Thermal Efficiency of Jet Engines given Effective Speed Ratio Calculator

✖Exit Velocity refers to the speed at which gases expand at the exit of nozzle of a engine.ⓘ Exit Velocity [V_e]			+10% -10%
✖Effective Speed Ratio is the ratio of aircraft forward speed to the aircraft jet velocity.ⓘ Effective Speed Ratio [α]			+10% -10%
✖Fuel Air Ratio is the ratio of mass flow rate of fuel to the mass flow rate of air.ⓘ Fuel Air Ratio [f]			+10% -10%
✖Fuel Calorific Value denotes the amount of heat energy released per unit mass of fuel upon complete combustion.ⓘ Fuel Calorific Value [Q]			+10% -10%

✖Thermal efficiency is a dimensionless performance measure of a device that uses thermal energy, such as an internal combustion engine, a steam turbine or a steam engine.ⓘ Thermal Efficiency of Jet Engines given Effective Speed Ratio [η_th]

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Thermal Efficiency of Jet Engines given Effective Speed Ratio Solution

STEP 0: Pre-Calculation Summary

Formula Used

Thermal Efficiency = (Exit Velocity^2*(1-Effective Speed Ratio^2))/(2*Fuel Air Ratio*Fuel Calorific Value)
η_th = (V_e^2*(1-α^2))/(2*f*Q)
This formula uses 5 Variables

Variables Used

Thermal Efficiency - Thermal efficiency is a dimensionless performance measure of a device that uses thermal energy, such as an internal combustion engine, a steam turbine or a steam engine.
Exit Velocity - (Measured in Meter per Second) - Exit Velocity refers to the speed at which gases expand at the exit of nozzle of a engine.
Effective Speed Ratio - Effective Speed Ratio is the ratio of aircraft forward speed to the aircraft jet velocity.
Fuel Air Ratio - Fuel Air Ratio is the ratio of mass flow rate of fuel to the mass flow rate of air.
Fuel Calorific Value - (Measured in Joule per Kilogram) - Fuel Calorific Value denotes the amount of heat energy released per unit mass of fuel upon complete combustion.

STEP 1: Convert Input(s) to Base Unit

Exit Velocity: 248 Meter per Second --> 248 Meter per Second No Conversion Required
Effective Speed Ratio: 0.4475 --> No Conversion Required
Fuel Air Ratio: 0.009 --> No Conversion Required
Fuel Calorific Value: 43510 Kilojoule per Kilogram --> 43510000 Joule per Kilogram (Check conversion here)

STEP 2: Evaluate Formula

Substituting Input Values in Formula

η_th = (V_e^2*(1-α^2))/(2*f*Q) --> (248^2*(1-0.4475^2))/(2*0.009*43510000)

Evaluating ... ...

η_th = 0.0628047697847238

STEP 3: Convert Result to Output's Unit

0.0628047697847238 --> No Conversion Required

FINAL ANSWER

0.0628047697847238 ≈ 0.062805 <-- Thermal Efficiency

(Calculation completed in 00.020 seconds)

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Thermal Efficiency of Jet Engines given Effective Speed Ratio Formula

LaTeX Go

Thermal Efficiency = (Exit Velocity^2*(1-Effective Speed Ratio^2))/(2*Fuel Air Ratio*Fuel Calorific Value)
η_th = (V_e^2*(1-α^2))/(2*f*Q)

What is thermal efficiency?

Thermal efficiency is a dimensionless performance measure of a device that uses thermal energy, such as an internal combustion engine, a steam turbine or a steam engine, a boiler, furnace.

How to Calculate Thermal Efficiency of Jet Engines given Effective Speed Ratio?

Thermal Efficiency of Jet Engines given Effective Speed Ratio calculator uses Thermal Efficiency = (Exit Velocity^2*(1-Effective Speed Ratio^2))/(2*Fuel Air Ratio*Fuel Calorific Value) to calculate the Thermal Efficiency, Thermal Efficiency of Jet Engines given Effective Speed Ratio is a measure of the efficiency of a jet engine, which is a critical parameter in aerospace engineering, defined as the ratio of the useful work output to the total energy input, indicating how effectively the engine converts energy into thrust. Thermal Efficiency is denoted by η_th symbol.

How to calculate Thermal Efficiency of Jet Engines given Effective Speed Ratio using this online calculator? To use this online calculator for Thermal Efficiency of Jet Engines given Effective Speed Ratio, enter Exit Velocity (V_e), Effective Speed Ratio (α), Fuel Air Ratio (f) & Fuel Calorific Value (Q) and hit the calculate button. Here is how the Thermal Efficiency of Jet Engines given Effective Speed Ratio calculation can be explained with given input values -> 0.062805 = (248^2*(1-0.4475^2))/(2*0.009*43510000).

FAQ

What is Thermal Efficiency of Jet Engines given Effective Speed Ratio?

Thermal Efficiency of Jet Engines given Effective Speed Ratio is a measure of the efficiency of a jet engine, which is a critical parameter in aerospace engineering, defined as the ratio of the useful work output to the total energy input, indicating how effectively the engine converts energy into thrust and is represented as η_th = (V_e^2*(1-α^2))/(2*f*Q) or Thermal Efficiency = (Exit Velocity^2*(1-Effective Speed Ratio^2))/(2*Fuel Air Ratio*Fuel Calorific Value). Exit Velocity refers to the speed at which gases expand at the exit of nozzle of a engine, Effective Speed Ratio is the ratio of aircraft forward speed to the aircraft jet velocity, Fuel Air Ratio is the ratio of mass flow rate of fuel to the mass flow rate of air & Fuel Calorific Value denotes the amount of heat energy released per unit mass of fuel upon complete combustion.

How to calculate Thermal Efficiency of Jet Engines given Effective Speed Ratio?

Thermal Efficiency of Jet Engines given Effective Speed Ratio is a measure of the efficiency of a jet engine, which is a critical parameter in aerospace engineering, defined as the ratio of the useful work output to the total energy input, indicating how effectively the engine converts energy into thrust is calculated using Thermal Efficiency = (Exit Velocity^2*(1-Effective Speed Ratio^2))/(2*Fuel Air Ratio*Fuel Calorific Value). To calculate Thermal Efficiency of Jet Engines given Effective Speed Ratio, you need Exit Velocity (V_e), Effective Speed Ratio (α), Fuel Air Ratio (f) & Fuel Calorific Value (Q). With our tool, you need to enter the respective value for Exit Velocity, Effective Speed Ratio, Fuel Air Ratio & Fuel Calorific Value and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

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