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

Answer:

a) 158.4 HP.

b) 1235.6 °F.

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:

$W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP$

Regards!