School of Engineering, FETBE

EM203/208 (January 2013 Semester) - Assignment 2

1. piston–cylinder device initially contains steam at 200 kPa, 200°C, and 0.5 m3. At this state,

a linear spring (F x) is touching the piston but exerts no force on it. Heat is now slowly

transferred to the steam, causing the pressure and the volume to rise to 500 kPa and 0.65 m3,

respectively. Show the process on a P-v diagram and determine (a) the final temperature, (b)

the work done by the steam, and (c) the total heat transferred.

2. Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2-kg

steam at 2 MPa and 300°C while Tank B contains 3-kg saturated liquid–vapor mixture with a

vapor mass fraction of 60 percent. Now the partition is removed and the two sides are allowed

to mix until the mechanical and thermal equilibrium are established.

If the pressure at the final state is 400 kPa, determine (a) the temperature and quality of the

steam (if mixture) at the final state and (b) the amount of heat lost from the tanks.

3. A steam turbine operates at steady state with a steam mass flow rate of 4600 kg/h,

producing a power output of 1000 kW. At the inlet, the pressure is 4 MPa, the temperature is

500C, and the velocity is 10 m/s. At the exit, the pressure is 0.1 bar, the quality is 0.9 (90%),

and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and

surroundings, in kW.

4. Refrigerant-134a is compressed in a compressor from an inlet state of 100 kPa, 10C to an

outlet state of 1200 kPa, 60C. The compressor is water cooled with a heat loss amounting to

40 kW. The shaft work input is 150 kW. What is the mass flow rate (in kg/s) of the refrigerant

through the compressor?

5. A small turbine is operated at part load by throttling a 0.25 kg/s steam supply at 1.4 MPa,

250°C down to 1.1 MPa before it enters the turbine and the exhaust is at 10 kPa. If the turbine

produces 110 kW, find the exhaust temperature and quality.

100 kPa10C

1200 kPa60C

1.4 MPa250C

1.1 MPa

10 kPa

110 kW

Solution

1.

(a) We take the contents of the cylinder as the system. This is a closed system since no mass

enters or leaves. Noting that the spring is not part of the system (it is external), the energy balance

for this stationary closed system can be expressed as

E¿−Eout⏟Net energy transfer by heat, work and mass

= ∆ E system⏟Change in internal, kinetic,potential, etc. energies

Q¿−W b , out=∆ U=m (u2−u1 ) (since KE = PE = 0)

Q¿=m ( u2−u1 )+W b ,out

The properties of steam are (Tables A-4 through A-6)

P1=20 0 kPaT 1=200 ° C }v1=1.08049 m3 /kg

u1=2654.6 kJ/kg

m=V 1

v1

= 0.5 m3

1.08049 m3 /kg=0.4628 kg

v2=V 2

m= 0.65 m3

0.4628 kg=1.4 04 5m3 /kg

P2=500 kPa

v2=1.4045 m3 /kg} T 2=1249℃u2=4575 kJ/kg

(by interpolationusing databelow )

Extract from 0.5 MPa section of Table A-6.

(b) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V

diagram will be a straight line.

The boundary work during this process is simply the area under the process curve, which is a

trapezoidal. Thus,

W b=P1+P2

2(V 2−V 1 )=

(200+500 ) kPa2

(0.65−0.5 ) m3=52.5 kPa∙ m3=52.5 k J

(c) From the energy balance we have

Q¿=m ( u2−u1 )+W b ,out= (0.4628 kg ) ( 4575−2654.6 ) kJ/kg + 52.5k J = 941.3 kJ

2. We take the contents of both tanks as the system. This is a closed system since no mass enters

or leaves. Noting that the volume of the system is constant and thus there is no boundary work,

the energy balance for this stationary closed system can be expressed as

E¿−Eout⏟Net energy transfer by heat, work and mass

= ∆ E system⏟Change in internal, kinetic,potential, etc. energies

−Qout=∆ U A+∆ UB=[m (u2−u1 ) ]A+[m (u2−u1 ) ]B (since W = KE = PE = 0)

The properties of steam in both tanks at the initial state are

P1 , A=2 M PaT1 , A=3 00 ° C }v1 , A=0.12551 m3 /kg

u1 , A=2773.2 kJ/kg

T1 , B=15 0° Cx1 , B=0.6 }v1 , B=v f +x1 , B v fg=0.001091+0.6 0 (0.39248−0.001091 )=0 .235 9m3 /kg

u1 , B=u f +x1 ,B u fg=631.66+0. 60 (19 27.4 )=1788.1 kJ/kg

The total volume f the system is

V=V A+V B=mA v1 , A+mB v1 , B= (2kg ) (0.12551 m3 /kg )+ (3 kg ) (0 . 235 9 m3 /kg )=0 . 95 87 m3

The specific volume at the final state is

v2=Vm

=0.9587 m3

5 kg=0.1917 m3 /kg

We can then determine the other properties at the final state as:

P1

V1

P2

V2

P2=400 kPa

v2=0.1917 m3 /kg} T2=143.61℃u2=u f +x2u fg=604.22+(0.413 ) (1948.9 )=1409.1 kJ/kg

x2 is obtained based on data below (taken from Table A-5):

x2=v2−v f

v g−v f

= 0.1917−0.0010840.46242−0.001084

=0.413

(b) Substituting,

−Qout=[m (u2−u1 ) ]A+ [m (u2−u1 ) ]B

¿ (2 kg ) (1409.1−2773.2 ) kJ/kg+(3 kg ) (1409.1−1788.1 ) kJ/kg

=−3865.2 kJ/kg

or, Qout=3865.2 kJ/kg

3.

There is only one inlet and one exit, and thus m1=m2=m. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

E¿− Eout⏟Rate of net energy transfer by heat, work and mass

= ∆ E system⏟Rate of change in internal,kinetic, potential, etc. energies

=0

m(h1+V 1

2

2 )−m(h2+V 2

2

2 )−W out−Qout=0

Thus

4600 kg/h4 MPa500C10 m/s

0.1 bar (10 kPa)90% quality

50 m/s

W out=1000 kW

steady

Qout=m [ (h1−h2)+(V 12−V 2

2

2 )]−W out

P1=4 M PaT 1=5 00 °C }h1=3446.0 kJ/kg

P2=10 kPax2=0.9 }h2=h f +x2 hfg=191.81+0.9 0 (2392.1 )=2344.7 kJ/kg

m=(4600kgh )( 1 h

3600 s )=1.28 kg /s

V 12−V 2

2

2=[ (10 m /s )2−(50 m /s )2

2 ]( 1 kJ/kg1000 m2 / s2 )=−1.2 kJ/kg

Substituting

Qout=(1.28kg/ s) [ (3446.0−2344.7 ) kJ/kg−1.2 kJ/kg ]−1000 k W=1101 kW

4.

There is only one inlet and one exit, and thus m1=m2=m. We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

E¿− Eout⏟Rate of net energy transfer by heat, work and mass

= ∆ E system⏟Rate of change in internal,kinetic, potential, etc. energies

=0

m h1+W ¿−m h2−Qout=0

Thus

100 kPa10C

steady

m=Qout−W ¿

h1−h2

P1=100 kPaT 1=−1 0 °C }h1=247.49 kJ/kg

P2=1200 kPaT 2=5 0 °C }h2=278.27 kJ/kg

Substituting

m=(40−150 ) k J/s

(247.49−278.27 ) k J/kg=3.57 k g/s

5.

There is only one inlet and one exit, and thus m1=m2=m. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

E¿− Eout⏟Rate of net energy transfer by heat, work and mass

= ∆ E system⏟Rate of change in internal,kinetic, potential, etc. energies

=0

m h2−m h3−W out=0 (assuming turbine is adiabatic)

h3=h2−W out

m

For the throttling process, h1 = h2. Hence

P1=1.4 MPaT 1=25 0 °C }h1=2927.9 kJ/kg=h2

Substituting

h3=2927.9 kJ/kg− 110 kJ /s0.25 kg /s

=2487.9 kJ/kg . Thus

1.4 MPa250C

1.1 MPa

10 kPa

110 kW

steady

P3=10 k Pah3=2487.9 kJ/kg}

Inspection of the data from Table A-5 shows that hf < h3 < hg. Hence state 3 is a saturated mixture

phase and thus T3 = Tsat @ 10 kPa = 45.81C.

The quality is

x3=h3−hf

hfg

=2487.9−191.812392.1

=0.96

Therefore

P3=10 k Pah3=2487.9 kJ/kg}T 3=45.81C

x3=0.96