Can someone show typical calculations for a Boeing 777 jet engine for the following criteria:
- Take off thrust of the engine at 170 mph take off speed and 45 degrees F
- Cruising thrust required at 500 mph at 35,000 feet altitude?
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Neville p
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2This might be too broad as takeoff thrust especially varies depending on runway length, takeoff weight, elevation and ambient conditions. – Ben Sep 24 '18 at 09:18
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@Ben, thrust is not depending on runway length. I really can't see how that could be possible – Federico Sep 24 '18 at 09:58
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3@neville, this looks a bit like a homework question. could you please show us what you have tried and why can't you answer? – Federico Sep 24 '18 at 09:59
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@Federico Derated takeoff - where less than maximum thrust is used for takeoff when it is not needed, to protect engine wear. Airbus uses a similar process called Flex Temp. – Ben Sep 24 '18 at 21:09
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Why would the thrust requirement of a 777 be dependent only on 500 MPH and 35,000 feet? Is the 777 lightly loaded at the end of a long journey after burning fuel, or is it at the beginning of the journey? Which engine variant is installed?
Thrust can be calculated from the TSFC and fuel flow, to give you an idea of how far apart the values can be just for 35,000 feet here are the fuel flow figures at two different weights (620,000 and 300,000 lb):
9,501 and 4,466 pounds of fuel per hour per engine$^1$. With a cruise TSFC value of 0.520 for the 777's GE90-85B$^2$, the thrust can be anywhere between 18,300 and 8,600 lbf, per engine.
Likewise for takeoff thrust, a lightly loaded 777 on a long runway and no obstacles will use a lower takeoff thrust than on a short runway with an obstacle to clear, so 45°F is not really the only variable.
What you should be asking is how altitude and Mach number affect the engine thrust, and without proprietary engine data from the manufacturer, be it GE, R-R, or P&W for the 777, I will only provide a theoretical equation$^3$:
$$ T = T_{{SL}_{max}} \delta_T \frac{\rho}{\rho_{SL}} (1+K_TM) $$
- $T$ = thrust
- $T_{{SL}_{max}}$ = max thrust at sea level, for the GE90-85B$^2$ that's 376,764 N
- $\delta_T$ = throttle setting, 0 $\lt$ $\delta_T$ $\le$ 1, a cruise value you can use is 0.8
- $\rho$ = air density, its ratio of 35,000 feet ISA to SL ISA is 0.30988, you can search for tables or calculators online and see how temperature (e.g., 45°F) affects it
- $K_T$ = a constant between 0.1 and 0.9
- $M$ = Mach number (a typical cruise value is 0.84).
An example at 35,000 feet, M0.84, and $K_T$ = 0.1 yields a thrust of 101,246 N (22,761 lbf), which for a simple equation is good enough when compared to the TSFC calculation at max weight above, and is 27% of the max takeoff thrust as can be expected. You can play around with the numbers in your favorite spreadsheet too.
$^1$ Boeing 777 Flight Manual
$^2$ http://www.jet-engine.net/civtfspec.html
$^3$ http://www.dept.aoe.vt.edu/~lutze/AOE3104/thrustmodels.pdf
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Thanks ymb1 for your expert answer and the equation which will help me a lot. – Neville p Sep 25 '18 at 18:02
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Assume a B-767 engine with thrust rating of 60,000 lb. I heard that the Bypass fan does 70% of the thrust (In latest 787 I believe it is 85% of the total thrust),could you please calculate the max. HP needed to drive the Bypass fan?I assume the remaining 30% of the thrust is provided by the exhaust nozzle by reaction principle. Please enlighten me on these equations. Thanks in advance. – Neville p Sep 25 '18 at 18:05
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Hi @Nevillep - Since you're new here, please understand that the comments are not for asking new questions. With that said, you are presenting an XY problem. Who said power can be calculated from a percentage of thrust? Thrust is in N, and power is in Nm/s. To get to Nm/s you either need to make assumptions about the fan and mass flow, or have access to engine telemetry and design parameters. – Sep 25 '18 at 20:42