PRELIMINARIES: LIFT AND DRAG (Text Chapter 4)
ENROUTE: THRUST REQUIRED VS. POWER REQUIRED CURVES (Text Chapter 5)
Thrust Required and Power Required Curves are derived from an aircraft’s total drag curve. Since there is a formulaic relationship between the two curves, they are in one sense two ways of visualizing the same information. However, the Thrust Required Curve is most relevant to jet propelled aircraft, which are thrust producers, while the Power Required Curve is more suitable for prop propelled aircraft, which are power producers. Moreover, different interpretations of the curves apply, depending on whether the curves reflect performance of a thrust producer on the one hand, or a power producer on the other.
What follows is divided into five sections:
1. Derivation of Thrust Required and Power Required Curves
2. Interpretation of the Curves for Thrust Producers
3. Interpretation of the Curves for Power Producers
4. Effects of Weight and Altitude Change on Thrust and Power Required Curves
5. Interpreting Effects of Weight and Altitude Change on Range and Endurance
In general, AS310 focuses on jet propelled heavy aircraft. However, the information that follows also touches on prop airplane performance.
1. Derivation of Thrust Required and Power Required Curves
Parasite + Induced Drag = Total Drag: Since dynamic pressure q = (V2/2, parasite drag increases exponentially as airspeed increases, especially when compressibility effects begin. At the same time, since high velocity 1G steady state flight is associated with a low AOA, induced drag decreases as airspeed increases. The scalar sum of parasite and drag for a given airspeed is known as total drag, and the total drag curve is just a plot of total drag vs. TAS over the airspeed range of the aircraft. A total drag curve is valid for a specific airplane in straight and level flight at fixed density altitude, gross weight, and aircraft configuration.
Figure 5.1depicts typical drag curves
[pic]
Figure 5.1. Typical Induced, Parasite, and Total Drag Curves
Thrust Required Curve. Let TR be the thrust required to sustain steady state 1G level flight at a given altitude and gross weight. If (and only if) the pitch angle is small, then thrust TR is approximately equal to drag D, and lift L is approximately equal to weight W. Under these circumstances, the Total Drag Curve is also the Thrust Required Curve; i.e., we may read TR on the vertical axis of the Total Drag Curve. Figure 5.2 (top half) depicts a typical Thrust Required Curve.
Again, bear in mind that any Total Drag / Thrust Required Curve is valid for a single given airplane operating in straight and level flight at fixed gross weight and in a given configuration. Changes in any of these three parameters—gross weight, altitude, or configuration—will also change the curve.
[pic]
Figure 5.2. Corresponding Thrust Required and Power Required Curves
Power vs. Thrust. We saw earlier that power is work / time = force x distance / time = thrust x distance / time = thrust x velocity. More formally, for steady state 1G level flight
P = T V, where P is power, T is thrust in pounds, and V is TAS in ft/sec.
Recall that the unit of power is ft-#/sec. Since horsepower is a more conventional measure of aircraft power, and since 1 hp = 550 ft-#/sec. of work, we can rewrite the power equation as
P = T V / 550, where T is in #, V in ft/sec, and P in hp.
If we want to express V in kts, we must multiply V by [pic], giving
P = TV/550 = [pic] = TV / 325.9 hp.
As an example, a airplane producing 10,000# of thrust at 350 kts is putting out
P = TV /325.9 = 10,000 (350) / 325.9 =10,739 hp of work.
Power Required Curve. These Power Required Curve is related to the Thrust Required Curve by the equation PR= TR V / 325.9. The two curves in one sense are simply two different ways of presenting the same information about an airplane’s performance.
However, since jets airplanes are thrust producers and prop airplanes are power producers, for performance analyses it is natural and customary to use the Thrust Required Curve for a jet and the Power Required Curve for a prop.
Figure 5.2 shows the Thrust Required Curve and corresponding Power Required (PR) Curve for an airplane. The latter curve can produced by taking points (V, TR) on the former curve and generating corresponding points (V, TR V / 325.9) = (V, PR) for the Power Required Curve. For example, the minimum point [(L/D)max] on the Thrust Required Curve is (160, 1250), producing the corresponding point on the Power Required Curve (160, 1250 (160) / 325.9) = (160, 614). Note that the derived point is not minimum on the Power Required Curve; however it does correspond to (L/D)max, as discussed below.
Alternately, the Thrust Required Curve in Figure 5.2 could be constructed by taking points (V, PR) on the Power Required Curve and producing corresponding points (V, 325.9 PR/V)., since T = 325.9 P/V. Thus, point (160, 614) on the Power Required Curve becomes point (160, 325.9 (614) / 160) = (160, 1250) on the Thrust Required Curve.
We will use versions of Figure 5.2 frequently in what follows. Table A presents three points on the TR Curve in Figure 5.2 and the corresponding three points on the PR Curve. These three points will be of interest again in Section 4. Note that the minimum point on the TR curve corresponds to the point on the PR curve where a tangent line (yellow) from the origin intersects the curve. More on this later.
|Points on TR Curve |Corresponding Points on PR Curve |
|(160, 1250) |(160, 1250 (160) / 325.9) = (160, 614) |
|(100, 1800) |(100, 1800 (100) / 325.9) = (100, 552) |
|(250, 1800) |(250, 1800 (250) / 325.9) = (250, 1381) |
Table A. Three Corresponding Points on TR and PR Curves of Figure 5.2
2. Interpretation of Thrust and Power Required Curves for Thrust Producers
This section applies to jet aircraft. Prop aircraft are discussed in the following section.
Thrust Required Curve Regions. The power required curve reveals that, except for the airspeed associated with (L/D)max (minimum thrust), two distinct airspeeds typically frequently correspond to the same thrust setting. That is, the same thrust setting will sustain two different airspeeds in steady state straight and level flight.
• The lower of these two airspeeds always lies in the region of reverse command, so called because any disturbance to steady state flight in this region which causes the aircraft to decrease velocity requires that more throttle be applied to reestablish steady state flight. For this reason, the region of reverse command has also been called the speed unstable region, the region of reverse command, or the back side of the thrust curve. In this region, thrust required for steady state flight increases as airspeed decreases, and decreases as airspeed increases.
• The higher of these two airspeeds lies in the speed stable region or region of normal command, where power required for steady state flight increases as airspeed increases.
• For flight at TASs near (L/D)max, it may be difficult to maintain airspeed without fairly frequent small throttle adjustments. This is reflected in the flat slope of the curve near this point: a small throttle change corresponds to a significant airspeed change. In short, it’s hard to get the throttle set precisely to TR in this regime, which is why this part of the curve is called the region of neutral airspeed stability.
Max Endurance Airspeed (VBE) on Thrust Required Curve. Assume W = L for 1 G steady state flight), and that D = TR (approximately true for small pitch angles only). Then, since the ratios of equal values are themselves equal, we may write
L / D = W / TR, or TR = W / (L/D).
Since weight is assumed constant, it follows that the minimum thrust required (TR)min will occur when flying at the airspeed (AOA) corresponding to (L/D)max. More formally,
(TR)min = [pic].
Since (TR)min is just the minimum point on the Thrust Required Curve, and since minimum thrust results in minimum fuel flow, it follows that the airspeed corresponding to this point is the airspeed for (L/D)max.
Important Point. The airspeed corresponding to (L/D)max occurs at the lowest point on the Thrust Required Curve, and—as we learned in AS310 Unit I—(L/D)max is the slope of a line on the Drag Polar drawn from the origin tangent to the curve.
Important Point Maximum time on station (best endurance or max endurance) is achieved when thrust is minimum. Thus the airspeed associated with (L/D)max is the best endurance airspeed. As seen above, this is the airspeed corresponding to the lowest point on the Thrust Required Curve.
Best Range Airspeed (VBR) on Thrust Required Curve. Best range requires that the ratio fuel flow / true airspeed (FF / V) be minimum. If we assume that Thrust Specific Fuel Consumption (TSFC) is constant over the airspeed range for a given altitude, TSFCmin results in FFmin. That is, under these assumptions, (TR / V) ( (FF / V), and
(FF / V)min = k (TR / V)min for some constant k.
Thus it suffices to minimize the ratio (TR / V) to get minimum fuel flow and maximum range. On the TR curve, (TR / V)min occurs at the point where a tangent line from the origin intersects the curve, because
• any point on the curve has the (x, y) coordinates (V, TR), and the slope of a straight line from the origin to that point is (y / (x = TR / V.
• we desire that (TR / V)—i.e. the slope of the tangent line—be minimum.
• a straight line drawn from the origin to any other point on the TR curve has a slope greater than the slope of the tangent line.
These ideas are illustrated in the following diagram.:
Max Endurance and Max Range Points on Thrust Required Curve for Jet Aircraft.
(V/PR)min on the Power Required Curve. Any point (V, PR) on the PR curve can be used to produce the ratio PR/V for that point, because the slope of a line drawn from the origin to any point on the curve is PR/V. To find the point where the ratio (PR/V) is minimum, draw a straight line from the origin tangent to the curve. The intersection of the tangent line with the curve intersects the point (PR/V)min, because the tangent line produces the minimum slope of any straight line from the origin that intersects the curve.
(L/D)max and Best Endurance Airspeed on the Power Required Curve. (PR/V)min on the PR curve gives VBE and (L/D)max. Since P = TV, (PR/V)min = (TR V/ V)min = (TR)min. But we have seen that (TR)min corresponds to (L/D)max, so the airspeed associated with the point (PR/V)min on the PR is the airspeed for best endurance (VBE) and best power off glide ratio (VBG).
These ideas are illustrated in the following diagram:
Max Endurance Point on Power Required Curve for Jet Aircraft
Summary of Important Points about Thrust Producers Performance. Assuming TSFC is constant with airspeed changes for fixed altitude:
• FF ( TR for Thrust Producers, i.e. for jet airplanes. This relationship is the basis for conclusions about max range and endurance airspeeds for jets.
• The minimum point on the TR curve corresponds to max endurance airspeed VBE. This airspeed is achieved at (L/D)max, and is also the airspeed for best power off glide ratio VBG.
• To find the point on the TR curve corresponding to max range airspeed VBR, draw a straight line from the origin tangent to the curve.
• To find the point on the PR curve corresponding to VBE, draw a straight line from the origin tangent to the curve. This airspeed is achieved at (L/D)max, and is also the airspeed for VBG.
• Ordinarily, the TR curve (as opposed to the PR curve) is used for performance analysis on jet propelled aircraft. On the TE curve, one can find both VBE and VBR for a jet; on the corresponding PR curve, one can find only the VBE airspeed.
• Interpretation of the TR and PR curves is different for jet aircraft than for prop aircraft.
3. Interpretation of Thrust and Power Required Curves for Power Producers
This section applies to normally aspirated, reciprocating powered prop aircraft. Jet aircraft are discussed in the previous section.
Prop aircraft are power producers, and their fuel flow is proportional to brake horsepower (BHP) developed. Contrast this with thrust producers (jets) where fuel flow is proportional to thrust developed for fixed altitude.
Brake Specific Fuel Consumption. Brake specific fuel consumption (BSFC) for prop aircraft which produce brake horsepower (BHP) is the counterpart of Thrust Specific Fuel Consumption (TSFC) for jet powered aircraft. BSFC is defined as fuel flow / brake horsepower.
BSFC = FF/BHP.
We will assume that BSFC for a given power setting is constant across the airspeed regime, and that propeller efficiency does not change. (This is a fairly naïve—but not absurd—assumption.). From this assumption, we may conclude
FF/V = k’ (PR / V) for some constant k’.
We also assume that that FF is proportional to PR for fixed altitude. It follows that VBR for a prop occurs at (PR/V)min on the PR curve, and that VBE occurs at (PR)min.
Max Endurance Airspeed on Power Required Curve. The maximum endurance airspeed VBE for a prop occurs at the minimum point on the PR curve. This is true since we are assuming that FF ( PR for fixed altitudes.
Important Point: The point (VBE, (PR)min) gives VBE for a power producer only. As we saw in the previous section, VBE for a thrust producer on the PR curve is the point intersected by a line drawn from the origin tangent to the curve. The reason for this is that—according to our assumptions—a thrust producer has minimum fuel flow at minimum thrust, while a power producer has minimum fuel flow at minimum power.
Max Range Airspeed on Power Required Curve. Max range airspeed VBR is achieved when fuel flow / V (FF/V ) is minimum. Since we assume FF ( PR for a power producer, we have seen that this occurs when PR / V is minimum, i.e., at (PR/V)min on the power required curve. We already know this point can be found by drawing a tangent from the origin to the curve, and that the airspeed corresponding to this point gives (L/D)max. These ideas are illustrated in the following diagram.
Max Endurance and Max Range Points on Power Required Curve for Prop Aircraft.
Significance of the (L/D)max Point on the TR and PR Curves. We saw in the previous section that the point on the PR curve corresponding to (PR/V)min is the point corresponding to Tmin on the TR curve, i.e., the minimum point on the TR curve. Both points occur for the velocity for (L/D)max. From this we also conclude that a prop gets max range at (L/D)max, while a jet gets best endurance at (L/D)max.
The following diagram depicts the point on the TR curve that gives VBR for a prop.
Max Range Point on Thrust Required Curve for Prop Aircraft
Summary of Important Points about Power Producers. Assuming constant propeller efficiency, and constant BSFC with airspeed changes—i.e. assuming FF (PR for constant altitude:
• The assumption that FF (PR for an unaspirated reciprocating engine leads to the generalizations that follow.
• The minimum point on the PR curve corresponds to the airspeed for maximum endurance speed VBE because it’s the point where minimum FF occurs..
• To find the point on the PR curve corresponding to max range speed VBR, draw a straight line from the origin tangent to the curve. This line has the slope (PR/V)min, hence intersects the point on the curve where FF/V is minimum, i.e. the best range airspeed point. The point corresponds to (L/D)max, so a prop airplane gets max range at (L/D)max.
• The point on the TR curve corresponding to maximum range airspeed for a prop is the minimum point on the curve., and also corresponds to (L/D)max.
• Ordinarily, the PR curve (as opposed to the TR curve) is used for performance analysis on prop propelled aircraft because both max endurance and max range airspeeds can be found, whereas only the max range airspeed can be found on the thrust required curve.
• Interpretation of the TR and PR curve is different for prop aircraft than for jet aircraft.
Summary of Interpretations of TR and PR Curves for Thrust and Power Producers.
This pages summarizes information presented in the previous two sections. Everything shown below is fundamental for answering quiz and test questions about the TR and PR curves.
|Interpretation of TR and PR Curves for Thrust Producers and Power Producers |
|TR in #, PR in HP, Velocity in KTAS |
|Assumptions: For Fixed Altitude, FF ( TR for Thrust Producer; FF ( PR for a Power Producer |
|TR |For Thrust Producers Assuming FF( TR |Power Producers Assuming FF( PR |
| |[pic] |[pic] |
|PR | | |
| |VBE Jet VBR Jet |VBE Prop VBR Prop |
|VBE best endurance KTAS; VBR = best range KTAS |
| |Thrust Producer (Jet) |Power Produce (normally aspirated prop) |
|VBE |Low Point on TR Curve |Low Point on PR Curve |
| |Tangent Point on PR Curve |Can’t Find on TR Curve |
|VBR |Tangent Point on TR Curve |Tangent Point on PR Curve |
| |Can’t Find on PR Curve |Low Point on TR Cur |
|(L/D)MAX |Best Endurance for a Jet |Best Range for a Prop |
|Fuel Flow |FF( TR |FF( PR |
|Importance |Has More Information for a Jet |Has More Information for a Prop |
| |Can Find both VBE and VBR |Can Find both VBE and VBR |
• Relationship between curves: P = (TV)/325.9; T = (325.9 P)/V; if know one curve, can construct the other.
• How the curve pairs are interpreted does not change the curves themselves or the relationship between them.
• Different interpretations for jets and props derives from fuel flow assumptions—proportional to thrust required for a jet; proportional to power required for a prop.
4. Effects of Weight and Altitude Change on Thrust and Power Required Curves
We will explain the effect of changes in weight W and density altitude (as reflected by density ratio () on the Thrust Required Curve. We have already seen that every point (V, TR) on this curve becomes the point (V, TR V / 325.9) = (V, PR) on the Power Required Curve. As a consequence, the explanation also implicitly indicates effects of weight and altitude changes on the Power Required Curve.
Weight Changes. Let (V1, T1) be any point on a Thrust Required Curve corresponding to aircraft gross weight W1. We saw earlier that for some other gross weight W2, V2 = [pic], or [pic]. Since D = TR = [pic], [pic] when CD and ( are constant. We may conclude that any point (V, TR) on the Thrust Required Curve for gross weight W1 becomes the point [pic]on the Thrust Required Curve for gross weight W2. For example, if we plot the curves for gross weights W1 = 15,000# and W2 = 22,500#, since 22,500 / 15,000 = 1.5, a point (V, TR) the curve for gross weight W1 is transformed to the point (V((1.5, TR(1.5) on the curve for gross weight W2.
Figure 5.4 depicts these two curves (and the corresponding Power Required Curves). Table A shows three examples of how points on the curve for 15,000# gross become corresponding points on the curve for 22,500# gross. Note that the points in the left column of Table B are also points in the left column of Table A, since Figures 5.2 and 5.4 depict the same PR curves for 15,000# gross.
|Points on 15,000 # Gross TR Curve |Corresponding Points on 22,500 # Gross TR Curve |
|(160, 1250) |(160(((22,500/15,000), 1250((22,500/15,000)) = (196, 1875) |
|(100, 1800) |(100(((22,500/15,000), 1800((22,500/15,000)) = (122, 2700) |
|(250, 1800) |(250(((22,500/15,000), 1800((22,500/15,000)) = (306, 2700) |
Table B. Examples of Point Relationships on TR Curves for 15,000# and 22,500# Gross in Figure 5.4.
Figure 5.4 reflects the fact that increasing gross weight shifts the Thrust Required Curve (and the Power Required Curve) up and to the right, while reducing gross weight shifts the curves down and to the left.
The quantitative change in the Power Required Curve with a change in gross weight may be analyzed as follows:
• Point (V, TR) on the Thrust Required Curve corresponds to point (V, PR) = (V, TR V / 325.9) on the Power Required Curve.
• With a change in weight from W1 to W2, point (V, TR) on the Thrust Required Curve for W1 becomes point (V ((W2 / W1), TR (W2 / W1) on Thrust Required Curve for W2.
• Thus, the corresponding point (V, PR) = (V, TR V / 325.9) on the Power Required Curve for W1 becomes point (V ((W2 / W1), [TR (W2 / W1)] [V ((W2 / W1)] / 325.9) = (V ((W2 / W1), TR V (W2 / W1)3/2 / 325.9) = (V ((W2 / W1), PR (W2 / W1)3/2) on the Power Required Curve for W2.
That is, for a weight change from W1 to W2, a point (V, PR) on the Power Required Curve for weight W1 becomes point (V ((W2 / W1), PR (W2 / W1)3/2) on the Power Required Curve for weight W2. As an example, the point (160, 614) on the Power Required Curve for W =15,000# in Figure 5.4 becomes point (160 (((22,500 / 15,000), 614 ((22,500 / 15,000)3/2 = (196, 1128) on the Power Required Curve for W = 22,500#.
Referring still to Figure 5.4, Table C exemplifies how three points on the PR Curve for 15,000# gross are transformed to three corresponding points on the PR Curve for 22,500# gross. Points in the left column of Table C are the same as points in the right column of Table A. The points (V, PR) in the left column are transformed by (V, PR) ( (V ((W2 / W1), PR ((W2 / W1)3/2 ) to produce points in the right column.
|Points on 15,000 # Gross PR Curve |Corresponding Points on 22,500 # Gross PR Curve |
|(160, 614) |(160(((22,500/15,000), 614((22,500/15,000)3/2) = (196, 1128) |
|(100, 552) |(100(((22,500/15,000), 552((22,500/15,000)3/2) = (122, 1014) |
|(250, 1381) |(250(((22,500/15,000), 1381((22,500/15,000)3/2) = (306, 2537) |
Table C. Examples of Point Relationships on PR Curves for 15,000# and 22,500# Gross in Figure 5.4.
[pic]
Figure 5.4. Effects of Changing Gross Weight on Thrust Required and Power Required Curves
Important Point: When W2 > W1, both the TR and PR curves move right by a factor of ((W2 / W1). The TR curve moves up by a factor of W2 / W1, while the PR curve moves up by a factor of (W2 / W1)3/2. That is, both curves move up faster than they move right when gross weight increases. We will see in the next section that these facts have important performance implications.
Altitude Changes. Let (V1, T1) be any point on a Thrust Required Curve corresponding to aircraft density altitude (1. We saw earlier that for some other density altitude (2, V2 = [pic], or [pic]. Since D = TR = [pic], [pic]. That is (TR)2 = (TR)1. We may conclude that any point (V, TR) on the Thrust Required Curve for density altitude (1 becomes the point [pic]on the Thrust Required Curve for density altitude (2.
Figure 5.8 depicts corresponding PR curves for SL and 22,000’ altitude, and Table D shows three examples of how points on the TR curve for (1 = 1.0 (SL) become corresponding points on the TR curve for (2 = 0.498 (22,000’). Note that the points in the left column of Table D are those in the left column of Table A, since Figures 5.2 and 5.8 depict the same TR SL curves. Points in right column of the table below are transformed from the left column by (V, TR) ( (V (((1 / (2), TR) using the equality ((1.0 / 0.498) = 1.417.
|Points on SL TR Curve |Corresponding Points on 22,000’ TR Curve |
|(160, 1260) |(160 ((1.0 / 0.498), 1260) = (227, 1260) |
|(100,1800) |(100 ((1.0 / 0.498), 1800) = (142, 1800) |
|(250,1800) |(250 ((1.0 / 0.498), 1800) = (354, 1800) |
Table D. Examples of Point Relationships on TR Curves for SL and 22,000’ in Figure 5.8
Note that Figure 5.8 reflects the fact that increasing altitude shifts the Thrust Required Curve laterally to the right, (and the Power Required Curve up and to the right), while decreasing altitude shifts the Thrust Required Curve to the left (and the Power Required Curve down and to the left.
The quantitative change in the PR Curve with a change in altitude may be analyzed as follows:
• Point (V, TR) on the TR Curve corresponds to point (V, PR) = (V, TR V / 325.9) on PR Curve.
• With a change in altitude from (1 to (2, point (V, TR) on the TR curve for (1 becomes point (V (((1 / (2), TR) on the TR Curve for (2.
• Thus, the corresponding point (V, PR) = (V, TR V / 325.9) on the PR Curve for (1 becomes point (V (((1 / (2), [TR V (((1 / (2) / 325.9) = (V (((1 / (2), PR (((1 / (2) on the PR Curve for (2.
Corresponding points on the SL and 22,000’ Power Required Curves are presented in Table E. Points in the left column of Table # are the same as points in the right column of Table A, since Figures 5.2 and 5.8 depict the same PR sea level curves. Points in right column of the table below are transformed by (V, PR) ( (V (((1 / (2), PR (((1 / (2).
|Points on SL PR Curve |Corresponding Points on 22,500’ PR Curve |
|(160, 619) |(160 ((1.0 / 0.498), 619 ((1.0 / 0.498)) = (227, 877) |
|(100, 552) |(100 ((1.0 / 0.498), 552((1.0 / 0.498)) = (142, 782) |
|(250, 1381) |(250 ((1.0 / 0.498), 1381 ((1.0 / 0.498)) = (354, 1957) |
Table E. Examples of Point Relationships on PR Curves for SL and 22,000’ in Figure 5.8
[pic]
Figure 5.8. Effects of Changing Altitude on Thrust Required and Power Required.
Important Point: When (1 > (2 (i.e., when altitude increases), both the TR and PR curves move right by a factor of (((1 / (2). The TR curve does not move up, while the PR curve moves up by a factor of (((1 / (2). That is, the TR curve moves right but not up when altitude increases, while the PR curve moves right and up by the same amount. In the next section, we see that these facts have important performance implications.
Summary of Quantitative Changes to TR and PR Curves with Gross Weight or Altitude Change. Table F summarizes the effects of changing gross weight or altitude on thrust or power required, and the corresponding steady-state straight-and-level A/Ss sustainable by this thrust or power.
| |Gross Weight Change W1 to W2 |Density Altitude Change (1 to (2 |Change W1 to W2 & (1 to (2 |
| | | | |
|TR Curve |V2 = V1 ((W2 / W1) |V2 = V1 (((1 / (2) |V2 = V1 ((W2 / W1) (((1 / (2) |
| |(TR)2 = (TR)1 (W2 / W1) |(TR)2 = (TR)1 |(TR)2 = (TR)1 (W2 / W1) |
| | | | |
|PR Curve |V2 = V1 ((W2 / W1) |V2 = V1 (((1 / (2) |V2 = V1 ((W2 / W1) (((1 / (2) |
| |(PR)2 = (PR)1 (W2 / W1)3/2 |(PR)2 = (PR)1 (((1 / (2) |(PR)2 = (PR)1 (W2 / W1)3/2 (((1 / (2) |
Table F. Summary of Gross Weight and Altitude Change Effects on TR and PR Curves
Table G indicates how the formulas in Table F, together with P = TV/325.9 and T = 325.9 P / V can be used to make calculations for solving problems about how gross weight and density altitude changes affect PR and TR. In the table, T means TR, and P means PR.
|V1, T1 |V2 = V1 ((W2 / W1) (((1 / (2) |V2, T2 |
| | | |
| |T2 = T1 (W2 / W1) | |
|P1 = T1 V1/ 325.9| |P2 = T2 V2/ 325.9|
| | | |
| | | |
|T1 = 325.9 P1 / | |T2 = 325.9 P2 / |
|V1 | |V2 |
|V1, P1 |V2 = V1 ((W2 / W1) (((1 / (2) |V2, P2 |
| | | |
| |P2 = P1 (W2 / W1)3/2 (((1 / (2) | |
Table G.
5. Interpreting Effects of Weight and Altitude Change on Range and Endurance.
Following are some conclusions we may draw from the preceding. Assertions about max range and endurance assume that for fixed density altitude, FF ( TR for a thrust producer, and FF ( PR for a power producer.
Jet Aircraft Change in Gross Weight:
• Max endurance and max range TASs increase with an increase in gross weight.
• The increase in max endurance and max range TASs is proportional to ((W2 / W1).
• TR and hence fuel consumption also increase for both TASs. EASs also increase.
• The increase in TR and FF is proportional to (W2 / W1).
• The ratio TR / V also increases.
• As a consequence, max range and max endurance time decrease. (This is intuitive.)
Prop Aircraft Change in Gross Weight:
• Max endurance and max range TASs increase with an increase in gross weight.
• The increase in max endurance and max range TASs is proportional to ((W2 / W1).
• PR and hence fuel consumption also increase for both TASs.
• The increase in PR and FF is proportional to (W2 / W1)3/2.
• The ratio PR / V also increases.
• As a consequence, max range and max endurance time decrease. (This is also intuitive.)
Jet Aircraft Change in Altitude:
• Max endurance and max range TASs increase with an increase in altitude.
• The increase in TAS is proportional to (((1 / (2). EAS does not change.
• TR remains constant: since EAS does not change with altitude, so DT does not change.
• TR does not change, but if altitude increases up to the tropopause, endurance time will also increase, because a jet engine gets better TSFC at higher altitudes. That is, the same thrust will be achieved with a somewhat lower fuel flow.
• Since TAS increases and TR remains constant, TR / V decreases.
• Thus max range increases by a factor of (((1 / (2). This increase is due solely to the increasing disparity between EAS and TAS, given by the equation TAS = EAS / ((. There will also be an increase in range due to better TSFC at higher altitudes.
Prop Aircraft Change in Altitude:
• Max endurance and max range TASs increase with an increase in altitude.
• The increase in max endurance and range TASs is proportional to (((1 / (2),
• PR and fuel consumption also increase for TASs.
• The increase in PR and FF is proportional to ( ((1 / (2).
• Since (PR)min increases, fuel flow for VBE also increases, and max endurance decreases.
• Since PR and TAS for max range increase at the same rate, the ratio PR / V remains constant, so max range remains unchanged. That is, low altitude means low airspeed and fuel consumption, and high altitude means high airspeed and fuel consumption, while max range no wind is unchanged. However, you do get to your destination faster at the higher altitude, assuming no adverse winds. On the other hand, it takes extra fuel to climb, so in fact range will theoretically be reduced by the amount of extra fuel it takes to climb to a higher altitude.
• Theoretically, best range for a normally aspirated reciprocating powered prop airplane is achieved at sea level.
Example Problem. A jet air transport at 200,000# gross requires 45,000# of thrust to maintain an optimum cruise airspeed of 450 KTAS at FL240. If gross weight decreases to 180,000#, what is the new cruise airspeed and thrust required to maintain it?
V2 = V1 ( (W2 / W1) = 450 ( (180,000 / 200,000) = 426.9 KTAS.
(TR)2 = (TR)1 (W2 / W1) = 45,000 # (180,000 / 200,000) = 40,500 #.
Example Problem. A prop airplane requires 575 HP at SL to cruise at 250 KTAS at 10,000 # gross weight. If gross weight increases to 11,500#, what is the new SL cruise airspeed and power required to maintain it?
V2 = V1 ( (W2 / W1) = 250 ( (11,500 / 10,000) = 268.1 KTAS.
(PR)2 = (PR)1 (W2 / W1)3/2 = 575 HP (11,500 / 10,000)3/2 = 709.1 HP.
Example Problem. A prop airplane has a maximum SL endurance airspeed of 175 KTAS at 375 HP at 7000# gross. What is the maximum endurance airspeed and power required at 10,000’ at the same gross weight? Assume a standard atmosphere.
From standard atmosphere table, (10,000’ = 0.7385.
V2 = V1 ( ((1 / (2) = 175 ( (1.0 / 0.7385) = 203.6 KTAS.
(PR)2 = (PR)1 ( ((1 / (2) = 375 HP ( (1.0 / 0.7385)= 436.4 HP.
Example Problem. A jet air transport at 100,000# gross requires 15,000# of thrust to maintain an optimum endurance airspeed of 350 KTAS at FL240. If gross weight increases to 110,000#, what is the optimum endurance airspeed at 10,000’, and thrust required to maintain it? Assume a standard atmosphere, and that TSFC is a function of TR only. Assuming no pitot static error or compressibility effect, what is the endurance airspeed in KIAS?
From standard atmosphere table, (24,000’ = 0.4642, and (10,000’ = 0.7385.
V2 = V1 (( (W2 / W1)) (( ((1 / (2)) = 350 (( (110,000 / 100,000) (( (0.4642 / 0.7385)= 291 KTAS.
(TR)2 = (TR)1 (W2 / W1) = 15,000 (110,000 / 100,000) = 16,500 #.
EAS = (TAS) (( = (291.0326442) ( 0.7385 = 250 KEAS = 250 KIAS by assumptions.
ENROUTE: JET AIRPLANE SPECIFIC RANGE AND ENDURANCE
Specific Range. Specific range SR for an airplane is defined as nautical miles traversed per pound of fuel expended. If we ignore winds aloft, then specific range may be defined as
SR = [pic].
For example, if TAS = 500 kts and fuel flow is 3000 # / hr for each of two engines, then
SR = [pic] = 0.0833 nm / # fuel.
The above figure is a no wind specific range for the data given. Of course, in flight planning we must always take winds aloft into account. Note: SR for jet aircraft is often cited in nm / 1000 # fuel. If SR = 0.0833 nm / # of fuel, it is also true that SR = 0.0833 (1000) nm / 1000 # of fuel = 83.3 nm / 1000# of fuel.
We have seen previously that under the assumption that fuel flow for a jet is proportional to thrust, we can find (fuel flow / TAS)min by drawing a line from the origin of the Thrust Required Curve tangent to the curve. The point on the curve intersected by the tangent line corresponds to max range airspeed, since
(fuel flow / TAS)min = (TAS / fuel flow)max = SRmax.
Specific Endurance. Specific endurance SE for an airplane is defined as hours of flight sustainable per pound of fuel; i.e., SE = hours / pound of fuel, which is just the reciprocal of fuel flow (FF):
SE = [pic].
Note: SE for jet aircraft is often cited in hours / 1000# of fuel. A jet that can hold for 30 minutes on 1000 # of fuel has a fuel flow of 2000 # hour, and a SE = 1 / 2000 = 0.5 / 1000 = .0005 hours / # fuel or 0.5 hours / 1000 # fuel.
Specific endurance will be maximum when fuel flow is minimum.
SEmax [pic].
Relationship Between Specific Range and Specific Endurance. The following derivation reveals that dividing specific range by velocity gives specific endurance. That is, for any TAS V, SE = SR / V.
[pic].
Consider a plot of specific range vs. TAS, as shown in Figure 15.3 (middle curve). The following is true:
• A line from the origin intersecting the curve at (V, SR) has rise SR and run V, so its slope is SR / V.
• (SR / V)max occurs when the line is tangent to the curve, since no line with steeper slope can be drawn to intersect the curve.
• That is, a straight line drawn from the origin of the SR vs. TAS plot tangent to the curve has slope (SR / V)max = SEmax, i.e., gives the best endurance TAS.
[pic]
Figure 15.3. Fuel Flow and Specific Range Relationships
Recall that if fuel flow for a jet is proportional to thrust, the lowest point on the TR curve is where FFmin occurs, giving the airspeed for SEmax. The point where a tangent line from the origin intersects the curve is where (FF/V)max occurs, giving the airspeed for SRmax.
Effect of Weight and Altitude Changes on Jet Aircraft Specific Range and Specific Endurance
The following is summarized from this earlier discussion of Thrust Required Curves:
• A point (V, TR) on the Thrust Required Curve corresponding to gross weight W1 becomes point [pic]on the Thrust Required Curve corresponding to gross weight W2.
• A point (V, TR) on the Thrust Required Curve corresponding to an altitude where density ratio is (1 becomes point [pic]on the Thrust Required Curve corresponding to an altitude where density ratio is (2.
Assuming FF is proportional to thrust required (FF ( TR), it follows that
• When gross weight changes from W1 to W2, best endurance and best range TASs change by a factor of [pic], and thrust and FF change by a factor of [pic].
• When density altitude changes such that (1 corresponds to the initial altitude and (2 corresponds to the new altitude, best endurance and best range TASs change by a factor of [pic], and thrust and FF do change.
Since SR is TAS / fuel flow (V / FF) and specific endurance SE is 1 / fuel flow (1/FF):
1. When gross weight changes from W1 to W2, SR changes by a factor of ((W1 / W2).
SR1 = [pic].
2. When gross weight changes from W1 to W2, SE changes by a factor of W1 / W2.
SE1 = [pic].
3. When altitude changes such that (1 becomes (2, SR changes by a factor of (((1 / (2).
SR1 = [pic].
4. When altitude changes such that (1 becomes (2, SE does not change.
SE1 = [pic].
(Note: Recall that TSFC is defined as (# fuel/hour consumed) per # of Thrust), i.e., TSFC = FF/T. This analysis does not consider changes in TSFC for a jet airplane as altitude changes.)
Table G summarizes these equations:
|Assumes No Change in |Gross Weight Change W1 to W2 |Density Altitude Change (1 to (2 |
|TSFC With Altitude | | |
|Specific Range |SR2 = SR1 ((W1 / W2) |SR2 = SR1 (((1 / (2) |
|Specific Endurance |SE2 = SE1 (W1 / W2) |SE2 = SE1 |
Table G. Effect of Gross Weight and Altitude Change on Jet Aircraft Specific Range and Endurance
(Equations Do Not Account for Changes in TSFC as Altitude Changes)
Important review point: when altitude changes for a jet airplane with no change in gross weight, the EAS and total drag/thrust required do not change. But the corresponding TAS varies according to the equation TAS = EAS/(( = TAS (SMOE). Also, TSFC changes. For example, if altitude increases, more TAS is achieved without a change in EAS and drag, and range improves. Range also improves because the EAS is achieved at a lower fuel flow, since TSFC improves with increased altitude (up to the tropopause). Similarly, endurance improves because less fuel flow is required to sustain max endurance EAS at the higher altitude.
A more sophisticated approach to predicting SR and SE changes would account for changes in TSFC in a jet airplane as altitude changes. Let TSFC1 be the thrust specific fuel consumption at the initial altitude, and TSFC2 be the thrust specific fuel consumption at the final altitude. Assuming a linear change in TSFC as altitude changes, SR and SE would be changed by a factor of (TSFC1)/ (TSFC2), in addition to the changes in SR due to the TAS change with EAS and DT constant.
As an example, suppose a jet airplane achieves SR1 at an altitude where the density ratio is (1 and thrust specific fuel consumption is TSFC1. At a higher altitude where the density ratio is (2, the TSFC2 is only 80% of the TSFC1 at the lower altitude. That is, TSFC2 = 0.8 TSFC1, or (TSFC2)/)TSFC2) = 0.8, and (TSFC1)/TSFC2) = 1.25. Then the improvement in specific range SR2 is more accurately predicted by the equation:
[pic].
That is, in addition to the higher TAS achieved at the higher altitude, there is a 25% increase in range due to improved TSFC.
Specific endurance would be improved in a similar fashion; i.e.
[pic]
For example if TSFCFL35,0 = 0.75 TSFCSL, then at 35,000’ your loiter time would improved by 1 / 0.75 = 84/3 (1.33 over your loiter time at sea level, , i.e., by about 33%. That is, if you had 60 minutes endurance at SL, you would have (4/3) 60 = 80 minutes endurance at 20,000’. (However, you might not save enough fuel in holding at 35,000’ to make up the fuel it takes to climb to the “better” altitude. On the plus side, however, if you did climb to 35,000’, you would ultimately be able to make an idle descent at best endurance speed from that altitude, saving significant fuel as a result. As we have said before, fuel management in jet powered airplanes is a complex subject, part science and part art.)
Table H modifies Table G to take into account TSFC changes with altitude.
| |Gross Weight Change W1 to W2|Density Altitude Change (1 to (2|Density Altitude Change (1 to (2 |
| | |TSFC Constant |TSFC Varies with Altitude |
|Specific |SR2 = SR1 ((W1 / W2) |SR2 = SR1 (((1 / (2) |SR2 = (TSFC1)/TSFC2) SR1 (((1 / (2) |
|Range | | | |
|Specific |SE2 = SE1 (W1 / W2) |SE2 = SE1 |SE2 = (TSFC1)/TSFC2) SE1 |
|Endurance | | | |
Table H. Effect of Gross Weight and Altitude Change on Jet Aircraft Specific Range and Endurance, Taking into Account that TSFC Varies as Altitude Changes.
Example Calculation. A jet fighter at 25,000# has a specific range of 0.30 nm / # fuel (or 300 nm / 1000 # fuel) at FL370. If gross weight decreases to 22,000#, what is specific range at FL 370? What is the specific range at 25,000# gross at FL240? Assume a standard atmosphere, and that TSFC is not affected by altitude.
From standard atmosphere table, (37,000 = 0.2844, and (24,000 = 0.4642.
SR2 = SR1 ( (W1 / W2) = 300 ( (25,000 / 22,000) = 319.8 nm / 1000 # fuel.
SR2 = SR1 ( ((1 / (2) = 300 ( (0.2844 / 0.4642) = 234.8 nm / 1000 # fuel.
Example Calculation. For the previous aircraft, assume TSFC37,000 = 0.7 TSFCSL and TSFC24,000 = 0.79 TSFCSL. Assume a standard atmosphere and solve for SR at 24,000. Also, suppose SE at FL240 is 0.5 hrs / 1000 # fuel. What is SE at FL 370? How many additional minutes holding time per 1000 # fuel are achieved by holding at FL 370, neglecting fuel needed for the climb? Do you think it would be a good idea to climb to FL370 from FL 240 to gain the additional holding time?
SR2 = (TSFC1 / TSFC2) SR1 ( ((1 / (2)) = (0.7 / 0.79) 300 ( (0.2844 / 0.4642)
= 208.1 nm / 1000 # fuel.
SE2 = (TSFC1 / TSFC2) SE1 = (0.79 / 0.7) 0.5 = 0.56 hours / 1000 # fuel.
(0.56 – 0.5) hours / 1000 # fuel * 60 min / hr = 4 minutes / 1000# fuel.
The improved specific endurance is probably not sufficient to recover the considerable fuel needed to make the climb unless you will be holding for a very long time.
Marginal Cost of a Pound of Fuel (Costs of Added Weight)
In economics, the marginal cost of the next unit produced on an assembly line is the increase in total cost associated with producing that unit. Since there is a large fixed cost associated with an assembly line, in general marginal cost decreases as units produced increases.
The marginal cost situation described above is reversed for carrying fuel, and may be stated as follows:
For each additional pound of fuel added to an airplane, the cost of carrying a pound of fuel increases, and specific range decreases accordingly. The reason for this is easily seen from the relationship:
[pic],
which tells us that if W2 > W1 (as is the case when adding fuel), then SR2 < SR1. That is, since increasing gross weight decreases specific range, carrying more fuel than is needed for a mission increases the total amount of fuel required for the mission. Of course, this observation applies to any increase in gross weight (e.g., cargo, souls on board, provisions, &c.), not just to gross weight increases due to adding fuel.
In airline parlance, this phenomenon is known as the effect of weight change on fuel burn out. Figure 15.5 presents this effect in table form for ten different air transport aircraft and various flight distances. Each entry in the table lies at the intersection of a distance row and aircraft column. The entry gives the additional fuel (in pounds) required to carry 100# additional weight the prescribed distance in the prescribed aircraft. Most efficient for a 2000 nm flight is the B747, which requires 15.0# of additional fuel for an 100# increase in gross weight. Least efficient for the same distance is the B727S, which requires 27.4#, an increase of (27.4 – 15.0) / 15, = 0.83, or 83%. That is, it takes 83% more fuel in the B727S than in the B747 to carry an extra 100# of weight 2000 nm. The B767 is almost as efficient as the B747, requiring only 15.7# of fuel extra.
Suppose jet fuel costs $1.50 per gallon, i.e., about $1.5 per gallon / 6.4 # per gallon = $0.234 / # of fuel. For the B747 as described above, the marginal cost of 100 pounds of fuel on a 2000 nm flight is just
15.0 # ($.234 / #) = $3.51, or $3.51 /100#, or $.0351 / #.
Suppose we added 1000 gallons more fuel than needed for each flight, i.e., about 6400# of extra fuel. Then the additional cost (in fuel) required to carry this unnecessary extra fuel would be approximately
6400# ($0.0351/ #) = $224.64.
If the aircraft flew the 2000 nm flight 350 times a year, the total cost would be about $224.64 (350) = $78,624.
Of course, this cost does not reflect the 1000 gallons ($1.50 / gal) = $1500 cost of extra fuel cost which is unnecessarily stored in the plane, instead of being invested at say 10%, which is possibly about what an airline pays for short term borrowed money during an inflationary period. If an airline had a fleet of 150 airplanes, each carrying 1000 # extra fuel, then the total lost investment might be as much as $150 (150) = $22,500 per year in addition to the (150) $78,624 (in fuel) = $11,793,600 needed to carry the extra fuel.
[pic]
Figure 15.5. Fuel in Pounds Required by Various Airplanes
to Carry an Additional 100 Pounds of Fuel Various Distances.
Figure 15.6 is a table which gives the estimated cost of carrying an extra pound of weight in an aircraft for one year. The figure comes from a flight attendant publication article devoted to alerting flight cabin personnel about the high cost of carrying unnecessary weight. For instance, the article calculated that carrying an extra case of Coke on each of its flights would cost approximately $41,400 per year.
Note: Figure 15.6 was created at a time when aviation fuel prices were much lower than they are today. For what it’s worth as a matter of interest, in the 1960s the U.S. Navy was paying 9.9 ¢per gallon for jet fuel.
[pic]
Figure 15.6. Marginal Yearly Cost of a Pound of Fuel for Various Aircraft Types and Fuel Costs.
Practical Approaches to Saving Fuel / Money. Pilots need not be overly concerned with the subtleties of calculating the cost of deviating from optimum performance operations. However, every pilot needs to know of how to operate his aircraft safely and efficiently. Also, any good employee is cost conscious, since every dollar spent needlessly is a dollar reduction in the company’s bottom line, and certainly company profits (or lack thereof) can affect employee salaries and job security. Consider the fact that both Eastern and Pan Am went out of business after they became unprofitable. In fact, we currently have two ex-Eastern pilots on the Air Science faculty.
Cruise parameters, for example, are typically promulgated in tables similar to those in Figures 15.7 and 15.8. The tables are for the B767 at an indicated Mach number of 0.80. Column 2 entries give IAS, standard TAT (Total Air Temperature), and TAS from top to bottom. For each table entry past column 2, the numbers from top to bottom denote EPR setting, Limiting TAT (actual Total Air Temperature) at which that EPR can be set (if applicable), and fuel flow in pounds per hour. For non-standard TAT, adjust the table entries as follows:
• Increase IAS by one kt for each degree centigrade above standard TAT.
• Decrease IAS by one kt for each degree centigrade below standard TAT.
• Increase fuel flow by 3% for each 10 degrees centigrade above standard TAT.
• Decrease fuel flow by 3% for each 10 degrees centigrade below standard TAT.
[pic]
[pic]
Figure 15.7 & 15.8
CLIMB
Best Angle and Rate of Climb Rates and Airspeeds. Best angle of climb gives the greatest gain in altitude for a given horizontal distance. The associated airspeed is
VX, best angle of climb airspeed.
Best rate of climb gives the greatest gain in altitude over a given time. The associated airspeed is
VY, best rate of climb airspeed.
Relationship of Excess Thrust to Best Angle of Climb. The figure below shows an aircraft in a steady state climb. We assume that thrust acts along the longitudinal axis. Angle c is the climb angle p is the pitch angle. Angle r, called the relative pitch angle, may be computed as r = p – c.
The x-axis by convention lies along the flight path. For steady state flight, F+x = F-x, and F+y = F-y, so
F+x = F-x: T cos r = D + W sin c
F+y = F-y: L + T sin r = W cos c
For simplicity, assume r = 0; i.e., assume thrust acts along the direction of flight. (This is not an unreasonable assumption for most jet transport aircraft.) In that case, since cos 0 = 1 and sin 0 = 0,
T = D + W sin c
L = W cos c.
From this we may conclude that T – D = W sin c, or
sin c = (T – D) / W.
If we assume a max power climb, then T is TA (thrust available). In addition, if the climb angle c is small, then W sin c ( 0, and D ( TR (thrust required). Assume D = TR. Then
c max = sin-1 [(TA – TR)max / W].
That is, angle of climb is maximum when thrust available exceeds thrust required by the maximum amount. This observation may be intuitive. Note: as indicated above, this result applies only for small climb angles, as for example climb angles in typical transport type airplanes.
Relation of Excess Power to Best Rate of Climb.
Next, we discuss maximum rate of climb. From sin c = (TA – TR) / W, by multiplying both sides of the equation by velocity V, we obtain
V sin c = V (TA – TR) / W = (V TA – V TR) / W = (PA – PR) / W,
since power P in ft-# / sec is just thrust T in times velocity V in ft / sec. But V sin c is just vertical speed or R/C (rate of climb), as shown in the vector diagram below.
Thus,
(V sin c) max =(R/C) max = (PA – PR)max / W,
That is, rate of climb is maximum when power available exceeds power required by the maximum amount. This may be less intuitive than previous observation that maximum angle of climb is related to excess thrust.
To sum up, for a fixed gross weight,
• Angle of climb is a function of excess thrust (TA – TR).
• Rate of climb is a function of excess power (PA – PR).
These two points may seem inconsistent, but they are not.
• Angle of climb is not a function of aircraft velocity. We are not interested in how fast we climb, but only how steeply. For example, rate of climb is not of paramount importance in clearing an obstacle at the takeoff end of the runway.
• Rate of climb is a function of velocity. To get maximum rate of climb, we must establish an angle of climb that, together with airspeed, gives us the maximum vertical velocity. Angle of climb, in itself, is not of paramount importance.
Example Calculation. Suppose a corporate jet airplane weighing 20,000# and developing 5,000# thrust at SL has best angle of climb airspeed VX = 165, and best rate of climb airspeed VY = 250. Total drag at VX is 3,700# and at VY is 4,000#. Determine maximum angle of climb and maximum rate of climb (in fps) possible on takeoff at SL. To convert nm/hr to ft/min, we must multiply by (6076 ft/nm) / (60 min/hr). As previously indicated, we assume that TR = D, a simplification that leads to only small inaccuracies at low climb angles.
R/C max = (PA – PR) / W = (TAV – TRV) / W = (TA - TR) V / W =
[(5,000 – 4,000) 250 (6076 / 60)] / 20,000 = 1,265.833334 fpm
C max = sin-1 [(TA – TR) / W] = sin-1 [(5,000 – 3,700) / 20,000] = sin-1 ((0.065000) = 3.726853142o
The rate of climb at the best climb angle is given by:
R/C = (PA – PR) / W = (TAV – TRV) / W = (TA - TR) V / W =
[(5,000 – 3,700) 165 (6076 / 60)] / 20,000 = 1086.085000 fpm.
Determining Best Angle of Climb Airspeed VX from the Thrust Required and Thrust Available Curves. Figure 10.3 gives plots of TA and TR for the B767 at 300,000 # gross at SL. Note that the TA curve does not conform to our usual assumption that thrust available is independent of airspeed.) Best angle of climb airspeed VX occurs at the point where airspeed V gives the maximum difference (TA – TR). The analysis in the text book asserts that this point occurs at the airspeed for (L/D)max, i.e., at the airspeed corresponding to the lowest point on the TR curve i.e., 156 kts. This leads to the following conclusions:
VX = 156 kts
C max = sin-1 [(TA – TR) / W] = sin-1 [(72000 – 34000) / 300000] = 7.3o
However, this analysis is true only if the slope of the TA curve is decreasing more rapidly than the slope of the TR curve in the airspeed region below 156 kts. To me, that doesn’t look to be the case, and I would guess VX for this plot is more like 150 kts. However, for these curves, it appears that about the same best angle of climb would result from airspeed in the range 140 to 160 kts. TAS.
[pic]
Figure 10.3. Typical TR and TA Curves for B767 at SL and 300,000# Gross.
Determining Best Rate of Climb Airspeed VY from the Power Required and Power Available Curves. Figure 10.4 gives the PR and PA curves corresponding to Figure 10.3. Best rate of climb airspeed VY occurs at the point where airspeed V gives the maximum difference (PA – PR). From Figure 10.4, this appears to occur around 180 KTAS, where PR = 20,000 HP and PA = 39,000 HP. Remembering that 1 HP = 33000 ft-#/min of work, we may calculate
VY = 180 kts
R/C max = [(PA – PR) / W] = (33000 ft-#/min) [(39000 – 20000) / 300000#] = 2090 ft/min.
The R/C at best angle of climb airspeed VX = 156 kts is
VX = 156 kts
R/C 156 kts = [(PA – PR) / W] = (33000 ft-#/min) [(35000 – 17000) / 300000#] = 1980 ft/min.
To the extent that Figures 10.3 and 10.4 accurately reflect B767 performance, we may conclude that the airplane’s performance does not differ much in the max climb angle / max climb rate airspeed range, and that the airspeed range itself does not differ much. We may conclude also from the figures that the B767 cannot climb at all at SL and 300,000# gross above about 270 KTAS.
[pic]
Figure 10.4. Typical PR and PA Curves for B767 at SL and 300,000# Gross.
Note: Thrust and power available in Figures 10.3 and 10.4 don’t obey the property that jet thrust is pretty much constant for a given rpm across the airspeed regime. This is probably because the curves don’t take into account the increase in intake air density rises as airspeed increases. However, this is not a significant anomaly, since the figures are simply meant to depict the varying difference between TA (PA) and TR (PR) at different airspeeds.
Time to Climb. Since thrust and power available and required vary as altitude varies, R/C also varies, indeed quite drastically for airplanes capable of operating at high altitudes. Then, time to climb in minutes from one altitude A1 to another altitude A2 (where A2 > A1) is simply
T = (A2 – A1) (ft) / (R/C)ave (ft/min),
where (R/C)ave is the average rate of climb during the climb. T can be determined in several ways, depending on the circumstances which apply:
• R/C is constant; then (R/C)ave = R/C.
• R/C varies linearly with altitude; then (R/C)ave = [(R/C)initial + (R/C)final] / 2.
• R/C varies non-linearly with altitude; then mathematical integration is required to find the time to climb, and ordinarily a Climb Table is used to represent the information thus obtained.
Figure 10.10 presents the climb airspeed schedule for the B767, and gives time, fuel, distance, and average TAS for climbs from SL to various cruise altitudes for various initial gross weights. Fuel adjustments for departure airports significantly above SL are included.
[pic]
Figure 10.10. B767 Enroute Climb Table.
To use the Climb Table:
1. Determine departure elevation, gross weight, and desired cruising altitude.
2. Let C be the column associated with gross weight, and R be the row associated with desired cruising altitude.
3. At the intersection of Column C with Row R, read four figures which (in order) represent time to climb in minutes, pounds of fuel required for the climb, no wind distance in nm during the climb (assuming an on-course climb), and average TAS in kts.
4. If departure field elevation is 2000’ or greater, subtract the indicated amount of fuel (bottom of chart) from fuel required.
During the climb, accelerate to and maintain 250 kts up to 10,000’; then maintain 290 KIAS to the crossover altitude (this is the company’s “compromise” best climb speed VY); thereafter, maintain MMO = 0.78 indicated Mach number.
Other important facts about the Climb Table:
• Blank blocks in the Climb Table suggest that the allowable airspeed range for the corresponding altitude/initial gross weight is deemed insufficient for safe operation.
• “Shaded” blocks in the Climb Table indicate optimum initial cruising altitudes for given initial gross weights.
• No time or distance adjustments are necessary when departing from elevations above SL, since from these elevations it takes longer to reach climb schedule airspeed owing to reduced thrust. Also, climbs in the lower atmosphere are accomplished expeditiously compared to comparable altitude changes in the upper atmosphere. That is, the B767—like all other jets—climbs fast at low altitudes and more slowly at high altitudes.
[pic]
Figure 10.11 (edited). B767 Climb Schedule EPR Settings.
To assist the flight deck crew in power control during climb, the B767 handbook also contains climb EPR settings for 250 kts, 290 kts, and 0.78 IMN at various altitudes. See Figure 10.11, which assumes that engine bleed packs are on and anti-ice systems off. (Bleed packs use air from the engine compressor section to run the cabin environmental control system.) Use the corrections in the bottom three rows of the middle table to adjust EPR for other circumstances. Note that the corrections are commensurate with the fact that taking air from the compressor lowers engine thrust, hence lowers EPR.
Note: The information in Figures 10.10 and 10.11 conceivably could be available to the crew via a computer rather than as printed matter. If a flight director computer is present, it will use this data to maintain proper power setting and climb airspeed schedule.
Example Lookup. Determine the climb EPR at 20,000’ with a total air temperature of 5o C with wing and engine anti-ice on. How much higher would the EPR read if bleed packs were off? (Note: the second question is academic. It is hard to imagine a situation where the captain would turn off the cabin environmental control system at 20,000’ without starting an immediate descent.)
From Figure 10.11, EPR with packs on and anti-ice off is 1.53. From this, subtract .02 for wing anti icing on, and .03 for engine anti-icing on, giving 1.47. With the bleed air packs off, add .01 to 1.47 to give 1.48.
Example Problem. Suppose 340,000# gross in a B767 at SL takeoff. Find climb parameters for cruise at FL180 and for optimum legal cruise FL. Also, find corresponding climb parameters for takeoff at 7000’ MSL elevation and climb to optimum cruise FL.
From Figure 10.10, climb to FL180 from SL takes 8 minutes, requires 3900# of fuel, and takes you 32 nm at an average TAS of 336 kts.
The optimum cruise altitude for 340,000# gross weight is FL330 or FL340. We choose FL330 because it is a legal cruising altitude. Climb from SL to this altitude takes 16 minutes, requires 6700# of fuel, and takes you 89 nm at an average TAS of 390 kts.
Takeoff from 7000’ MSL and climb to optimum cruise altitude of FL330 takes 16 minutes, requires (6700 – 1050) = 5650# of fuel, and takes you 89 nm at an average TAS of 390 kts. This is the same as takeoff from SL, except that fuel is less. The amount of fuel to be subtracted is interpolated between 6000’ and 8000’: i.e., by computing (900 + 1200) /2 = 1050. Effects of higher takeoff elevation on time and distance are negligible.
Cruise Climb. (Text, pp. 15.12 –15.14) We have seen previously that jet airplane range improves with altitude. However, initial high gross weight makes high altitude cruise impracticable. For example, from Figure 10.10, the B767 optimum cruise altitude at takeoff gross weight of 400,000# is FL 300 or FL 310, while the optimum cruise altitude at 300,000# takeoff gross is FL 360 or FL 370. This implies that as an airplane gets lighter from burning fuel on a long flight, it might be able to extend range by climbing to a higher altitude.
Such an incremental climb is called a cruise climb, because the aircraft climbs gradually or in incremental altitude steps (as cleared by ATC) as it proceeds closer and closer to its destination.
Suppose we know the “optimum” cruise altitude for a given gross weight W1, with (1 the density ratio at that altitude. The best cruise altitude with density ratio (2 and gross weight W2 is calculated as follows.
1. We know that as gross weight increases, range decreases, and vice versa.
2. Also, as altitude increases, range increases, and vice versa.
3. This implies that we should make W1(2 = W2(1.
Example Problem. Suppose we are cruising at optimum altitude FL330 at 300,000# gross. What would be the optimum weight for cruising at FL370, the next higher legal flight level in a standard atmosphere?
(33,000 = 0.3345 and (37,000 = 0.2844. Thus W2 = W1 (2 / (1. = 300,000 (0.2844) / 0.3345 = 255,067 #.
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Power Required Curve
PR
PR max endurance
(PR/V)min
Best endurance velocity VBE
TAS
Thrust Required Curve
TR for max range
TR TR for max endur
Max range velocity VBR
TAS
Max endurance velocity VBE
Power Required Curve
PR PR for max endurance max endurance velocity
PR for max range
(PR/V)min
max range velocity
TAS
Thrust Required Curve
TR TR for max range
Max range velocity
TAS
V V sin c
c
V cos c
p = pitch angle
c = climb angle
r = p – c = relative pitch angle
T sin r
T cos r
c
W cos c
r
c
flight path
(x-axis)
p
T
D
W
L
W sin c
Thrust Required Curve
TR for max range (SRmax)
TR TR for max endurance (SEmax)
Max range velocity (SRmax)
TAS
Max endurance velocity (SEmax)
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