Question 7.229E: A small air turbine with an isentropic efficiency of 80% sho...

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.

 

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

C.V. Turbine actual energy Eq.4.13:

w = h _{ i }- h _{ e , ac }=120

Table 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   R

C.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]=0.2328

 

P _{ 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}

 

1
2
F.5.1
F.5.2

Related Answered Questions