ME 200 Fall 2xxx

Final Exam NAME:______________________________________ PUID:_______________________________________ Circle your section: aaa bbbb cccc dddd eeee 7:30AM 10:30AM 1:30PM 3:30PM 4:30PM Instructions: This is a closed book/note exam. You may use a calculator. You must start from the most basic form of the equations and simplify accordingly to receive full credit. Show all your work! Keep your eyes on your own paper. If you are caught cheating you will get a zero for the exam and your name will be turned over to the Dean of Students. Score:

PROBLEM TOTAL POSSIBLE

YOUR SCORE

1 50

2 25

3 25

4 25

5 25

TOTAL 150

1 Short Questions a) Score 2 points each by entering T for True and F for false in the boxes in front of

each of the ten statements T The force exerted on the wall of an evacuated tank at 5 feet in a diving pool equals half the force exerted on the wall at 10 feet from the surface of water in the same pool.

Two degrees (Celsius) above normal body temperature measured in Europe is more serious than 2 degrees (Fahrenheit) above normal body temperature measured in the U. S. A bicyclist travelling a distance of 5 miles on a flat terrain at 10 miles/hour expends as much energy as she would have travelling the same distance at 20 miles/hour.

Routine processes require that a HVAC unit transfer 20 kW of heat from a perfectly insulated home maintained at 70 F to the surroundings at an average temperature of 80 F. The HVAC unit consumes less than 1 kW power. The quality of a liquid-vapor mixture is defined as the mass of the liquid divided by the mass of the mixture. The quality of a liquid-vapor mixture is defined as the mass of the liquid divided by the mass of the mixture. The entropy of air at 20 C heated to 40 C increases by 0.07 kJ/kg-K in a constant pressure heating process. The second law of thermodynamics requires that entropy of a system must always increase. The second law of thermodynamics also requires that the entropy of a system plus the entropy of the surroundings can increase or decrease.

Known: A Rankine cycle operates with steam as the working fluid. The boiler and condenser

pressures are specified, and the mass flow rate is given.

Find: Determine the output of the turbine; the power input to the pump; the thermal efficiency of

the plant.

Sketch:

Assumptions:

(1) Each component of the cycle is analyzed as a control volume at steady state

(2) All processes of the working fluid are internally reversible

(3) The turbine and pump operate adiabatically

(4) Kinetic and potential energy effects are negligible

Analysis:

From the table, at pressure 60 bar, the temperature of saturated liquid-vapor mixture is 275.6 oC;

at pressure 0.06 bar, the temperature of saturated liquid-vapor mixture is 36.16 oC.

So T1=(275.6+100) oC=375.6

oC, T3=36.16

oC.

State 1: T1=375.6 oC, p1=60 bar, by interpolation,

60 bar

0.06 bar

h1=3112.5 kJ/kg, s1=6.4383 k J / k g K

State 2: 2 1

6 .4 3 8 3 /s s kJ kg K , which is greater than 2 f

s and smaller than 2 g

s , so state 2

is in mixture state. T2=36.16 oC.p2=0.06 bar, by interpolation,

h2=1867.3 kJ/kg.

State 3: p3=0.06 bar, saturated liquid,

3 3

31 .0 0 6 4 1 0 /v m kg

, h3=151.53 kJ/kg

State 4: p4=60 bar,

4 3 3 4 3( )h h v p p =151.53 kJ/kg+(

3 31 .0 0 6 4 1 0 /m kg

)(60 bar-0.06 bar)(

1 0 0 kP a

b a r)

=157.56 kJ/kg

(i) Power output of turbine:

1 2( ) (1 8 / ) (3 1 1 2 .5 / 1 8 6 7 .3 / )

tW m h h kg s k J kg k J kg

=22413.6 kW

(ii) Power input to the pump:

4 3

( ) (18 / )(157 .56 / 151 .53 / )p

W m h h kg s kJ kg kJ kg

=108.54 kW

(iii) 1 4

( )in

Q m h h

Thermal efficiency:

1 2 4 3

1 4

( ) ( ) [(3 1 1 2 .5 1 8 6 7 .3) (1 5 7 .5 6 1 5 1 .5 3)] /

( ) (3 1 1 2 .5 1 5 7 .5 6 ) /

t p

in

W W h h h h kJ kg

Q h h kJ kg

=0.419 (41.9%)

(iv) Differences with Carnot cycle

a) This cycle has a lower thermal efficiency than the Carnot cycle having the same maximum and

minimum temperature.

b) Heat transfer across the system boundary of Carnot Cycle is isothermal (constant temperature),

and heat transfer across the system boundary in a Rankine cycle is isobaric (constant pressure).

Known: An air-standard Otto cycle with a given value of compression ratio executed with

specified conditions at the beginning of the compression stroke and a specified maximum

temperature during the cycle

Find: Determine the thermal efficiency of the cycle; heated added assuming variable specific heat;

percentage differences between the heat added and the heat rejected if the specific heat is constant.

Sketch:

Assumptions:

(1) The air in the piston–cylinder assembly is the closed system.

(2) The compression and expansion processes are adiabatic.

(3) All processes are internally reversible.

(4) The air is modeled as an ideal gas.

(5) Kinetic and potential energy effects are negligible.

Analysis:

(a) Shown in the “Sketch” part.

At T1=300K, u1=214.07, 1

6 2 1 .2r

v .

For isentropic compression process 1-2,

2

2 1

1

6 2 1 .27 7 .6 5

8r r

Vv v

V

By interpolation we get T2=673.09 K, and u2=491.22kJ/kg . With the idea gas equation

2 1

2 1

1 2

6 7 3 .0 9(1 0 0 )( )8 1 7 9 4 .9 1

3 0 0

T V Kp p kP a kP a

T V K

Process 2-3 is a constant volume process, thus

3

3 2

2

1 4 0 0(1 7 9 4 .9 1 )( ) 8 3 7 6 .2 5

3 0 0

T Kp p kP a kP a

T K

At T3=1400K, u3=1113.52 kJ/kg, vr3=8.919

For the isentropic expansion process 3-4

4

4 3

3

8 (8 .9 1 9 ) 7 1 .3 5r r

Vv v

V

Interpolation in the table gives T4=694.32K, u4=507.86kJ/kg.

(b)

The thermal efficiency is

4 1 4 1

2 3 3 2

/ (5 0 7 .8 6 2 1 4 .0 7 ) /1 1 1 5 2 .8 %

/ (1 1 1 3 .5 2 4 9 1 .2 2 ) /

Q m u u kJ kg

Q m u u kJ kg

(c)Heat added:

2 3 3 2/ 1 1 1 3 .5 2 / 4 9 1 .2 2 / 6 2 2 .3 /Q m u u kJ kg kJ kg kJ kg

(d)For heat added,

2 3 3 2/ 6 2 2 .3 /Q m u u kJ kg

If a constant specific heat was used(v

c is determined by interpolation),

2 3 3 2 3 2( / ) ( ) ( ) (0 .8 6 3 / )(1 4 0 0 6 7 3 .0 9 )

vQ m u u c T T kJ kg K K K

=627.32 kJ/kg

The percentage difference is (6 2 7 .3 2 6 2 2 .3 ) /

1 0 0 % 0 .8 1 %6 2 2 .3 /

kJ kg

kJ kg

For heat rejected,

4 1 4 1/ (5 0 7 .8 6 2 1 4 .0 7 ) / 2 9 3 .7 9 /Q m u u kJ kg kJ kg

If a constant specific heat was used (v

c is determined by interpolation),

4 1 4 1 4 1( / ) ( ) ( ) (0 .7 4 1 / )(6 9 4 .3 2 3 0 0 ) 2 9 2 .1 9 /

vQ m u u c T T kJ kg K K K kJ kg

The percentage difference is ( 2 9 2 .1 9 2 9 3 .7 9 ) /

1 0 0 % 0 .5 4 %2 9 3 .7 9 /

kJ kg

kJ kg

Known: Steam goes through a diffuser at steady state from a specified inlet state to a specified

exit state.

Find: The exit pressure of the diffuser; the isentropic efficiency.

Sketch:

Assumptions:

(1) Change of potential energy between the inlet and outlet can be neglected

(2) Steady state

(3) Adiabatic reversible process, and 0C V

W

Analysis:

(i) For state 1, T1=200oC, p1=0. 3MPa, so h1=2865.5 kJ/kg, s1=7.3115 /kJ kg K

Since the process is isentropic, s2= s2s= s2a=7.3115 /kJ kg K

The energy rate balance for the control volume enclosing the diffuser reads:

2 2

2 1

2 _ 1

2 2s a

V Vh h

2 2

1 2

2 _ 1

2 2s a

V Vh h

2 2

3(7 9 5 / ) (7 0 / )2 8 6 5 .5 1 0 /

2 2

3 1 7 9 .0 6 /

m s m sJ kg

k J kg

So T2=360oC, and

2 _s ah =3179.06 kJ/kg.

From the table, it can be found that the exit pressure is 1 MPa.

(ii) For the diffuser, if all the kinetic energy was extracted,

2

1

2 _ 1

2s id ea l

Vh h

2

3(7 9 5 / )2 8 6 5 .5 1 0 /

2

m sJ kg

=3

3 1 8 1 .5 1 1 0 /J kg

= 3 1 8 1 .5 1 /kJ kg

3 1 7 9 .0 6 / 2 8 6 5 .5 /9 9 .2 3 %

3 1 8 1 .5 1 / 2 8 6 5 .5 /

a c tu a l

d if fu se r

id ea l

h kJ kg k J kg

h k J kg k J kg

Known: A turbo-supercharger consisting of an exhaust gas-driven turbine directly connected to

an air compressor.

Find: (i) turbine exit temperature and power output (ii) compressor exit pressure and temperature

(iii) repeat (i) and (ii) by considering the isentropic efficiencies

Assumptions:

(1) Steady state

(2)Turbine and compressor are reversible and adiabatic

(3) Kinetic energy and potential energy are negligible

(4) Air can be modeled as ideal gas

Analysis:

(i) state 1: p1=100 kPa, T1=30oC(303.15K), h1=303.36 kJ/kg, pr1=1.438

state 3: p3=170 kPa, T3=850oC (1123.15 K), pr3=181.81, h3=1187.95 kJ/kg

state 4: 4 4

3 3

r

r

p p

p p

4

4 3

3

1 0 0(1 8 1 .8 1) 1 0 6 .9 5

1 7 0r r

p kP ap p

p kP a

47 1 0 .8 3 (9 8 3 .9 8 )

oT C K

41 0 2 7 .7 8 /h kJ kg

Considering all the components as a whole system to be the control volume, we can get the net

energy output from the engine:

4 1( ) (0 .1 / ) (1 0 2 7 .7 8 / 3 0 3 .3 6 / )Q W m h h kg s k J kg kJ kg

72 .4 kW

(ii) Take the engine as the control volume,

3 2

2 3

( )Q W m h h

Q Wh h

m

7 2 .41 1 8 7 .9 5 /

0 .1 /

4 6 3 .9 5 /

kWkJ kg

kg s

kJ kg

2 24 6 1 .8 9 , 6 .3 3 9

rT K p

2

2 1

1

6 .3 3 9(1 0 0 ) 4 4 0 .8 2

1 .4 3 8

r

r

pp p kP a kP a

p

(iii) 3 4

( ) 1 1 8 7 .9 5 / 1 0 2 7 .7 8 / 1 6 0 .1 7 /t

s

Wh h kJ kg kJ kg kJ kg

m

( ) (0 .8 5 )(1 6 0 .1 7 / ) 1 3 6 .1 4 /t t

t s

W WkJ kg kJ kg

m m

( ) (0 .8 )(1 3 6 .1 4 / ) 1 0 8 .9 1 /C t

C

W WkJ kg kJ kg

m m

2 _ 13 0 3 .3 6 / 1 0 8 .9 1 / 4 1 2 .2 7 /

C

a

Wh h kJ kg kJ kg kJ kg

m

4 _ 31 1 8 7 .9 5 / 1 3 6 .1 4 / 1 0 5 1 .8 1 /

t

a

Wh h kJ kg kJ kg kJ kg

m

2 _ 4 _4 1 1 .1 3 , 1 0 0 5 .0 5

a aT K T K

Engine power:

3 2 _( ) (0 .1 / )(1187 .95 / 412 .27 / ) 77 .57

aQ W m h h kg s kJ kg kJ kg kW