Question

Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow rate is 2500 lbm/hr. Determines:

a) The work produced, hp.
b) The exit temperature, oF.

Answer 1

a) 158.4 HP.

b) 1235.6 °F.

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

`m_(in)h_(in)=W_(out)+m_(out)h_(out)`

Whereas the mass flow is just the same, which means we have:

`W_(out)=m_(out)(h_(out)-h_(in))`

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

`h_(in)=1860.7BTU/lbm\\\\s_(in)= 2.2096BTU/lbm-R`

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

`s_(out)= 2.2096BTU/lbm-R`

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

`h_(out)=1659.15BUT/lbm`

Then, the isentropic work (negative due to convention) is:

`W_(out)=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_(out)=-503,875BTU`

And the real produced work is:

`W_(real)=0.8*-503875BTU\\\\W_(real)=-403100BTU`

Finally, in horsepower:

`W_(real)=-403100BTU/hr*(1HP)/(2544.4336BTU/hr) \\\\W_(real)=158.4HP`

Regards!