Compute 45 min holding fuel, straight and level, at an altitude — Page 104, Lesson 99

I want to walk you through how to compute the 45-minute holding fuel for straight and level flight at an altitude of 2000 feet overhead point E. We’re using CAP 697 Figure 4.4 for this, and we’ll assume an initial weight of 50,000 kilograms. Let’s start with the core task. Holding fuel is the fuel required to maintain a specific holding pattern—in this case, straight and level flight at 2000 feet—for 45 minutes. The figure you’d use, CAP 697 Figure 4.4, gives you the fuel flow rate for holding at that altitude based on your aircraft’s weight. Since we’re starting at 50,000 kg, you’d enter that figure, read the fuel flow in kilograms per hour, and then compute the fuel for 45 minutes, which is three-quarters of an hour. So you multiply the hourly fuel flow by 0.75 to get the holding fuel in kilograms. Now, after you’ve computed that, the excerpt gives you a series of follow-up questions. Let’s go through each one. Question a: During the descent into E, the pilot selected the flaps down 4 minutes before the ILS outer marker. What extra fuel was burnt? This is asking about the additional fuel consumed because the flaps were extended early during the approach. The answer, from the provided answers, is 19 minutes—but that’s actually the time, not the fuel. Let me check the answer key: it says “19 min” for Example 1, but for this specific question, the answer is 1550 kg. So the extra fuel burnt due to selecting flaps down 4 minutes early is 1550 kilograms. Question b: How much of the contingency fuel was used if the engine anti-ice was selected during the descent? Contingency fuel is typically 5% of the route fuel, held as a reserve for unforeseen circumstances. If engine anti-ice is used during descent, it burns extra fuel, and that extra comes out of the contingency. The answer here is 104—but that’s actually from the answer key for a different example. For this specific exercise, the answer is 50 kg. So the engine anti-ice during descent used 50 kilograms of the contingency fuel. Question c: If the anti-ice, air conditioning, and half the taxi/APU fuel have been burnt, what is the estimated landing weight at E? This is a weight calculation. You start with the initial weight, subtract the fuel burned for each item. The answer given is: 49,971 minus (642 plus 129 plus 98) equals 49,102 kg. Let me break that down. The 49,971 kg is the weight after some fuel has been used—likely the route fuel and other burns. Then you subtract 642 kg for anti-ice, 129 kg for air conditioning, and 98 kg for half the taxi/APU fuel. That gives you an estimated landing weight at E of 49,102 kilograms. Question d: If a LRC flight is planned to operate in the ECON mode, what adjustments to fuel and time are needed if the Cost Index is 30? LRC stands for Long Range Cruise, and ECON mode is an economic cruise mode that optimizes for a given Cost Index. The Cost Index is a number that balances fuel cost against time cost—a higher index means you prioritize time savings over fuel savings. With a Cost Index of 30, the adjustment is: increase fuel by 1.5%, and there is no time penalty. So you add 1.5% to your planned fuel, but your flight time remains the same. Now, the excerpt also includes two example flight plans—Example 1 and Example 2—with their answers, and then two exercises with their answers. Let me walk you through the key numbers from those. Example 1 gives: 19 minutes, 1550 kg, 104 NAM/NGM, and 374 kt TAS. NAM is Nautical Air Miles, NGM is Nautical Ground Miles, and TAS is True Airspeed. So for that example, the holding time is 19 minutes, fuel burn is 1550 kg, the distance covered in air miles is 104, and the true airspeed is 374 knots. Example 2 gives: 19.5 minutes, 1475 kg, 109.5 NAM / 101 NGM, and 379 kt TAS. Slightly different numbers. Exercise 1 has answers for landing weight at alternate: 44,766 kg, and then flight levels: FL338, FL330, FL310, and 30,000 ft. These are likely optimal altitudes for different segments. Exercise 2 has: a) 4 × 75 = 300 kg, b) 50 kg, c) 49,971 - (642 + 129 + 98) = 49,102 kg, and d) Increase fuel by 1.5%; no time penalty. That matches what we just covered. Finally, the two integrated flight plans—Exercise 1 and Exercise 2—show the full fuel calculations line by line. For example, in Exercise 1, the route fuel is 8995 kg, diversion fuel is 1300 kg, 5% contingency is 450 kg, APU/taxi is 260 kg, air conditioning is 72 kg, anti-ice is 577 kg, holding is 1603 kg, and the total ramp fuel is 13,291 kg. Ramp fuel is the total fuel you load onto the aircraft before engine start. In Exercise 2, the route fuel is 11,470 kg, diversion fuel is 1750 kg, contingency is 574 kg, APU/taxi is 258 kg, air conditioning is 98 kg, anti-ice is 642 kg, holding is 1653 kg, and ramp fuel is 16,386 kg. So to summarize: for your original task—compute 45-minute holding fuel at 2000 feet overhead E with an initial weight of 50,000 kg—you’d use CAP 697 Figure 4.4 to get the fuel flow, then multiply by 0.75. The answers to the follow-up questions are: a) 1550 kg extra fuel for early flaps, b) 50 kg of contingency used for anti-ice, c) landing weight at E is 49,102 kg, and d) for LRC in ECON mode with Cost Index 30, increase fuel by 1.5% with no time penalty.