Thermodynamics help please -- Air passing through a gas turbine system

[clark123]

Summary:: NO TEMPLATE BECAUSE THIS HOMEWORK PROBLEM WAS MISPLACED IN A REGULAR FORUM

Cant do part c, using the steady flow equation I am confused how to continue. Please help!

Mainly confused as to what heat transfer loss represents in the steady flow equation and where to go to find the power.

Air passes through a gas turbine system at a rate of 4.5kg/s. It enters the turbine system with a velocity of 90 m/s and a specific volume of 0.85m3 /kg. It leaves the turbine system with a specific volume of 1.45m3 /kg. The exit area of the turbine system is 0.038m2 . In its passage through the turbine system, the specific enthalpy of the air is reduced by 200kJ/kg and there is a heat transfer loss of 40kJ/kg.

Determine: a. The inlet area of the turbine in m 2
b. The exit velocity of the air in m/s
c. The power developed by the turbine system in kW.

 

[anorlunda]

I believe that heat transfer loss refers to heat leaking into the environment.

Like many thermodynamics problems, the place to start is to make an energy balance. Can you do that and post it here? Energy in, energy out, energy lost.

 

[Chestermiller]

You are learning about the open system (control volume) version of the 1 st law. Please write down that equation as it applies to this steady state problem.

 

[clark123]

I'd believe the equation is Q - W = M[(Ho-Hin)+(Co^2-Cin^2)/2]
I can interpret it normally but just confused.

 

[Chestermiller]

I'd believe the equation is Q - W = M[(Ho-Hin)+(Co^2-Cin^2)/2]
I can interpret it normally but just confused.

Good. Now, how is mass flow rate related to density, velocity, and area?

 

[clark123]

Mass flow rate is the product of all 3 combined I’m pretty sure.

 

[Chestermiller]

Mass flow rate is the product of all 3 combined I’m pretty sure.

Good. If you apply this to the exit cross section, what do you get for the exit velocity?

 

[clark123]

171.71m/s

 

[Chestermiller]

Good. So using the equation in post #4, what do you get for W?

 

[clark123]

That's my main point of confusion, is the heat transfer loss of -40kJ/Kg represented by Q in the formula? Or is it something else?

 

[Chestermiller]

That's my main point of confusion, is the heat transfer loss of -40kJ/Kg represented by Q in the formula? Or is it something else?

It represents Q/M

 

[clark123]

Okay didn't realize that, would you be able to explain why please?

 

[clark123]

 

[clark123]

There is the worked solution they give you, do not understand it at all.

 

[Chestermiller]

Okay didn't realize that, would you be able to explain why please?

It's just the way they chose to do it. Nothing wrong with that.

 

[Chestermiller]

There is the worked solution they give you, do not understand it at all.

Yes. They did a nice job. That was the answer I got.

 

[clark123]

If possible, would you be able to do some more detailed workings out with explanations as I cannot see where they have gone so quickly, take a photo and put it in here? No worries if not!

 

[Chestermiller]

If possible, would you be able to do some more detailed workings out with explanations as I cannot see where they have gone so quickly, take a photo and put it in here? No worries if not!

If I take your equation in post #4 and solve for $\dot{W}$, I get:
$$\dot{W}=\dot{M}\left[\frac{\dot{Q}}{\dot{M}}-(h_{out}-h_{in})-\frac{(c_{out}^2-c_{in}^2)}{2}\right]$$
with $$\frac{\dot{Q}}{\dot{M}}=-40\ \frac{kJ}{kg}$$
$$(h_{out}-h_{in})=-200\ \frac{kJ}{kg}$$
and $$\frac{(c_{out}^2-c_{in}^2)}{2}=\frac{(171.71^2-90^2)}{2}=10692\ \left(\frac{m}{s}\right)^2=10692\ \frac{J}{kg}=10.692\ \frac{kJ}{kg}$$