Question 3 of 7: Question 3 (Part A, Question A3): Hydrogen Pipeline — Fanno-Flow Booster Spacing, Compressor Power and Cooler Duty

Question 3 (Part A, Question A3): Hydrogen Pipeline — Fanno-Flow Booster Spacing, Compressor Power and Cooler Duty (14 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The maximum booster spacing (the choking length), the shaft power of each booster compressor, the heat each cooler must reject, and the static pressure and temperature of the hydrogen delivered at Calgary.

Approach. Adiabatic pipe flow with friction is Fanno flow: compute the station Mach number from continuity, read the choking length from the Fanno function, then treat each booster as an adiabatic compressor (total-to-total) followed by an isobaric aftercooler, and march the leftover length to find the delivery state.

1. Fix the station-outlet state and its Mach number. From the ideal-gas law and continuity, \[\begin{aligned}\rho_1&=\frac{P_1}{RT_1}=\frac{10^{6}}{(4124)(243.15)}=0.9973\ \text{kg/m}^3,\\V_1&=\frac{\dot m}{\rho_1 A}=\frac{74.3}{(0.9973)(3.1416)}=23.72\ \text{m/s}.\end{aligned}\] With \(a_1=\sqrt{\gamma R T_1}=\sqrt{(1.4)(4124)(243.15)}=1184.8\ \text{m/s}\) (hydrogen is very "fast"), the Mach number is \[\boxed{M_1=\frac{23.72}{1184.8}=0.02002}\] Because \(M_1\) is so small the stagnation state is barely distinguishable from the static one: \(T_{o1}=243.17\) K and \(P_{o1}=1.00028\) MPa.
2. Evaluate the Fanno function at the station outlet. For adiabatic flow with friction in a constant-area duct, \[\frac{fL^{*}}{D}=\frac{1-M^2}{\gamma M^2}+\frac{\gamma+1}{2\gamma}\ln\!\left[\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}\right].\] Substituting \(M_1 = 0.020016\) and \(\gamma=1.4\) gives \(1782.20-6.55=1775.65\). The first term dominates completely at low Mach number, where it reduces to \(1/(\gamma M^2)\).
3. Convert to the maximum booster spacing. \(L^{*}\) is the length that would drive the flow to \(M=1\); no longer run is possible without reducing the mass flow, so it is the maximum spacing: \[L_{\max}=\frac{D}{f}\left(\frac{fL^{*}}{D}\right)=\frac{2.0}{0.03}(1775.65)=1.1838\times10^{5}\ \text{m}\] \[\boxed{L_{\max}=118.4\ \text{km}}\]
4. Find the state at the end of a full-length run. At the choking point the Fanno reference relations give \[\begin{aligned}\frac{T_1}{T^{*}}&=\frac{\gamma+1}{2+(\gamma-1)M_1^2}=1.1999,\\\frac{P_1}{P^{*}}&=\frac{1}{M_1}\sqrt{\frac{\gamma+1}{2+(\gamma-1)M_1^2}}=54.73,\end{aligned}\] so \(T^{*}=202.64\ \text{K}\) and \(P^{*}=18.27\ \text{kPa}\). The corresponding stagnation pressure is \(P_o^{*}=P^{*}(1.2)^{3.5}=34.59\ \text{kPa}\), while \(T_o\) is unchanged at 243.17 K because the pipe is insulated. Friction has destroyed 96.5 per cent of the stagnation pressure over one span.
5. Size the compressor. Each booster must lift the total pressure from \(P_o^{*}=34.59\) kPa back to \(P_{o1}=1000.3\) kPa, a total-to-total ratio of \[\Pi=\frac{P_{o1}}{P_o^{*}}=\frac{1000.3}{34.59}=28.92 .\] The isentropic total enthalpy rise is \[\begin{aligned}\Delta h_{o,s}&=c_p T_{o,\text{in}}\left(\Pi^{\frac{\gamma-1}{\gamma}}-1\right)\\&=(14\,350)(243.17)\left(28.92^{0.2857}-1\right)\\&=(14\,350)(243.17)(1.6151)=5636\ \text{kJ/kg},\end{aligned}\] and dividing by the total-to-total efficiency, \[\Delta h_o=\frac{\Delta h_{o,s}}{\eta_{tt}}=\frac{5636}{0.85}=6630\ \text{kJ/kg}.\]
6. Compressor power. \[\boxed{\dot W_c=\dot m\,\Delta h_o=(74.3)(6630)=4.926\times10^{5}\ \text{kW}=493\ \text{MW}}\] The gas leaves the compressor at \(T_{o,\text{out}}=243.17+6630/14.35=705.2\ \text{K}\).
7. Cooler duty. The aftercooler must return the gas to the station outlet condition, i.e. back to \(T_o=243.17\) K. Since it starts and ends at the same stagnation temperature it entered the booster with, the heat removed is exactly the work put in: \[\dot Q=\dot m\,c_p\,(T_{o,\text{out}}-T_{o,\text{in}})=(74.3)(14.35)(705.2-243.17)\] \[\boxed{\dot Q=4.926\times10^{5}\ \text{kW}=493\ \text{MW rejected per station}}\] This identity \(\dot Q=\dot W_c\) is not a coincidence: over one complete station-to-station cycle the gas returns to its initial state, so the first law demands that every joule of shaft work eventually leave as heat.
8. March the route and find the delivery state. With \(L_{\max}=118.4\) km, the 300 km route needs \(300/118.38=2.53\) spans, so two intermediate boosters are installed (at 118.4 km and 236.8 km) and the final leg is \(L_f=300-2(118.38)=63.25\ \text{km}\). Over that leg the Fanno function falls by \(fL_f/D=(0.03)(63\,247)/2.0=948.7\), leaving \[\begin{aligned}\left.\frac{fL^{*}}{D}\right|_{\text{Calgary}}&=1775.65-948.70=826.94\\&\Rightarrow\ M_3=0.02927 .\end{aligned}\]
9. Convert the exit Mach number to static conditions. Using the same Fanno reference relations with \(T^{*}=202.64\) K and \(P^{*}=18.27\) kPa, \[\begin{aligned}T_3&=\frac{T_o}{1+\tfrac{\gamma-1}{2}M_3^2}=243.13\ \text{K},\\P_3&=P^{*}\frac{1}{M_3}\sqrt{\frac{\gamma+1}{2+(\gamma-1)M_3^2}}=683.7\ \text{kPa}.\end{aligned}\] \[\boxed{P_3=0.684\ \text{MPa},\qquad T_3=243.1\ \text{K}=-30.0\,{}^{\circ}\text{C}}\] The temperature barely moves because the flow is so slow that the static and stagnation temperatures are nearly equal and the duct is insulated; essentially all of the friction shows up as pressure loss, not heating.