A small air turbine with an isentropic efficiency of 80% should produce 120 Btu/lbm of work. The inlet temperature is 1800 R and it exhausts to the atmosphere. Find the required inlet pressure and the exhaust temperature.
A small air turbine with an isentropic efficiency of 80% should produce 120 Btu/lbm of work. The inlet temperature is 1800 R and it exhausts to the atmosphere. Find the required inlet pressure and the exhaust temperature.
C.V. Turbine actual energy Eq.4.13:
w = h _{ i }- h _{ e , ac }=120Table F.5: h _{ i }=449.794 Btu / lbm
\Rightarrow h _{ e , ac }= h _{ i }-120=329.794 Btu / lbm , \quad T _{ e }= 1 3 4 9 RC.V. Ideal turbine, Eq.7.27 and energy Eq.4.13:
w _{ s }= w / \eta _{ s }=120 / 0.8=150= h _{ i }- h _{ e , s } \Rightarrow h _{ e , s }=299.794 Btu / lbm
From Table F.5: T _{ e , s }=1232.7 R , s _{ Te }^{ o }=1.84217 Btu / lbm R
Entropy Eq.7.9: s _{ i }= s _{ e , s } adiabatic and reversible
To relate the entropy to the pressure use Eq.6.19 inverted and standard entropy from Table F.5:
P _{ e } / P _{ i }=\exp \left[\left( s _{ Te }^{ o }- s _{ Ti }^{ o }\right) / R \right]=\exp \left[(1.84217-1.94209) \frac{778}{53.34}\right]=0.2328P _{ i }= P _{ e } / 0.2328=14.7 / 0.2328= 6 3 . 1 4 \text { psia }
If constant heat capacity was used
\begin{array}{l}T _{ e }= T _{ i }- w / C _{ p }=1800-120 / 0.24=1300 R \\T _{ e , s }= T _{ i }- w _{ s } / C _{ p }=1800-150 / 0.24=1175 R\end{array}
Eq.7.9 (adibatic and reversible) gives constant s and relation is Eq.6.23
P _{ e } / P _{ i }=\left( T _{ e } / T _{ i }\right)^{ k /( k -1)} \Rightarrow P _{ i }=14.7(1800 / 1175)^{3.5}=65.4 psia
…………………………………..
Eq.4.13 : q+h_{i}+\frac{ V _{i}^{2}}{2}+g Z_{i}=h_{e}+\frac{ V _{e}^{2}}{2}+g Z_{e}+w
Eq.7.27 : \eta_{\text {turbine }}=\frac{w}{w_{s}}=\frac{h_{i}-h_{e}}{h_{i}-h_{e s}}
Eq.7.9 : s_{e}=s_{i}+\sum \frac{q}{T}+s_{\text {gen }}
Eq. 6.19 : s_{2}-s_{1}=\left(s_{T 2}^{0}-s_{T 1}^{0}\right)-R \ln \frac{P_{2}}{P_{1}}
Eq.6.23 : \frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}