Pressure Loss in Gas and Steam Pipes
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) |
8 q
N4 2 pu 7r
When both sides of the equation are multiplied by p we have
.11 |
Pdp = |
D5tt2 P |
IdUD^ |
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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) |
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) |
A P _
Apr 1 + JT-ISpTPx
And
Ap2 1 + J — 2Ap2/p2
These equations can be used when the Mach number is small, and the acceleration effect is ignored.
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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