Thermal, steady-flow, table, continuity

Click For Summary
SUMMARY

The discussion focuses on a thermodynamics problem involving steam expansion through a nozzle, specifically at 2 MPa and 208°C. The key equations utilized include the Steady-flow Energy Equation and the Continuity Equation. The participant initially miscalculated the exit velocity using incorrect enthalpy values from the steam tables, leading to a discrepancy in results. The correct exit velocity is approximately 400 m/s, which highlights the importance of using accurate data from the steam tables.

PREREQUISITES
  • Understanding of the Steady-flow Energy Equation
  • Familiarity with the Continuity Equation in fluid dynamics
  • Knowledge of steam tables for thermodynamic properties
  • Basic principles of thermodynamics, particularly regarding enthalpy
NEXT STEPS
  • Review the Steady-flow Energy Equation applications in nozzle design
  • Study the Continuity Equation and its implications in fluid flow
  • Learn how to accurately read and interpret steam tables
  • Explore the effects of pressure and temperature on steam properties
USEFUL FOR

Students and professionals in mechanical engineering, particularly those specializing in thermodynamics and fluid mechanics, will benefit from this discussion.

scrubber
Messages
20
Reaction score
0

Homework Statement



Steam at 2 MPa and 208°C enters a nozzle with 20m/s. During the expansion process, its enthalpy drops to 2.86 MJ/kg because of the losses encountered.

a) Determine the exit velocity from the nozzle.
b) If the mass flow rate is 1kg/s, determine the flow area at the nozzle inlet.

Homework Equations



Steady-flow Energy Equation: Q-W=m{(hf-hi)+1/2(Uf^2-Ui^2)+g(Zf-Zi)},
difference between rate of heat and work=difference between rate of energy at inlet and exit

For Nozzle, heat, work, and PE can be neglected:
(hi-hf)=1/2(Uf^2-Ui^2)

Continuity equation: mi=mf=constant, mass flow rate is constant
m=ρ*A*U, ρ=density, A=flow area, U=fluid velocity

steam tables: http://enpub.fulton.asu.edu/ece340/pdf/steam_tables.PDF

The Attempt at a Solution



a.) checking the steam tables, the enthalpy of sat. liquid is 1236kJ/kg,
(1236-2.86)=1/2(Uf^2-20*2)
Uf=53.54m/s, which is not correct. The given answer is around 400m/s.
I don't know why ;(

I will try b) after I understand a).
Please help me with part a!
 
Physics news on Phys.org
1MJ = 1000kJ =1,000,000 J

How did you get the 1236 number?
 
paisiello2 said:
1MJ = 1000kJ =1,000,000 J

How did you get the 1236 number?
Looks like it came from the 280C row instead of 208C.
Scrubber, should you be using the sat liquid or sat vapor value? (I genuinely don't know.). Even if you use sat vap at 280, it's still less than 2.86MJ/kg, so it doesn't seem to fit.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K