Question #297673

A turbine operating under steady flow conditions receives steam at the





following state : pressure 13.8 bar ; specific volume 0.143 m3/kg ;





internal energy 2590 kJ/kg ; velocity 30 m/s. The state of the steam





leaving the turbine is : pressure 0.35 bar ; specific volume 4.37 m3/kg





; internal energy 2360





kJ/kg ; velocity 90 m/s. Heat is lost to the surroundings at the rate of





0.25 kJ/s. If the rate of steam flow is 0.38 kg/s, what is the power





developed by the turbine ?





1
Expert's answer
2022-02-15T16:54:02-0500

m˙(u1+P1V1+C122)Q=m˙(u2+P2V2+C222)+w˙\dot{m}(u_1+P_1V_1+\frac{C_1^2}{2})-Q=\dot{m}(u_2+P_2V_2+{\frac{C_2^2}{2}})+\dot{w}



0.38[2590×103+(13.8×105)(0.143)+3022]0.25×1030.38 \begin{bmatrix} 2590×10^3+(13.8×10^5)(0.143)+\frac{30}{2}^2 \\ \end{bmatrix}-0.25×10^3

=1059110.2=1059110.2


0.38[2360×103+(0.35×105)(4.37)+9022]+w˙0.38\begin{bmatrix} 2360×10^3+(0.35×10^5)(4.37)+\frac{90}{2}^2 \\ \end{bmatrix}+\dot{w}

=956460+w˙=956460+\dot{w}


1059110.2=956460+w˙1059110.2=956460+\dot{w}

w˙=102650.2\dot{w}=102650.2

=102.65kW=102.65kW







Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS