Question

Air is entering a 4200-kW turbine that is operating at its steady state. The mass flow rate is 20 kg/s at 807 C, 5 bar and a velocity of 100 m/s. This air then expands adiabatically, through the turbine and exits at a velocity of 125 m/s. Afterwards the air then enters a diffuser where it decelerates isentropically to a velocity of 15 m/s and a pressure of 1 bar. Using the ideal gas model, determine, (a) pressure and temperature of the air at the turbine exit, in units of bar and Kelvin. (b) Entropy production rate in the turbine in units of kW/k, and (c) draw the process on a T-s Diagram.

Answer 1

Answer:

a)$T_2=868.24 K$ ,$P_2=2.32 bar$

b) $s_2-s_1=0.0206$KW/K

Explanation:

P=4200  KW ,mass flow rate=20 kg/s.

Inlet of turbine

 $T_1$=807°C,$P_1=5 bar,V_1=100 m/s$

Exits of turbine

 $V_2=125 m/s$

Inlet of diffuser

$P_3=1 bar,V_3=15 m/s$

Given that ,use air as ideal gas

R=0.287 KJ/kg-k,$C_p$=1.005 KJ/kg-k

Now from first law of thermodynamics for open system at steady state

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here given that turbine is adiabatic so Q=0

Air treat ideal gas   PV=mRT, Δh=$C_p(T_2-T_1)$

$w=\dfrac{P}{mass \ flow\ rate}$

$w=\dfrac{4200}{20}$

w=210 KJ/kg

Now putting the values

$1.005\times (273+807)+\dfrac{100^2}{2000}=1.005T_2+\dfrac{125^2}{2000}+210$

$T_2=868.24 K$

Now to find pressure

We know that for adiabatic $PV^\gamma =C$ and for ideal gas Pv=mRT

⇒$\left (\dfrac{T_2}{T_1}\right )^\gamma=\left (\dfrac{P_2}{P_1}\right )^{\gamma -1}$

$\left(\dfrac{868.24}{1080}\right )^{1.4}=\left (\dfrac{P_2}{5}\right )^{1.4-1}$

$P_2=2.32 bar$

For entropy generation

$s_2-s_1=1.005\ln\dfrac{868.24}{1080}-0.287\ln\dfrac{2.32}{5}$

$s_2-s_1=0.00103$KJ/kg_k

$s_2-s_1=0.00103\times 20$KW/K

$s_2-s_1=0.0206$ KW/K