Question

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s and leaves at 375°C and 400 kPa while losing heat at a rate of 27 kW. For an inlet area of 800 cm^2, determine the velocity and the volume flow rate of the steam at the nozzle exit. Use steam tables.

Answer 1

Answer:

Using the equation of continuity:

A

1

v

1

=

A

2

v

2

0.08

(

10

)

=

A

2

(

225

)

A

2

=

3.55

×

10

−

3

m

2

Q

2

=

A

2

v

2

Q

2

=

3.55

×

10

−

3

×

225

Q

2

=

0.798

m

3

/

s

Explanation:

Steady Flow Energy Equation:

The steady flow energy equation is a representation of the first law of thermodynamics. It is the conservation of energy law for an open system. A nozzle is an open system in the context of thermodynamics. It is used to produce a high velocity by reducing its pressure.

The steady flow energy equation can be given by the following formula:

h

1

+

1

2

v

2

1

+

g

z

1

+

q

=

h

2

+

1

2

v

2

2

+

g

z

2

+

w

where 'h' is enthalpy, 'v' is velocity, 'z' is height, 'q' is the heat and 'w' is work.

h

=

C

p

d

T

Answer and Explanation:

Given:

initial temp,

T

1

=

400

0

C

initial Pressure,

p

1

=

800

k

P

a

Initial Velocity,

v

1

=

10

m

/

s

Final temp,

T

2

=

300

0

C

Final Pressure,

p

2

=

200

k

P

a

Rate of heat loss, Q = 25 KW

Inlet Area,

A

1

=

800

c

m

2

As per the steady flow energy equation:

h

1

+

1

2

v

2

1

+

g

z

1

+

q

=

h

2

+

1

2

v

2

2

+

g

z

2

+

w

Since, there is external work, w= 0. Also, consider there is a negligible change in KE.

h

1

+

1

2

v

2

1

+

q

=

h

2

+

1

2

v

2

2

h

1

−

h

2

+

1

2

v

2

1

+

q

=

1

2

v

2

2

C

p

(

T

1

−

T

2

)

+

1

2

(

10

)

2

+

25000

=

1

2

v

2

2

2

(

400

−

300

)

+

50

+

25000

=

1

2

v

2

2

2

(

400

−

300

)

+

50

+

25000

=

1

2

v

2

2

25250

=

1

2

v

2

2

v

2

≈

225

which is the answer.

Using the equation of continuity:

A

1

v

1

=

A

2

v

2

0.08

(

10

)

=

A

2

(

225

)

A

2

=

3.55

×

10

−

3

m

2

Now, volume flow rate,

Q

2

=

A

2

v

2

Q

2

=

3.55

×

10

−

3

×

225

Q

2

=

0.798

m

3

/

s