Question 8 of 28 (Marks: 2) (CFPH_009)

Given the following data:

Ground Speed...................................... 92 kt
Planned Track......................................255 Degrees Magnetic
GPWT extract for cruise altitude...........25 025 +12
Magnetic Variation................................5W
ISA deviation.........................................+10
QNH.......................................................1007 hPa
Cruise SAR .......................................... 0.47 nm/lb

Your calculation of the trip fuel flow in lb/hr (to the nearest whole lb) is -

For the life of me, I can't get the provided answer. What am I missing here?

Cheers

Suggest you include the answer with your question so we know what the target might be ?

Of course.

Your answer of 164 lb/hr was incorrect.

The correct answer is between 247 lb/hr and 251 lb/hr.

Part 61 MOS Schedule 3 Unit 1.9.5 COPC 3.1.1 Fuel Flow from SAR

Several distractors in the question intended to confuse you a bit. However, it is a dead simple question providing you understand what you are doing.

SAR = TAS/FF (nm/hr)/(lb/hr), the units cancelling to give nm/lb.

W/V = 255/25, roughly as the GPWT is rounded off to the nearer 10 degrees.

TAS = 92 + 25 = 117 kt

FF = TAS/SAR = 117/0.47 = 249 lb/hr which is bang in the middle of the answer range specified.