Instruments Exam - Help!
Thread Starter
Joined: Apr 2002
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From: London
Instruments Exam - Help!
Hi,
I understand how to work out the following questions using maths, just about but can anybody explain how you work this question out on the CRP5?
If pressure altitude is 40000 ft amsl, indicated TAT is -29, what is the Mach Number assuming ISA conditions.
a) 0.7
b) 0.75
c) 0.8
d) 0.85
I'm struggling to remember all the formulae for these exams so was just wondering if there is a quick way on the CRP5 to work this out.
Please can anyone help?
Many Thanks
I understand how to work out the following questions using maths, just about but can anybody explain how you work this question out on the CRP5?
If pressure altitude is 40000 ft amsl, indicated TAT is -29, what is the Mach Number assuming ISA conditions.
a) 0.7
b) 0.75
c) 0.8
d) 0.85
I'm struggling to remember all the formulae for these exams so was just wondering if there is a quick way on the CRP5 to work this out.
Please can anyone help?
Many Thanks
Joined: Aug 2001
Posts: 775
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From: Dorset
To solve this problem you must use the equation for the relationship between SAT and TAT. This is:
SAT = (TAT / (1 +(0.2 x k x Mach no squared))
This can be rearranged to give:
Mach no = Square root of ( (TAT/ (k x SAT) )-1 ) / 0.2)
k is the ram recovery factor of the temp probe, which is one for a perfect probe. If we assume that k is 1 we can ignore it in the above equation.
We must be careful to note that SAT and TAT must be in degrees absolute.
At 40000 ft in the ISA SAT is approximately -56.6. Adding 273 to this gives 216.5 absolute.
The question states that TAT is -29. Adding 273 to this gives TAT = 244 absolute.
Putting these values into the above equation gives:
Mach no = Square root ((244/216.5) -1) / 0.2)
This simplifies to give mach 0.7969 or approximately Mach 0.8
It is probably possible to do the above on A CRP5 but I have not tried to do so.
The process probably involve working out the ram rise, which is TAT - SAT. Then using this in the CRP5 to get the Mach number that would give you that rise.
SAT = (TAT / (1 +(0.2 x k x Mach no squared))
This can be rearranged to give:
Mach no = Square root of ( (TAT/ (k x SAT) )-1 ) / 0.2)
k is the ram recovery factor of the temp probe, which is one for a perfect probe. If we assume that k is 1 we can ignore it in the above equation.
We must be careful to note that SAT and TAT must be in degrees absolute.
At 40000 ft in the ISA SAT is approximately -56.6. Adding 273 to this gives 216.5 absolute.
The question states that TAT is -29. Adding 273 to this gives TAT = 244 absolute.
Putting these values into the above equation gives:
Mach no = Square root ((244/216.5) -1) / 0.2)
This simplifies to give mach 0.7969 or approximately Mach 0.8
It is probably possible to do the above on A CRP5 but I have not tried to do so.
The process probably involve working out the ram rise, which is TAT - SAT. Then using this in the CRP5 to get the Mach number that would give you that rise.
Joined: Aug 2001
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From: Dorset
XYZ
When you took your exam you should certainly have been aware of the equation:
SAT = (TAT / (1 +(0.2 x k x Mach no squared))
It has always been part of the syllabus.
Past exams have often required students to identify the correct equation, but have rarely if ever required them to use it.
If the question posted in this string is a real JAR Instruments exam question then either there is a simple method of solving it using the CRP5, or else the CAA have xepanded their question bank and now require use of the equation. This is in line with their poilicy of trying to prevent students from passing simply by memorising the answers to feedback questions.
When you took your exam you should certainly have been aware of the equation:
SAT = (TAT / (1 +(0.2 x k x Mach no squared))
It has always been part of the syllabus.
Past exams have often required students to identify the correct equation, but have rarely if ever required them to use it.
If the question posted in this string is a real JAR Instruments exam question then either there is a simple method of solving it using the CRP5, or else the CAA have xepanded their question bank and now require use of the equation. This is in line with their poilicy of trying to prevent students from passing simply by memorising the answers to feedback questions.
Joined: May 1999
Posts: 1,846
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From: Bristol, England
The CRP5 solution:
TAT is -29, ISA is -56.5, temperature rise is 27.5º
On the CRP5 look at the blue temperature rise scale, against 27.5º read off a TAS of 500KT. Set the Mach number index in the AIRSPEED window to -56.5º. Against the outer scale TAS of 500KT read off the Mach number on the inner scale of M0.88.
I guess the answer is different because the blue temperature rise scale uses a recovery factor that is less than 1.0. I don't know what the factor is or whether it is typical of a modern system.
We are not aware of any CQB questions like this.
TAT is -29, ISA is -56.5, temperature rise is 27.5º
On the CRP5 look at the blue temperature rise scale, against 27.5º read off a TAS of 500KT. Set the Mach number index in the AIRSPEED window to -56.5º. Against the outer scale TAS of 500KT read off the Mach number on the inner scale of M0.88.
I guess the answer is different because the blue temperature rise scale uses a recovery factor that is less than 1.0. I don't know what the factor is or whether it is typical of a modern system.
We are not aware of any CQB questions like this.
Thread Starter
Joined: Apr 2002
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From: London
Thanks to everyone that replied.
Alex, that's what I kept getting with the CRP5 method (M0.88) which was different to the equation answer and that's what confused me initially
I can stop worrying a bit now!!
Thanks again
Alex, that's what I kept getting with the CRP5 method (M0.88) which was different to the equation answer and that's what confused me initially
I can stop worrying a bit now!!
Thanks again
Joined: Aug 2001
Posts: 775
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From: Dorset
ALEX
You are correct in concluding that the k factor used in the CRP5 is not 1. I worked it out once (a few months ago) and it turned out to be about 0.86.
The answer 0.8 can also be found by using the table at Figure 4.3 in the CAP 698.
I also agree (as I stated previously) that questions requiring the use of the equation have not been in previous exams. But we must be cautious here. The examiners are constantly devising new questions. Relying entirely on feedback inevitably leads to disappointment when new questions suddenly appear. The cry "That exam was right off the wall, there were no feedback questions" is heard all to frequently after exams.
You are correct in concluding that the k factor used in the CRP5 is not 1. I worked it out once (a few months ago) and it turned out to be about 0.86.
The answer 0.8 can also be found by using the table at Figure 4.3 in the CAP 698.
I also agree (as I stated previously) that questions requiring the use of the equation have not been in previous exams. But we must be cautious here. The examiners are constantly devising new questions. Relying entirely on feedback inevitably leads to disappointment when new questions suddenly appear. The cry "That exam was right off the wall, there were no feedback questions" is heard all to frequently after exams.
Joined: Jun 2002
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From: UAE
Keith.Williams I am of course fully aware of the equation but the way it was put forward does look somewhat confusing.
I suppose I should of used :
to show a touch of humour was intended?
YYZ
PS YYZ Not XYZ.... see, confusing
I suppose I should of used :
to show a touch of humour was intended?YYZ
PS YYZ Not XYZ.... see, confusing
Super-Friendly Aviator
Joined: Jun 2004
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From: Reigate, UK
Erm...how does one work out a Mach number simply from being given a temperature and an altitude? A plane at at this altitude (and subject to this temperature) can be flying at any speed it likes...unless this "Mach number" the question talks about is a measure of the speed of sound at this altitude and temperature, as a fraction of its sea level value? I can't see any other explanation.
I'm not at the ATPL stage yet, having just finished uni, but this question sounds a bit odd...or am I not well-versed enough in ATPL phraseology!?
V1R
I'm not at the ATPL stage yet, having just finished uni, but this question sounds a bit odd...or am I not well-versed enough in ATPL phraseology!?
V1R
Joined: May 1999
Posts: 1,846
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From: Bristol, England
The temperature given was a TAT (Total Air Temperature) not what you would get sitting in a balloon holding a thermometer, that would be a Static Air Temperature. The TAT is what is actually sensed by an aircraft temperature probe. It is artificially high because of friction heating and the like as the aircraft rushes through the air. A moving aircraft might detect, for instance, a temperature of +20ºC when the actual temperature is +10ºC.
The temperature rise between SAT and TAT is proportional to the mach number. This means that, if we know the temperature rise in the sensing system, we can work backwards to find the Mach Number.
The temperature rise between SAT and TAT is proportional to the mach number. This means that, if we know the temperature rise in the sensing system, we can work backwards to find the Mach Number.
Super-Friendly Aviator
Joined: Jun 2004
Posts: 424
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From: Reigate, UK
Thanks Alex - that makes sense. Thought it might be a case of my limited understanding of the terms!
So you know what the SAT should be at 40,000 ft because you assume ISA. They tell you what the TAT is and the difference between the TAT and SAT will be related to the airspeed. Makes sense, cheers
V1R
So you know what the SAT should be at 40,000 ft because you assume ISA. They tell you what the TAT is and the difference between the TAT and SAT will be related to the airspeed. Makes sense, cheers
V1R
