Question

A turbofan operates at 25,000 ft and moves at 815 ft/s. It ingests 1.2 times the amount of air into the fan than into the core, which all exits through the fan exhaust. The fuel-flow-to-core airflow ratio is 0.0255. The exit densities of the fan and core are 0.00154 and 0.000578 slugs/ft3, respe~tively. The exit pressures from the fan and core are 10.07 and 10.26 psia, respectively. The developed thrust is 10,580 !bf, and the exhaust velocities from the fan and core are 1147 and 1852 ft/s, respectively. (a) Find the ingested air mass flow rate for the core and TSFC. (b) What are the exit areas of the fan and core nozzles

Answer 1

The ingested air mass flow rate for the core and the Thrust Specific Fuel Consumption (TSFC) can be found using thrust, fuel-flow-to-core airflow ratio, and velocity equations. The exit areas of the fan and core nozzles can be calculated using the exhaust densities and velocities along with the total air mass flow rate.

Step-by-step explanation:

To address part (a) of the student's question: To find the ingested air mass flow rate for the core, we need to first determine the total mass flow rate through the engine. Since the fuel-flow-to-core airflow ratio is given as 0.0255, and all the air ingests exits through the fan exhaust, we can establish a relationship between the mass flow rate of air (μair) and the mass flow rate of the fuel (μfuel) as μfuel = 0.0255 μair. Using the thrust equation T = (μair + μfuel)(ve, core - v0) + μfan (ve, fan - v0) + (pe, core - p0)Ae, core + (pe, fan - p0)Ae, fan, and the given thrust of 10,580 lbf, we can solve for μair.

To calculate the Thrust Specific Fuel Consumption (TSFC), we use the formula TSFC = μfuel / T.

For part (b), finding the exit areas of the fan and core nozzles involves using the given exhaust densities and velocities, along with the mass flow rate of air. The exit area A can be found using the continuity equation A = μ / (ρv), where μ is the mass flow rate, ρ is the density, and v is the velocity at the exit.

Answer 2

a)

Mass flow rate of core =
`m_(e)` = 60.94 Kg/s

Mass flow rate of fan =
`m_(s)` = 73.12 kg/s

TSFC = 3.301 x
`10^(-5)`

b)

Exit Area of Fan =
`A_(e)` = 0.3624
`m^(2)`

Exit Area of Core =
`A_(s)` = 0.2635
`m^(2)`

Step-by-step explanation:

Data Given:

Height = 25000 ft

Vehicle velocity =
`u_(a)` = 815 ft/s = 248.41 m/s

`m_(s) = 1.2m_(e)`

`m_(f)` = 0.0255
`m_(e)`

Where,

`m_(s)` = Mass flow rate of fan

`m_(e)` = Mass flow rate of core

F = Thrust

Density of core =
`D_(e)` = 0.000578 slugs/
`ft^(3)` = 0.2979 kg/
`m^(3)`

Density of fan =
`D_(s)` = 0.00154 slugs/
`ft^(2)` = 0.7937 kg/
`m^(3)`

Ambient Pressure of Fan =
`P_(s)` = 10.07 Psi = 69430.21 Pa

Ambient Pressure of core =
`P_(e)` = 10.26 Psi = 70740.2 Pa

Thrust = F = 10580 lbf = 47062.2 N

Velocity of fan =
`u_(s)` = 1147 ft/s = 349.6 m/s

Velocity of core =
`u_(e)` = 1852 ft/s = 564.5 m/s

At the height of 25000 ft, P = 37600
`P_(a)`

Now,

we have:

`m_(e)` =
`u_(e)` x
`D_(e)` x
`A_(e)`

Plugging in the values, we get:

`m_(e)` = 168.16
`A_(e)` Equation 1

And,

`m_(s)` =
`D_(s)` x
`A_(s)` x
`u_(s)`

`m_(s)` = 277.5
`A_(s)` Equation 2

As, we know,

`m_(s) = 1.2m_(e)`

`m_(s)` = 277.5
`A_(s)`

And now for Thrust, we have:

F =
`A_(e)` x (
`P_(e)` -
`P_(a)` ) +
`A_(s)` x (
`P_(s)` -
`P_(a)` ) +
`m_(e)`x (
`u_(e)` -
`u_(a)` ) +
`m_(s)` x (
`u_(s)` -
`u_(a)` ) Equation 3

Now, substitute equation 1 and 2 in equation 3, we get:

Exit Area of Fan =
`A_(e)` = 0.3624
`m^(2)`

Exit Area of Core =
`A_(s)` = 0.2635
`m^(2)`

Mass flow rate of core =
`m_(e)` = 60.94 Kg/s

Mass flow rate of fan =
`m_(s)` = 73.12 kg/s

TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate / Thrust

TSFC =
`m_(f)`/F

And,

`m_(f)` = 0.0255
`m_(e)`

`m_(e)` = 60.94

`m_(f)` = 0.0255 x 60.94

`m_(f)` = 1.55397

TSFC =
`m_(f)`/F

TSFC = 1.55397/47062.2

TSFC = 3.301 x
`10^(-5)`

Low TSFC = High efficiency

High TSFC = Low efficiency

a)

Mass flow rate of core =
`m_(e)` = 60.94 Kg/s

Mass flow rate of fan =
`m_(s)` = 73.12 kg/s

TSFC = 3.301 x
`10^(-5)`

b)

Exit Area of Fan =
`A_(e)` = 0.3624
`m^(2)`

Exit Area of Core =
`A_(s)` = 0.2635
`m^(2)`