ATMOS 1. c. This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a pressure altitude of 30000 ft and a mach number of 0.84, the TAT in the standard atmosphere is -12.
ATMOS 2. a.
This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a pressure altitude of 30000 ft and a mach number of 0.84, The TAT in the standard atmosphere is -12. But the indicated TAT is -10. The difference between thes th ese e tw two o fi figu gure res s is a me meas asur ure e of th the e te temp mper erat atur ure e de devi viat atio ion n in th that at pa part rtic icul ular ar atmosphere. This is found by subtracting the ISA TAT from the indicated TAT. This is -10(-12), which gives a temperature deviation of +2. The atmosphere used in this question is therefore 2 degree warmer than the internation international al standard atmosphere.
ATMOS 3. c.
This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a pressure altitude of 40000 ft and an indicated TAT of-29, the Mach number in the standard atmosphere would 0.8.
ATMOS 4. d.
This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a mach number of 0.88 and an ISA TAT -4, the pressure altitude in the standard atmosphere atmosphere must be 28000 ft.
ATMOS 5. b.
This question must must be addressed in two stages. First, the temperature temperature deviation must be found from figure 4.3 on page 51 of CAP 698. This temperature deviation must then be used in the standard equation to calculate density altitude.
From figure 4.3, with a pressure altitude of 30000 ft, and a mach number of 0.82 the ISA TAT must be-14. But the indicated TAT is -10. The temperature deviation is equal to the indicated TAT minus the ISA TAT, which in this case is -10-(-14), which is +4. This temperature deviation of +4 can now be used to calculate density altitude using the following equation: Density alt= Pressure Pressure alti + (118 x 4) = 30472 ft which is option b.
ATMOS 6. d.
This question can be answered using figure 4.2 on page 50 of the CAP 698. From this it can be seen that if QNH is 999 hPa, a correction of +400 ft must be added to the elevation to give the pressure altitude. This gives a pressure altitude of 25400 ft at an elevation of 25000 ft.
ATMOS 7 a.
This question can be answered using figure 4.2 on page 50 of CAP 698. The question states that at an elevation of 22000 ft the pressure altitude is 22800 ft. This means that a correction of +800 ft has been added to the elevation. From figure 4.2 it can be seen that a correction of +800 ft equates to a QNH of between 983 and 986 hPa. Option a, 985 hPa is therefore the most accurate.
ATMOS 8. a.
This question can be answered using figure 4.1 onpage 49 of the CAP 698. A wind blowing from 330 approaches runway 04 from the left at an angle of 70 degrees. So a wind of 30 Kts. Is blowing from the left at an angle of 70 degrees.
From figure 4.1 this resolves to give a headwind component of approximately 10 Kts and a crosswind from the left of approximately 28 Kts.
ATMOS 9. c.
This question can be answered using figure 4.1 onpage 49 of the CAP 698. A wind blowing from 330 approaches runway 09 from the left at an angle of 120 degrees. Its will there fore comprise of components of a tailwind and crosswind from the left. From figure 4.1 it can be seen that the maximum permissible crosswind component of 10 Kts. Produced by a wind approaching at 120 degrees, equates to a wind of approximately 11.5 degrees. The most accurate option in this question it therefore option c, 12 Kts.
ATMOS 10. b.
This question can be answered using figure 4.1 onpage 49 of the CAP 698. A wind blowing from 330 approaches runway 09 from the left at an angle of 120 degrees. Its will there fore comprise of components of a headwind and crosswind from the left. From figure 4.1 it can be seen that a wind of 30 Kts approaching from an angle of 20 degrees gives a headwind of approximately 28 Kts and a crosswind of approximately 10.5 Kts from the left.
ATMOS 11. a.
A QNH of 900 mb (or hPa) is outside of the range covered in figure 4.2 on page 50 of the CAP 698, so the problem must be solved using the standard equation: Pressure Altitude = Elevation +(30x(1013-QNH)) So in this question pressure altitude = 4000 ft + (30x(1013-900))
Which resolves to give Pressure altitude = 7390 ft.
ATMOS 12. b.
This problem can be solved using the standard equation: Pressure Altitude = (30 x (1013 -QFE)) So in this question Pressure altitude = (30 x (1013- 1020))
Which resolves to give Pressure altitude = -210 ft.
ATMOS 13. a.
This problem can be solved using the standard equation: Pressure Altitude = Elevation + (30 x (1013-QNH)) Rearaanging this gives Elevation = Pressure altitude = - (30 x (1013-QNH))
Which resolves to give Elevation = 3700 – 390 Which = 3310 ft.
ATMOS – 14 d.
When calculating take-off and landing performance prior to the commencement of a flight the actual temperatures at the take-off site are known, whereas only forecast temperatures are available for the destination. Take-off performance calculations are therefore based on actual temperatures, whilst landing calculations are based on forecast or predicted temperatures.
ATMOS 15 b.
In the international Standard Atmosphere (ISA) the temperature at mean sea level is 150C and the lapse rate is 1.98 0C per 100- ft increase in altitude. At a pressure altitude of 5000 ft amsl, the temperature should therefore be 15 0 C- (5 x 1.98). This means that the temperature at 5000 ft amsl pressure altitude should be 5.1 0C. The temperature deviation is the actual temperature minus the ISA standard temperature. The temperature deviation in this case is therefore =10 0C- 5.10C. that is + 4.9 0C.
ATMOS 16 a.
The density of the air at any point in the atmosphere is dependent upon air pressure, temperature and humidity. The density altitude at any given point in the atmosphere at any given time, is the altitude at which the prevailing density
would occur in the
international standard atmosphere. It can be calculated using the equation. Density altitude = Pressure altitude plus ( 118 x the temperature deviation)
ATMOS 17 c.
The pressure altitude at any point can be found by setting 1013.2 mp (hPa) on the altimeter subscale.
ATMOS 18 a.
Density altitude is the altitude at which the prevailing air density would occur in the international standard atmosphere (ISA). It can be calculated from pressure altitude and temperature using the standard equation:
Density altitude = Pressure altitude plus +( 118 x the temperature deviation) The sea level temperature in the ISA is 15 0 C and the standard temperature lapse rate is 1.980C per 1000 ft. The temperature in the ISA at a pressure altitude of 5000 ft should therefore be 150C – (5 x 1.980C).
The temperature deviation is the actual temperature minus the ISA standard temperature, which in this case is (25 0C) – ( 5 x 1.98 0C).
Inserting these figures into the original equation gives:
Density altitude = 5000 + 118 (25 – ( 15 – ( 5 x 1.98))) = 7348.2 ft.
ATMOS 19 b.
QFE is the air pressure at the field elevation. This problem can be solved using the standard equation:
Pressure altitude = 30 x ( 1013 QFE)= 30 x (1013 – 1022) That is the pressure altitude is -270 ft amsl.
ATMOS 20 a.
This problem can be solved using the two standard equations:
Pressure altitude = Elevation + ( 30 x ( 1013 – QNH))
And Pressure altitude = 30 x ( 1013 – QFE))
Combining the two equations gives:
Elevation + (30 x (1013 – QNH)) = ( 30 x (1013 – QFE)) Inserting the data provided in the question gives:
4500 + ( 30 x (1013 – 1000)) = ( 30 x (1013 – QFE))
This simplifies to give 4890 = (30x (1013 – QFE))
This can be rearranged to give:
( 4890 / 30 ) -1013 = -QFE
Which is -850 = -QFE
So QFE = 850 hPa.
Pressure altitude is the altitude at which the existing air pressure would occur in the international standard atmosphere. It is the altimeter indication when 1013.25 hPa is set on the altimeter sub-scale.
ATMOS 22. a
Density is a measure of how tightly packed the molecules of a material are, or how much mass of material can be held in a given volume of space. Increasing altitude causes air pressure to decrease. Decreasing air pressure causes air to expand such that it is less tightly packed into a given volume. This decreases density. Increasing temperature causes air to expand, reducing the mass that can be contained within a given volume.
This reduces density. Water vapour is less dense than air so increasing humidity causes density to decrease. Air density is therefore decreases by increasing humidity, increasing altitude and increasing temperature.
ATMOS 23 c.
QNH is the ambient air pressure at mean sea level . QFE is the ambient air pressure at the airfield elevation . The two are related in that any increase in one results in an equal increase in the other, Option d is therefore incorrect in that an increase in QNH would increase QFE. The elevation of an airfield is its vertical distance above mean sea level. Elevation does not change with varying atmosphere condition. Option c is there fore correct.
ATMOS 24 c.
QNH is the ambient air pressure at mean sea level. QFE is the ambient air pressure at the airfield elevation. The two are related in that any increase in one results in an equal increase in the other. Option d is therefore incorrect in that an increase in QNH would increase QFE . Option b is incorrect because a change in QFE must cause a change in QNH. The elevation of an airfield is its vertical distance above mean sea level. Elevation does not change with varying
atmosphere condition. Option a is therefore incorrect.
Option c, increase QNH, is therefore correct.
ATMOS 25 a.
EAS is the Equivalent Airspeed. The movement of an aircraft through the air causes the air pressure close to it to increase. At low speed this increased pressure is not significant but as speed increases the magnitude of the pressure rise increases. At high speed the pressure rise is sufficient to compress the air slightly, thereby increasing the dynamic pressure. This process is called adiabatic compression. If left uncorrected this would cause the airspeed indicator to become progressively less accurate. Over indicating as speed increases. The EAS is the Calibrated Airspeed corrected for the effects of this adiabatic compression of the air.
ATMOS 26 c.
An airspeed indicator gives an indication proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so climbing at constant IAS or CAS means climbing at constant dynamic pressure. But dynamic pressure is equal to ½ pV 2, where p is air density and V is TAS. So when climbing at constant IAS, the value of TAS must increase to compensate for decreasing density, in order to maintain a constant dynamic pressure and airspeed indication. TAS therefore increases as altitude increases in a constant IAS climb.
ATMOS 27 d.
Mach number is the speed of an aircraft as a fraction of the local speed of sound. Mach 1 for example means a speed equal to the local speed of sound, whereas mach 0.5 means a speed only half of the local speed of sound. The speed of sound in the atmosphere is not constant but is proportional to the square root of absolute temperature ( LSS = 38. 94
TM
Absolute temperature). This means that if air temperature decreases, the local
speed of sound also decreases. But if the local speed of sound decreases, then the TAS equating to any given mach number also decreases. Decreasing temperature therefore causes the TAS at any given mach number to decrease. As pressure altitude increases up to the tropopause at 36000 ft. the air temperature decreases at a rate of approximately 1.98 degrees per 1000 ft. Above 36000 ft the temperature remains constant. This means that as pressure altitude increases up to 36000 ft. the local speed of sound decreases, then remains constant at higher altitudes. So when climbing at constant Mach number TAS will decrease up to 36000 ft. then remain constant.
ATMOS 28 d.
As pressure altitude increases up to the tropopause at 36000, the air temperature decreases at a rate of approximately 1.98 degrees per 1000 ft. Above 36000 ft the temperature remains constant. So as altitude increases the temperature. Decreases up to 36000 ft then remains constant above this altitude.
ATMOS 29 d.
An airspeed indicator gives an indication (IAS) proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so descending at constant IAS means descending at constant dynamic pressure. Airspeed indicators are susceptible to individual instrument errors and errors due to minor inaccuracies in the pressure sensing systems. The Calibrated Airspeed (CAS) is the IAS corrected for these minor errors. So descending at constant IAS or CAS means descending at constant dynamic pressure.
ATMOS 30 a.
Density altitude is the altitude at which the prevailing air density would occure in the international Standard Atmosphere (ISA). As altitude increases, the air pressure decreases allowing the air to expand. This expansion causes density to decreases, so density decreases with increasing altitude. This means that anything which causes the density to decrease causes the density altitude to increase. Increasing air temperature also causes the air to expand, thereby reducing its density and increasing the density altitude, So increasing temperature decreases density but increases density altitude. ATMOS 31 a.
In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature along, so decreasing temperature causes the local speed of sound decrease Mach number represents TAS as a fraction of LSS, so as the LSS decreases. The Mach number at any given TAS increases. So when descending through an inversion at any given TAS, the temperature and LSS decrease, causing the mach number to increase. This is the opposite to what happens in a normal atmosphere. So option a is correct and option b and C are incorrect. As altitude changes, the main factor controlling the ration of TAS : CAS is the ambient air pressure descending whether in an inversion or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to increase. So option d is incorrect.
ATMOS 32 b.
In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alonge, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of an inversion at any given TAS, the temperature and LSS increase, causing the mach number to decrease. This the opposite to what happens in a normal atmosphere. So option b is correct and options a and c incorrect.
As altitude changes, the main factor controlling the ratio of TAS: CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the CAS at any given TAS to decrease. So when climbing at constant TAS the CAS decreases. So option d is incorrect.
ATMOS 33. d.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, kthe pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS, the TAS decreases. So option d is correct and a is incorrect. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any given TAS, the temperature and LSS decrease, causing the mach number to increase.
But when descending through an inversion at constant CAS, the overall effect on Mach number is a balance between decreasing TAS and decreasing LSS. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion.
Of the options offered in this question, option d, decreasing TAS is therefore the only one that is definitely true .
ATMOS 34 a.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS, the TAS increases. So option a is correct and d incorrect.
In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature along, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any given TAS, the temperature and LSS increase, causing the mach number to decrease. But when climbing through an inversion at constant CAS, the overall effect on Mach number is a balance between increasing TAS and increasing LSS. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option a increasing TAS is therefore the only one that is definitely true.
ATMOS 35. d.
In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any given mach number, the temperature and LSS increase, causing the TAS to increase. So option d is correct and option b and c incorrect.
As altitude changes, the main factor controlling the ration of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant TAS the CAS decreases. But when climbing through an inversion at constant mach number, the overall effect on CAS is a balance between increasing TAS due to increasing LSS and decreasing CAS : TAS ratio due to decreasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, increasing TAS is therefore the only one that is definitely true.
ATMOS 36 b. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any given mach number, the temperature and LSS decrease, causing the TAS to decrease. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant TAS the CAS increases. But when descending through an inversion at constant mach number, the overall effect on CAS is a balance between decreasing TAS due to decreasing LSS and increasing CAS : TAS ratio due to increasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option b, decreasing TAS is therefore the only one that is definitely true.
ATMOS 37 d.
In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any given mach number, the temperature and LSS increase, causing the TAS to increase. Option d is there fore correct and option c incorrect .Also because temperature and LSS both increase when climbing in an inversion, option b is incorrect. As altitude changes, the main factor controlling the ration of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant TAS the CAS decreases. But when climbing through an inversion at constant mach number, the overall effect on CAS is a balance between increasing TAS due to increasing LSS and decreasing CAS : TAS ratio due to decreasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, increasing TAS is therefore the only one that is definitely true.
ATMOS 38 d.
In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any given mach number, the temperature and LSS decrease, causing the TAS to decrease. Option d is there fore correct and option c incorrect .Also because temperature and LSS both decrease when descending in an inversion, option b is incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the
pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant TAS the CAS decreases. But when descending through an inversion at constant mach number, the overall effect on CAS is a balance between increasing TAS due to increasing LSS and decreasing CAS : TAS ratio due to increasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, decreasing TAS is therefore the only one that is definitely true.
ATMOS 39 a.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS, the TAS increases. So option a is correct and options b and c are incorrect.
In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any given TAS, the temperature and LSS increase, causing the mach number to decrease.
But when climbing through an inversion at constant CAS, the overall effect on Mach number is a balance between decreasing mach number due to increasing LSS and decreasing CAS : TAS. Ratio due to decreasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option a increasing TAS is therefore the only one that is definitely true.
ATMOS 40 b.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS, the TAS decreases. So option b is correct and options a and c are incorrect. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any TAS, the temperature and LSS decrease, causing the mach number to increase.
But when descending through an inversion at constant CAS, the overall effect on Mach number is a balance between increasing Mach number due to decreasing. LSS and increasing CAS : TAS ratio due to increasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option b decreasing TAS is therefore the only one that is definitely true.
ATOMS 41 c.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS,
so in an isothermal layer the Mach number at any given TAS is constant. So option c is correct and options a and b are incorrect.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to increase. So when descending at constant TAS the CAS increases. So option d is incorrect.
ATOMS 42 c.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So option c is correct and options a and b are incorrect.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to decrease. So when climbing at constant TAS the CAS decreases. So option d is incorrect.
ATMOS 43 b.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS, the TAS decreases. So option d is incorrect.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. But when
descending at constant CAS the TAS decreases so the mach number decreases. Option b is therefore correct and options a and c incorrect.
ATMOS 44 a.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS, the TAS increases. So option d is incorrect.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. But when climbing at constant CAS the TAS increases so the mach number increases. Option a is therefore correct and options b and c incorrect.
ATOMS 45 c.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when climbing at constant Mach number in an isothermal the TAS remains constant. Option c is therefore correct and options a and b are incorrect.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the CAS at any given TAS to decrease. So when
climbing at constant Mach number and constant TAS in an isothermal, CAS decreases. So option d is incorrect.
ATOMS 46 c.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when descending at constant Mach number in an isothermal the TAS remains constant. Option c is therefore correct and option a and b incorrect.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to increase. So when descending at constant Mach number and constant TAS in an isothermal, CAS increases. So option d is incorrect.
ATOMS 47 b.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when climbing at constant Mach number in an isothermal the TAS remains constant. Option d is therefore incorrect.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the CAS at any given TAS to decrease. So when
climbing at constant Mach number and constant TAS in an isothermal, CAS decreases. So option b is correct and a and c incorrect.
ATOMS 48 c.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when descending at constant CAS in an isothermal the LSS remains constant. Option c is therefore correct and options a and b incorrect.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS in an isothermal, TAS decrease. So option d is incorrect.
ATOMS 49 a.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when climbing at constant CAS in an isothermal the LSS remains constant.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the TAS
at any given CAS to increase. So when
climbing at constant CAS in an isothermal, TAS increases. But LSS is constant in an isothermal, so increasing TAS causes Mach number to increase. So option a is correct and option b, c and d, are incorrect.
ATOMS 50 b.
As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS in an isothermal, TAS decreases. So option b is correct and option a and c incorrect.
An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when descending at constant CAS in an isothermal the decreasing TAS causes Mach number to decrease. Option d is therefore incorrect.
ATMOS 51 b.
This problem can be solved using the following equation: Pressure altitude = Elevation + (30 x ( 1013 – QNH)) Inserting the figures provided in the question gives: Pressure altitude = 4000 ft + ( 30 x (1013 – 1025)) Which is 4000 ft + -360 ft = 3640 ft.
ATMOS 52. a
In the international standard atmosphere (ISA) the relationship between altitude, temperature, pressure and density are fixed. As altitude decreases in the ISA, both the air pressure and temperature increase. The increasing pressure tends to compress the air increasing its density, whilst the increasing temperature. Tends to expand the air, decreasing its density. The effects of the pressure change are however the dominant factor so as pressure and temperature increase in the ISA, so does the air density.
ATMOS 53. c.
In the international standard atmosphere (ISA) the relationship between altitude, temperature, pressure and density are fixed. As altitude decreases in the ISA, both the
air pressure and temperature increase. The increasing pressure tends to compress the air thereby increasing its density, whilst the increasing temperature tends to expand the air, thereby decreasing its density. The effect of the pressure change is however the dominant factor so as pressure and temperature increase in the ISA, so does the air density. But in a non-standard atmosphere these relationships are not fixed. Metrological conditions can cause temperature and pressure to vary such that density might increase or decrease. Depending upon the relative magnitudes of the variation in each. If the increase in pressure is of
a greater proportion than that
of temperature then the
density will increase. Conversely, if the increase in temperature is of a greater proportion than that of pressure, the density will decrease.
ATMOS 54 d.
As atmospheric pressure increases, the air is compressed causing its density to increase. As air temperature increases, the air expands causing its density to decrease. In the ISA where both pressure and temperature increase together, the pressure change is the dominant factor, so density increases. If however pressure remains constant whilst temperature increase, then the air will expand and air density will decrease. Calibrated Airspeed (CAS) is proportional to dynamic pressure ( ½ρV 2 ), where ρ is air density and V is the True Airspeed (TAS). This means that any given value of ½ρV 2 will occur at the same CAS at all altitudes. If air density decreases, the TAS equating to any given CAS must increase, so that the increasing V 2 balance the decreasing ρ, such that ½ρV2 remains constant. This in turn means that if density decreases, then the ratio of CAS:TAS decreases. So if pressure remains constant whilst temperature increases the density will decrease, causing the CAS:TAS ratio to decrease.
ATMOS 55. c.
In the International Standard Atmosphere (ISA) the relationships between altitude, temperature, pressure and density are fixed. As altitude increases in the ISA, both the air pressure and temperature decrease. The decreasing pressure tends to expand the air thereby decreasing its density, whilst the decreasing temperature tends to compress the air thereby increasing its density. The effect of the pressure change is however the
dominant factor, so as altitude increases, the pressure temperature and density all decrease at a fixed rate. The pressure at any altitude and the temperature at any altitude are therefore fixed.
In the real atmosphere the rates of change of pressure and temperature with changes in altitude are not fixed. It is therefore possible that the pressure and density at any given altitude do not conform to the ISA predictions.
Pressure altitude is the altitude at which the prevailing pressure at any point in a real atmosphere would occur in the ISA. It is the altitude indicated on a barometric altimeter when 1013.25 hPa is set on the subscale. Elevations in the ISA are vertical distances above msl. Elevation in the ISA is therefore the same as pressure altitude.
ATMOS. 56 a.
In the International Standard Atmosphere (ISA) the relationships between altitude, temperature, pressure and density are fixed. As altitude increases in the ISA, both the air pressure and temperature decrease. The decreasing pressure tends to expand the air thereby decreasing its density, whilst the decreasing temperature tends to compress the air thereby increasing its density. The effect of the pressure change is however the dominant factor, so as altitude increases, the density decrease. Aircraft performance is directly proportional to power available. Power available is directly proportional to the mass flow of air passing through the engines. As air density decreases the mass flow of air passing through the engines decreases, causing power available and aircraft performance to decrease. This is the primary factor responsible for the reduction in aircraft performance as altitude increases so option a is the most appropriate in this question.
A secondary factor is that as altitude increases, the lift and drag forces at any given IAS remain constant but the TAS increases. But power required is equal to TAS multiplied by drag, so as altitude increases, the power required for any given flight condition increases. This increasing power requirement, coupled with decreasing power available increases the rate of reduction in performance.
ATMOS 57. b.
In the ISA the relationships between altitude, pressure, temperature and density are fixed such that the density at any given pressure altitude is always the same.
In the real atmosphere the rates of change of pressure and temperature with changes in altitude are not fixed. It is therefore possible that the density at any given altitude does not confirm to the ISA predictions. Pressure altitude is the altitude at which the prevailing pressure at any point in a real atmosphere would occur in the ISA. It is the altitude indicated on a barometric altimeter when 1013.25 hPa is set on the subscale. The density at any given pressure altitude is not however constant in a real atmosphere. The density altitude is the pressure altitude at which the prevailing density would occur in the ISA. Option b is therefore correct. Options a and c are incorrect because they use the term “conditions” but do not state what these conditions are ( temperature, pressure, density, humidity). This option is therefore too loosely worded to be accurate. Option d is also incorrect because dividing actual density by density lapse rate will not give the density altitude. It would be more accurate to say that: Density alt = ( ambient density – ISA msl density ) / density lapse rate. ATMOS. 58. a.
This problem can be solved using the following equation: Density alt = pressure alt + (118.6 x temperature deviation) Where temp deviation = (ambient temp – (15 – (1.98 x pressure alt / 1000 ft))) degrees. At a pressure altitude of 10000 ft and an ambient temperature of 12 0 C, Temp deviation = (12 – (15 –(1.98 x 10))) which is 12 + 4.8 or 16.8 Inserting this value into the original equation gives: Density altitude = 10000 ft + (118.6 x 16.8) which is 11992.4. The closest option is therefore option a, 11993 ft. It should be noted that in some examinations a less accurate of Density alt = Pressure alt + (120 x temp deviation) is sometimes used.
ATMOS. 59. b.
This problem can be solved using the following equation: Density alt = pressure alt + (118.6 x temperature deviation) Where temp deviation = (ambient temp – (15 – (1.98 x pressure alt / 1000 ft))) degrees. At a pressure altitude of 0 ft and an ambient temperature of 12 0 C, Temp deviation = (10 – (15 –(1.98 x 0))) which is -5 Inserting this value into the original equation gives: Density altitude = 0 ft + (118.6 x -5) which is -593 ft. It should be noted that in some examinations a less accurate of Density alt = Pressure alt + (120 x temp deviation) is sometimes used.
ATMOS. 60. c.
This problem can be solved using the following equation: Pressure altitude = Elevation + (30 x(1013-QNH)) Inserting the values given in the question gives: Pressure altitude = 1000 ft + (30 x ( 1013 -1025)) which is 640 ft. It should be noted that the problem can also be solved using the following equation: Pressure altitude = 30 x ( 1013 – QFE) Which gives pressure altitude = 30 x ( 1013 – 991.67) = 639.9 ft.
