HEM_example_solutions

HEM_example_solutions - Working with the HEM 1 1. (Solved...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Working with the HEM 1 1. (Solved on the board) Compute the pressure gradient for a saturated 1 kg s- 1 steam-water flow at 7 MPa and a quality of 0.1 in a smooth 2.5 cm diameter pipe using the homogeneous equilibrium model. f TP is to be taken as f lo Assume the flow direction is upward, there is no change in quality with axial distance, and ignore the effects of gas compressibility. Well need to know viscosities and densities for both phases to complete this problem. These can be obtained from steam tables: f = 740 kg m- 3 (1) v = 36 . 5 kg m- 3 (2) f = 9 . 55 10- 5 Pa s (3) v = 1 . 90 10- 5 Pa s (4) The flow area is also required: A = D 2 4 (5) A = (0 . 025 m ) 2 4 = 0 . 0004909 m 2 (6) As are the mass flux of the mixture, G m , and the mixture density, m : G m = m A (7) G m = 1 kg s- 1 . 0004909 m 2 = 2037 kg m- 2 s- 1 (8) m = 1 x v + 1- x f (9) m = 1 . 1 36 . 5 + 1- . 1 740 = 252 . 8 kg m- 3 (10) In 11-78, the second term in the numerator contains dx/dz ; it is zero since there is no phase change. The denominator becomes 1 when compressibility is ignored. The tube is round; the hydraulic diameter is simply the diameter, D . Upflow implies that cos ( ) is 1. The pressure drop relation then simplifies to:- dp dz HEM = f TP D G 2 m 2 m + m g (11) We start by computing Re lo , the Reynolds number for single-phase liquid at the same mass flux in the same tube: Re lo = GD l (12) Re lo = 2037 . 025 9 . 55 10- 5 = 5 10 5 (13) Working with the HEM 2 This Reynolds number is well-suited to use of the McAdams correlation: f lo = 0 . 184 Re- . 2 lo (14) f lo = 0 . 184 5 10 5- . 2 = 0 . 0132 (15) This is then substituted into Equation 11 along with the other known values:- dp dz HEM = . 0132 . 025 2037 2 2 252 . 8 + 252 . 8 9 . 81 (16)- dp dz HEM =- dp dz HEM,fric- dp dz HEM,grav (17)- dp dz HEM = 4333 Pa m- 1 + 2480 Pa m- 1 (18)- dp dz HEM = 6813 Pa m- 1 (19) 2. Use Equation 11-79b to compute the friction factor, employing each of the three estimates for TP (Equations 11-80a through 11-80c), and then use this to com- pute the pressure gradient. How important is the selection of the friction factorpute the pressure gradient....
View Full Document

Page1 / 6

HEM_example_solutions - Working with the HEM 1 1. (Solved...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online