11/2011

(26)

Verify :

Eqn 5 Eqn 6

Assume atmospheric pressure : P 101.325 kPa

R 8.314 J/mol-K

Plugging values into Eqn 6 yields : V1 24.05 L/mole

V2 23.23 L/mole

Answers :

Since the molar volume of both the indoor and outdoor air exceed 20 L/mole, it is accurate to 2 significant figures to treat the air as an ideal gas.

Since Tout < Tdew, the glasses SHOULD have fogged when they came into the house if the humidity were really 55%.

However, since the glasses did not fog, we can conclude that the actual relative humidity was less than 55% and the humidity gauge is not correct.

We cannot consider humid air to be a diatomic gas, so the criterion for ideal gas behavior is :

We need to verify the validity of the Ideal Gas EOS by determining the molar volume of the humid air in both the indoor and outdoor states. V 20 L / mole P V R T V R T P  

Dr. Baratuci - ENGR 224

hw1-sp11.xlsm, WB-4

4/11/2011