Question

Consider now a turbine(Q=0) that operates with N2 steam with T1 = 753.15K and P1 = 6 bar and P2 = 1 bar. The efficiency is 80% and the molar flowrate is (note not mass flow rate) is 200 mol/s. The inlet velocity is 30 m/s and the exit velocity is 120 m/s. The inlet is 6 m above the base of the turbine and the exit is 2 m above the base.  

a. Determine the output of the turbine with units.

b. what is the exhaust temperature and quality (if saturated) for 100% efficiency?

Answer 1

a) The given component is steam. We will make use of steam property table to obtain the value of enthalpy

At T = 753.15 K, P1 = 6 bar

H1 = 3440.3825 KJ/Kg

NOW P2 =1 bar

Now q = 0 , Hence the process is an adiabatic process

For an adiabatic process

gamma for steam = 1.3

T2 = 498.09033 K

Now from steam table at this condition H2 =2904.3932 KJ/Kg

Flow rate = 200 mol/s = 200*18 = 3600 g/s =3.6 Kg/s

Now we will do enthalpy balance

we will substitute all the known values and simplify the above equation

3.6*( 3440.3825 + 0.5*30² + 2) = 3.6*(2904.3932 + 0.5*0.5*200² +0) + 0 -W

W = -68450.4 J/S = -68.45 KW

Now efficiency is 80 %

W = 0.8*(-68.45) =-57.76 KW

b)

For this we will make use of steam table and obtain the value of specific enthalpy required to solve the problem

For state 1

P = 6 bar

T =753.15

at this pressure and temperature

H =3440.382KJ/Kg

s =7.94763 KJ/KgK

Now here q is zero and hence the process is adiabatic

Now if the efficiecy is 100 % it implies that the process can be assumed to be reversible

A reversible adiabatic process is nothing but isentropic process

hence the entropy will not change

Hence we will look at State in steam table Having P = 1 bar and s =7.94763 KJ/KgK

T = 225 oC superheated, q = 1