Question 4.P.6: Nitrogen at 12 MN/m² pressure is fed through a 25 mm diamete...
Nitrogen at 12 MN/m² pressure is fed through a 25 mm diameter mild steel pipe to a synthetic ammonia plant at the rate of 0.4 kg/s. What will be the drop in pressure over a 30 m length of pipe assuming isothermal expansion of the gas at 300 K? What is the average quantity of heat per unit area of pipe surface that must pass through the walls in order to maintain isothermal conditions? What would be the pressure drop in the pipe if it were perfectly lagged? μ = 0.02 mNs/m².
At high pressure, the kinetic energy term in equation 4.55 may be neglected to give:
\left(P_2^2-P_1^2\right) / 2 P_1 v_1+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0
Specific volume at entry of pipe, v_1=(22.4 / 28)(300 / 273)(0.1013 / 12)
=0.00742 m ^3 / kg
Cross-sectional area of pipe, A=(\pi / 4)(0.025)^2=0.00049 m ^2
∴ G / A=(0.4 / 0.00049)=816 kg / m ^2 s.
Reynolds number, d(G / A) / \mu=0.025 \times 816 /\left(0.02 \times 10^{-3}\right)=1.02 \times 10^6
If e/d = 0.002 and R e=1.02 \times 10^6, R / \rho u^2=0.0028 from Fig. 3.7.
Substituting: \left(12^2-P_2^2\right) 10^{12} /\left(2 \times 12 \times 10^6 \times 0.00742\right)=4(0.0028)(30 / 0.025)(816)^2
and: P_2=11.93 MN / m ^2
and: pressure drop =(12.0-11.93)=0.07 MN / m ^2 \equiv \underline{\underline{70 kN / m ^2}}
The heat required to maintain isothermal flow is given in Section 4.5.2 as G \Delta u^2 / 2.
The velocity at the high pressure end of the pipe = volumetric flow/area
=(G / A) v_1=(816 \times 0.0072)=6.06 m / s
and the velocity in the plant is taken as zero.
Thus: G \Delta u^2 / 2=0.4 \times(6.06)^2 / 2=7.34 W
Outside area of pipe =(30 \times \pi \times 0.025)=2.36 m ^2.
Heat required =(7.34 / 2.36)=\underline{\underline{3.12 W / m ^2}}
This low value of the heat required stems from the fact that the change in kinetic energy is small and conditions are almost adiabatic. If the pipe were perfectly lagged, the flow would be adiabatic and the pressure drop would then be calculated from equations 4.77 and 4.72. The specific volume at the low pressure end v_2 to be calculated from:
8\left(R / \rho u^2\right)(l / d)=\left[\frac{\gamma-1}{2 \gamma}+\frac{P_1}{v_1}\left(\frac{A}{G}\right)^2\right]\left[1-\left(\frac{v_1}{v_2}\right)^2\right]-\frac{\gamma+1}{\gamma} \ln \left(\frac{v_2}{v_1}\right)
(equation 4.77)
For nitrogen, \gamma=1.4 and hence:
8(0.0028)(30 / 0.025)=\left[\frac{1.4-1}{2 \times 1.4}+\frac{12 \times 10^6}{0.00742}\left(\frac{1}{816}\right)^2\right]\left[1-\left(\frac{0.00742}{v_2}\right)^2\right]
-\frac{1.4+1}{1.4} \ln \left(\frac{v_2}{0.00742}\right)
Solving by trial and error, v_2=0.00746 m ^3 / kg.
Thus:
\frac{1}{2}\left(\frac{G}{A}\right)^2 v_1^2+\left(\frac{\gamma}{\gamma-1}\right) P_1 v_1=\frac{1}{2}\left(\frac{G}{A}\right)^2 v_2^2+\left(\frac{\gamma}{\gamma-1}\right) P_2 v_2 (equation 4.72)
Substitution gives:
(816)^2(0.00742)^2 / 2+[1.4 /(1.4-1)] 12 \times 10^6 \times 0.00742
=(816)^2(0.00746)^2 / 2+[1.4 /(1.4-1)] P_2 \times 10^6 \times 0.00746
and: P_1=11.94 MN / m ^2
The pressure drop for adiabatic flow =(12.0-11.94)=0.06 MN / m ^2 or \underline{\underline{ 60 kN / m ^2}}