Pressure Loss in Gas and Steam Pipes

Dp =

подпись: dp =When the gas compressibility no longer can be bypassed, the pressure loss equation is written in a differential form

(4.58)

Where dl is the differential pipe length and D is the diameter. On the basis that pm A = qm or vm= qm /pA,

(4.59)

подпись: (4.59)8 q

N4 2 pu 7r

When both sides of the equation are multiplied by p we have

.11

Pdp =

D5tt2 P

IdUD^

Pressure Loss in Gas and Steam Pipes

(4.60)

 

For an ideal gas,

PV = NT or pv = p/p=T/M,

Or when the gas follows the formula pv = h*/Z,, where I is the process factor and b* = h ~hQ, the deviation from the enthalpy of the reference state, Eq.

(4.60) , can easily be integrated, giving

(4.61)

подпись: (4.61)Pl^ = p.^^UL+DyA

2 P 7T2 Ds{ ^ .

Where p/p = constant.

If the pressures and densities are known we can solve for either qm or D from this equation.

If either px or p2 is known, simplification of the left side of Eq. (4.61) gives

2 2 .

P±Z_P2 = Јi-i_l. (pl-p2) = pAp = pxApx = p2Ap2, (4,62)

Where Ap^ and Ap2 are auxiliary quantities, which can be solved from Eqs.

(4.61) and (4.62). The real pressure loss can then be solved with the equations

(4.63)

(4.64)

подпись: (4.63)
(4.64)
A P _

Apr 1 + JT-ISpTPx

And

AЈ =

Ap2 1 + J — 2Ap2/p2

These equations can be used when the Mach number is small, and the acceler­ation effect is ignored.

Example 2

Calculate the final pressure of air flowing in a 200 m long pipe, when the initial pressure of air is 800 kPa and the mass flow is 2.5 kg s-1, The pipe di­ameter is 100 mm, the roughness is 0.03 mm, and the sum of the single resis­tances is 6.5. The air temperature is 300 K, the molar mass 28.96 kg kmoH, and the dynamic viscosity 1.85 x 10~5 kg s-1m-1.

First calculate

■hr — — 300 • 8.314 _ o/ n r,7 r,,,“1

P ~M 28796………………… — 86.13 kj kg.

Determination of friction factor Ј:

Re = ilia =—————- 4_2J>——— = = | 72 . 1Q6

77- 0.1 • 1.85 • 10~"

And

K = 0.03 mm ‘% = 0.0154.

Equation (4.61) gives

P~P2 — p.. JL. ‘iml T PV’ 2 n5

Z 7T U

J

= 86.13 x 103 • 0.81 • ^(0.0154 • 200 + 0.1 ■ 6.5)Pa2 0.15

= 1,626 x 101 !Pa2

Pressure loss from Eqs. (4.62) and (4,63):

Ap] = Јf—^ = M26 x 10“ = l01 7 kPa 2’P l 2 • 800 x 10J

Ap i _ 101.7 _njiy pi 800 ‘

APi -_________ 2______

1.073

P’ 1+ /i-2-^l 1 +VI -(2-0.127)

■V Pi Ap = (1.073 ■ 101.7) kPa = 109 kPa

Final pressure is

P2 = Pi — Ap = (800 — 109) = 691 kPa.

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