P a g e | 1

MSE 321 (Fall 2013)

Dr. Peyman Taheri

MSE 321

Engineering Thermodynamics & Heat Transfer

Assignment 3 Due Date: Oct. 4

Turn in your homework problems in order, and always write down all the steps that led you to the answer. If you do not turn in a problem, write the number and letter, e.g. 2(b), and write ‘no answer’. When substance properties are needed, use the table in the textbook. Please staple your homework.

Problem 1 A piston–cylinder device with a set of stops initially contains 0.3 kg of steam at 1 MPa and 400°C. The location of the stops corresponds to 60 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is (a) 1 MPa and 250°C and (b) 500 kPa. (c) Also determine the temperature at the final state in part b.

Solution (a) Since weight of the piston and the outside pressure (Patm) are not changing, during free movement of the piston, the pressure inside the cylinder is constant (1 MPa). The specific volumes for the initial and final states are (Table A-6)

1 23 31 2

1 2

1 MPa 1 MPa0.30661 m /kg 0.23275 m /kg

400 C 250 C

P P

T T

ü ü= =ï ïï ïé ù é ù = =ý ýê ú ê úë û ë ûï ï= = ï ïþ þ

When piston sits on the steps the specific volume is,

310.6 0.6 0.30661 0.183966 m /kg é ù= = ´ = ê úë û

Since pressure at state 2 is equal to pressure at state 1 and 2 10.6 > , we can conclude that at state 2, the

piston is above the steps, and the boundary work due to the piston movement is

31 2( ) (0.3 kg)(1000 kPa)(0.30661 0.23275)m /kgbW mP = - = - = 22.16 kJ

(b) Since pressure is changed in state 2, then the piston movement has been affected by the steps and the volume of the cylinder at state 2 is 60% of initial volume. Then, the boundary work becomes

31 1( 0.60 ) (0.3 kg)(1000 kPa)(0.30661 0.60 0.30661)m /kgbW mP = - = - ´ = 36.79 kJ

The temperature at the final state is,

P a g e | 2

MSE 321 (Fall 2013)

Dr. Peyman Taheri

2

232

0.5 MPa

0.183966 m /kg

PT

ü= ïï = ýï= ïþ151.8 C (Table A-5)

Note: The reason for pressure decrease is the condensation of steam inside the cylinder as the result of heat removal Q.

Problem 2 A mass of 2.4 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during this process. Plot the process in P - diagram.

Solution The process must be considered as a quasi-equilibrium process for an ideal gas (air). The boundary work is determined from its definition to be

( )( )( )2

2 1,ou 1 1 1

1 21

150 kPaln ln 2.4 kg 0.287 kJ/kg K 285 K ln

600 kPab t

PW Pd P mRT

P

æ ö æ ö æ ö÷ ÷ç ç ÷ç÷ ÷= = = = ⋅ =-ç ç ÷ç÷ ÷ ÷ç ç ç÷ ÷ç ç è øè ø è øò 272 kJ VV V V

The negative sign indicates that work is done on the system (work input).

Problem 3 2-kg of saturated liquid R-134a with an initial temperature of −10°C is contained in a well-insulated, weighted piston–cylinder device. This device contains an electrical resistor to which 10 volts are applied causing a current of 2 amperes to flow through the resistor. Determine the time required for the refrigerant to be converted to a saturated vapor, and the final temperature.

P a g e | 3

MSE 321 (Fall 2013)

Dr. Peyman Taheri

Solution The compression or expansion process is quasi-equilibrium. We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this “stationary” closed system can be expressed as,

in( Q=0 (insulated)

out(insulated) 0Q

=- in out system) ( )W W U+ - =

Note that in the above equation sign convention is not applied, but if one wants to use the formula with sign convention should consider electrical work as input (with negative sign) and boundary work as output (with positive sign).

( )in, electrical out, boundary system 2 1 (since KE PE 0)W W U m u u - = = - = =

Combining the boundary work ( )2 1bW mP = - with internal energy u, gives the enthalpy,

in, electrical 2 1 2 1( ) fgW H H m h h mh= - = - =

Also

in, electrical

in, electrical @ 10 C

(10 V)(2 A) 20 Wand

205.96 kJ/kg (Table A-11)fgfg

W IW t mh

h

-

ì = = =ïï= íï =ïî

V

Substituting,

(0.020 kJ/s) (2 kg)(205.96 kJ/kg) 20,596 st t = = =5.72 h

The temperature remains constant during this phase change process,

2 1T T= =- 10 C

Problem 4 One ton of liquid water at 80°C is brought into a well-insulated and well-sealed room which has the dimensions of 4m 5m 6m´ ´ . The air in the room is initially at 22°C and 100 kPa. Assuming constant

specific heats for air and water at room temperature, determine the final equilibrium temperature in the room.

Solution The thermal properties of water and air are constant since specific heats are assumed to be constants (at room temperature).

Room temperature is a general term describing common indoor temperatures. It is usually 20 °C (68 °F or 293K) or 25 °C (77 °F or 298K). [For more info see Wikipedia]

The gas constant of air is approximated R = 0.287 kPa m3/kg K (Table A-1). The specific heat of water at

room temperature is cP = 4.18 kJ/kgC (Table A-3).

P a g e | 4

MSE 321 (Fall 2013)

Dr. Peyman Taheri

The volume and the mass of the air in the room are

( )( )( )( )

[ ]3

3 1 1air 3

1

100 kPa 120 m4 5 6 120 141.7 kg

0.2870 kPa m /kg K 295 K

Pm m

RTé ù= ´ ´ = = = =ê úë û ⋅ ⋅

VV

Taking the contents of the room, including the water, as our system, the energy balance can be written as,

inQ=0 (insulated)

out(insulated) 0Q

=-( ) inW+ out0

W=

-( ) system water air00U U U

== = +

or ( ) ( )2 1 2 1water air0 0Pmc T T mc T Té ù é ù= - + - =ë û ë ûV

Substituting,

0 (1000 kg)(4.180 kJ/kg C)( 80) C (141.7 kg)(0.718 kJ/kg C)( 22) C 0f fT T= ⋅ - + ⋅ - =

It gives,

Tf = 78.6C

where Tf is the final equilibrium temperature in the room.

Problem 5 Air is contained in a variable-load piston device equipped with a paddle wheel. Initially, air is at 500 kPa and 27°C. The paddle wheel is now turned by an external electric motor until 50 kJ/kg of work has been transferred to air. During this process, heat is transferred to maintain a constant air temperature while allowing the gas volume to triple. Calculate the required amount of heat transfer, in kJ/kg.

Solution Air can be treated as an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 132.5 K and 3.77 MPa. Also, the kinetic and potential energy changes are negligible. Constant specific heats is used for air.

Take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as,

in outQ Q-( ) ( ) ( ) ( )in out system in 2 1pw bW W U Q W W mc T T+ - = + - = -V1 20 (since )T T= =

Then, in inin unit mass formb pw b pwQ W W q w w= - = -

Using the boundary work relation on a unit mass basis for the isothermal process of an ideal gas gives

( ) ( )2

1

ln ln 3 (0.287 kJ/kg K)(300 K)ln 3 94.6 kJ/kgbw RT RTæ ö÷ç ÷= = = ⋅ =ç ÷ç ÷çè ø

VV

Substituting into the energy balance equation (expressed on a unit mass basis) gives

in 94.6 50b pwq w w= - = - = 44.6 kJ / kg