Next, we consider unsteady state

(1)

여재익, [email protected], (02) 880-9334 - 2006 Spring - 11

Next, we consider unsteady state Next, we consider unsteady state

• Unsteady Process

• filling closed tanks with a gas or liquid.

• discharge from closed vessels.

Basic assumptions are as follows:

• The control volume remains constant in relation to the coordinate frame.

• The state of the mass within the control volume may change with time,

• The state of the mass crossing each of the areas of flow on the control surface is constant with time, although the mass flow rate may vary with time.

=0

− +

∑

^e

∑

ⁱ

cv m m

dt

dm ^or

( m

₂

− m

₁

)

^cv

+ ∑ m

^e

− ∑ m

ⁱ

= 0

^(*)

The first law becomes

Or

cv e e e e cv

i i i i

cv

V gz W

h dt m

gz dE h V

m

Q + ∑ ( + 2

²

+ ) = + ∑ ( + 2

²

+ ) +

cv cv e

e e e

i i i i cv

W V gz

u m V gz

h m

V gz h m Q

+ +

+

− + + +

+ +

= + + +

∑

2 )]

( 2 )

( [ 2 )

(

2 ) (

1 2 1 1 1 2 2 2 2 2 2

---(**) To make any sense to these equations, consider example.

• Example 5-1

Steam at a pressure of 1.4Mpa, 300°C, is flowing in a pipe. An

evacuated tank is connected to this pipe through a valve. The valve is opened, the tank fills with steam until the pressure is 1.4 Mpa, and then the valve is closed. The process takes place adiabatically.

Kinetic energies and potential energies are negligible. Determine the final temperature of the steam.

0 0

=

= PE KE Q

δ δ δ

= 0 m

e

• Solution to 5-1

1st law for this unsteady process is

The first law becomes

using continuity (*)

then the first law gives

cv e

e i

i

cv

m h m h m u m u W

Q + ( ) = ( ) +

₂ ₂

−

₁ ₁

+ 0

, 0

₁

=

= m m

_e

2 2

u m h m

_i _i

=

0 )

( m

₂

− m

₁

+ m

_e

− m

_i

=

1

2

m

m =

u

2

h

_i

=

(2)

Solution continued…

That is, the final internal energy of the steam in the tank is equal to the enthalpy of the steam entering the tank.

Look up superheated steam tank at

C T

MPa

p = 1 . 4 , = 300 °

^{, we find}

kJkg h_i =3040.4

4 2

.

3040

u

h

_i = =

T

₂ ^p² ⁼¹^.⁴^MPa

C T

₂

~ 450 °

Then use to find state from the table at

• Example 5-2

Consider the same tank as in 5-1. Let volume , saturated vapor at . The valve is then opened and steam from the line at 1.4 Mpa, 300°C flows into the tank until the pressure is 1.4 Mpa.

Calculate the mass of steam that flows into the tank.

4 3

. 0 m V= kPa

p₁=350

= 0 m

e

• Solution to 5-2

First consider mass balance (*)

1^stLaw (**) becomes,

From saturated steam table at of saturated vapor. Thus

1

0

2

− m + m

_e

− m

_i

= m

1 1 2

2

u m u

m h

m

_i _i

= −

m kg v

kPa

p

₁

= 350 ,

_g

= 0 . 52

³

kg kg m

m v

m V

g

77 . 52 0

. 0

4 . 0

3 3 1

1

= = =

u kg kJ u

_g

= 2548 =

Solution 5-2 continued…

Again from the superheated steam table, as before at

(Trial & Error) Assume

Look up steam (superheated) vapor table at and

Must use 1st law (**) to check if guess was good.

kg kJ h

C T

MPa

p = 114 , = 300 ° ,

_i

= 3040 . 4

C T

₂

= 342 °

C

T₂=342° p₂=1.4MPa kg

kJ u

kg m

v₂=0.1974 ³ , ₂=2855.8 -> kg

v

m V 2.026

2 2

2 = =

T2

3823 74 . 3818

) 2548 ( 77 . 0 ) 8 . 2855 ( 026 . 2 4 . 3040 ) 77 . 0 026 . 2 (

)

( ₂ ₁ ₂ ₂ ₁ ₁

≈

−

≈

−

+

=

−m h mu mu

m _i

(3)

OK guess was good!

Using C (*)

kg m m m_i

256 . 1

77 . 0 026 . 2

1 2

=

−

=

−

=

• Example 5-3 (steady state case)

The mass rate of flow into a steam turbine is , and the heat transfer from the turbine is . The following data are known for the steam entering and leaving the turbine.

8 . 5 kW

s kg

5 . 1

Determine the Power ( ) of the turbine.

W

• Solution to 5-3

This is a steady state problem.

Thus c

1^stLaw (**) becomes

For , use superheated vapor table at

i e

i e cv

m m or

m m m m

=

− +

− ) 0

( ₂ ₁

cv e e e e cv i i

i

cv V gz W

h t m d E gz d h V

m

Q + + + = + + + )+

( 2 2 )

(

2 2

1

hi p_i=2MPaandT_i =350°C Æ h_i =3137kJkg

In the case of exit, use saturate vapor

kJ kg h

h

x MPa p

g e

e e

5 . 2675

% 100 ,

1 . 0

=

>

−

=

Upon substitution, kW W =655.7

• TA will solve extra practice problems from Chapter 5.

• We will next study the Second Law of Thermodynamics!

• GOOD LUCK, folks!

(4)

• Homework Set #4

• due 4/6 (We will have Midterm #1 on the same day)

• 5-3, 5-4, 5-6, 5-9, 5-15, 5-20, 5-26, 5-27, 5-31, 5-33