Calculating Gas Turbine Shaft Power

1. Dec 31, 2008

Pud

1. The problem statement, all variables and given/known data

I have been asked to calculate a range of values relating to a Gas Turbine engine, and while I have been able to find the majority of the values I am almost certain that the Thermal Efficiency value of 1.64% I have come up with is incorrect.

This relies on Shaft power, fuel mass flow rate, and the Lower Calorific Value in the formula shown below.

LCV = 42,000 kJ/kg (Given value)
Fuel mass flow rate = 0.00821256 kg/sec (Calculated)
Shaft power = 5.66 kW (Calculated, but not confident of this result)

N = 3040 rpm (Given value)
F (brake load)= 50 N (Given Value)

2. Relevant equations

The formula for Thermal Efficiency that I have is: $$\eta$$thermal = $$\dot{W}$$Shaft/($$\dot{m}$$fuel $$\times$$ Lower Calorific Value)

The formula for Shaft Power (in kW) I have is: $$\dot{W}$$Shaft = (F * N) / 26850

(Not sure where 26850 comes from, I was given this equation. If someone can explain where that number comes from I would be very grateful.)

The mass flow rate of fuel I worked out as follows:

2 litres of fuel used in 207 seconds (Given values) = 9.6618E-3 l/sec

0.0096618 $$\div$$ 1000 = 9.6618E-6 m3/sec

Relative density of fuel = 0.85 (Given value)

Mass flow rate = 9.6618E-6 $$\times$$ 850 = 0.00821256 kg/sec

There are a whole range of other values (Temps/Pressures/Cp(g)/Cp(a)/Gammaa/Gammag etc) which I haven't included as I don't believe them to be relevant.

If necessary I will add these as requests.

Thanks

2. Jan 1, 2009

nucleus

Pud
Your turbine speed is in RPM and your fuel flow is in kg/sec, so you have to get Wf in kg/min.
I have an example of this but it isn’t in metric but your formulas appear to be correct. The number 26850 probably comes from correcting for units.