Question

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s, and the exit conditions are 30 kPa, 92 percent quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine 

a. (3) Change in kinetic energy (-23.4 kJ) 

b. (4) Power output (12.12 MW) 

c. (3) Turbine inlet area (0.012966 m²)

Answer 1

Sol: tinut inlet condition Pi = 4Mla T = Sốc m=laegls Wout V = 8omis I outlet outlet condition P2 = 30kla X =0.92 V2 = somis (a) change in Kinetic energy :- AK.E = 2m V8? - 1m2 = n(+va)(v=vp) Ake = 1x12 x630) x (130) = -23400 AK.E = -23.4K) (b) Power output Pout = wilhi-ha) - ake from steam property table ! @ inlet :- Pi=AMPa Ti = 500°c ho = 3445.8 ; P= 0.0864 med den @ outlet :- P2 = 30kla x2 = 0.92 the = 2434.7 kg;

+ % = 12x (344.8 – 2431 4) - (-2314) 1000 1ooo Pout = 12.12 MW] C) inlet area of turbine : in= AXVI A. = mv, A, = 12X0.0864 = 0.01296m2 80 (A = 0.01296 m²

Answer 2

Assumptions Potential energy changes are negligible. Process is steady-flow. The device is adiabatic and thus heat transfer is neglegible. som steam table- Palomp 20, -0.029782m74 T= 950 y 7h, - 3242-4 kJ 1211 and P,=10MPa T,= 450°C V=gomis P₂ = lokpal ha hot hanty P2=10kPa % =0.92 V2 = som h = 0.92 h -1.97.81+0.42%2392- = 2392.5 kJ/kg 2 2950 7/kg = -1.950 kJ change in kinetic energy - (Perky) Ak = V2-7,2 = (58)² – (80)2. 2 6 . & in = Eout m[h + V ] = Wewnt in /ha + V2 ] Woul = - in [ha hu) + V 1, 2 ] = {23925 - 3242.4 let (-95) -851-603 45 Powegowford Wood 10222.19 kJ = 10.2 M1 - 10:2 MW

Steam som table The inlet Area of the turbine - (from mas flow rate) mo, = AV, A = MV - (12 kyl ) (0.029782 milky ) ( 80 mi) - 0.00447m2

Answer 3

Concepts and reason

Thermodynamic System:

It describes the amount of matter with in a boundary that can be described using thermodynamic variables like temperature, pressure, enthalpy, entropy.

Open thermodynamic system:

It refers to the physical system that has both mass and energy interaction with other systems as well as surroundings.

Steady flow:

The steady state flow condition of a fluid flow referring to the properties of the fluid at any point does not vary with time. The properties like velocity, pressure, temperature and mass flow rate does not change with respect to time

Turbine:

Turbine is a work producing device open system. The working fluid enters the turbine system and causes the turbine blade to rotate, which is coupled with the alternator to produce work.

First law of thermodynamics:

For open system, the total energy of the fluid entering the system is equal to the total energy leaving the fluid the system.

Fundamentals

Kinetic energy:

The energy possessed by the body by its motion with a given velocity.

$K.E = \frac{1}{2}m{V^2}$

Here, the kinetic energy is $K.E$ , mass of the body is $m$ , and velocity of the body is $V$ .

Potential energy:

The energy possessed by the body by its position of height with respect to a reference.

$P.E = mgh$

Here, potential energy is $P.E$ , acceleration due to gravity is $g$ , and height of the body is $h$ .

Energy equation for open system per unit time:

${\dot m_i}\left( {{h_i} + \frac{{V_i^2}}{2} + g{z_i}} \right) + \dot Q = {\dot m_o}\left( {{h_o} + \frac{{V_o^2}}{2} + g{z_o}} \right) + \dot W$

Here, inlet mass flow rate per unit time is ${\dot m_i}$ , inlet enthalpy is ${h_i}$ , inlet velocity is ${V_i}$ , acceleration due to gravity is $g$ , elevation at inlet is ${z_i}$ , heat interaction per unit time with the system is $\dot Q$ , outlet mass flow rate per unit time is ${\dot m_o}$ , outlet enthalpy is ${h_o}$ ,outlet velocity is ${V_o}$ , elevation at outlet is ${z_o}$ , and work interaction per unit time with the system is $\dot W$ .

Steady flow Energy equation for open system per unit time:

$\begin{array}{l}\\{{\dot m}_i} = {{\dot m}_o} = \dot m\\\\\dot m\left( {{h_i} + \frac{{V_i^2}}{2} + g{z_i}} \right) - \dot m\left( {{h_o} + \frac{{V_o^2}}{2} + g{z_o}} \right) + \dot Q - \dot W = 0\\\\\dot m\left( {\left( {{h_i} - {h_o}} \right) + \left( {\frac{{V_i^2}}{2} - \frac{{V_o^2}}{2}} \right) + \left( {g{z_i} - g{z_o}} \right)} \right) + \dot Q - \dot W = 0\\\end{array}$

The mass flow rate at a state is given by

$\dot m = \rho AV$

Here, the density is $\rho$ , area of cross section is $A$ and the velocity of flow is $V.$

Turbine Output power $\left( {\bf{W}} \right)$ .

Output power of the turbine is calculated by the product of mass flow rate and the workdone by the turbine.

${\bf{W}} = \dot mW$

From thermodynamic table “properties of superheated water vapor”, obtain the following properties at a pressure of 4 MPa and temperature $500^\circ {\rm{C}}$ .

$\begin{array}{l}\\{\rm{specific volume }}{v_1} = 0.08644{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/kg}}\\\\{\rm{specific enthalpy }}{{\rm{h}}_1} = 3446{\rm{ kJ/kg}}\\\end{array}$

Obtain the following properties at “saturated steam” from property tables, at 30 kPa pressure.

$\begin{array}{l}\\{\rm{enthalpy of saturated liquid }}{h_f} = 289.27{\rm{ kJ/kg}}\\\\{\rm{enthalpy of vapor }}{{\rm{h}}_{fg}} = 2335.3{\rm{ kJ/kg}}\\\end{array}$

Write the formula for the enthalpy.

$\begin{array}{l}\\{h_2} = {h_f} + {x_2}{h_{fg}}\\\\{h_2} = 289.27\frac{{{\rm{kJ}}}}{{{\rm{kg}}}} + \left( {0.92 \times 2335.3\frac{{{\rm{kJ}}}}{{{\rm{kg}}}}} \right)\\\\{h_2} = 2437.746{\rm{ kJ/kg}}\\\end{array}$

Write the formula for the change in kinetic energy $\left( {\Delta K.E} \right)$ .

$\begin{array}{c}\\\Delta K.E = \frac{{{V_2}^2 - {V_1}^2}}{2}\\\\\Delta K.E = \frac{{{{\left( {50\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2} - {{\left( {80\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}}}{2} \times \left( {\frac{{1\;{\rm{J}}}}{{1000{\rm{ kJ}}}}} \right)\\\\\Delta K.E = - 1.95{\rm{ kJ/kg}}\\\end{array}$

Write the Energy equation for open system per unit mass for steady state process.

$\begin{array}{l}\\\left( {{h_i} + \frac{{V_i^2}}{2} + g{z_i}} \right) + Q = \left( {{h_o} + \frac{{V_o^2}}{2} + g{z_o}} \right) + W\\\\Q - W = {h_o} - {h_i} + \frac{{V_o^2 - V_i^2}}{2} + g\left( {{z_o} - {z_i}} \right)\\\end{array}$

Since the adiabatic process does not involve any heat transfer $Q = 0$ and there is no change in potential energy $g\left( {{z_o} - {z_i}} \right) = 0$

$\begin{array}{c}\\ - W = {h_o} - {h_i} + \frac{{V_o^2 - V_i^2}}{2}\\\\ - W = \left( {2437.746 - 3446} \right)\frac{{{\rm{kJ}}}}{{{\rm{kg}}}} + \left( { - 1.95\frac{{{\rm{kJ}}}}{{{\rm{kg}}}}} \right)\\\\ = - 1008.25 + \left( { - 1.95} \right)\\\\W = 1010.2{\rm{ kJ/kg}}\\\end{array}$

Calculate the power output $\left( {\bf{W}} \right)$ for the entire system.

$\begin{array}{c}\\{\bf{W}} = \dot mW\\\\ = 12{\rm{ }}\frac{{{\rm{kg}}}}{{\rm{s}}} \times {\rm{1010}}{\rm{.2 }}\frac{{{\rm{kJ}}}}{{{\rm{kg}}}}\\\\{\bf{W}} = 12122.4{\rm{ kW}} \times \frac{{1{\rm{ MW}}}}{{1000{\rm{ kW}}}}\\\\{\bf{W}}{\rm{ = 12}}{\rm{.12 MW}}\\\end{array}$

Calculate the turbine inlet area $\left( {{A_1}} \right)$ using the mass flow rate equation.

$\begin{array}{l}\\\dot m = \rho {A_1}{V_1}\\\\\dot m = \frac{{{A_1}{V_1}}}{{{v_1}}}\\\\{A_1} = \frac{{\dot m{v_1}}}{{{V_1}}}\\\\{A_1} = \frac{{\left( {12\frac{{{\rm{kg}}}}{{\rm{s}}}} \right)\left( {0.08644\frac{{{{\rm{m}}^{\rm{3}}}}}{{{\rm{kg}}}}} \right)}}{{80{\rm{ }}\frac{{\rm{m}}}{{\rm{s}}}}}\\\\{A_1} = 0.013{\rm{ }}{{\rm{m}}^2}\\\end{array}$

Ans:

The change in kinetic energy $\left( {\Delta K.E} \right)$ is $- 1.95{\rm{ kJ/kg}}$ .

The power output of the turbine $\left( {\bf{W}} \right)$ is 12.12 MW.

The area of the turbine inlet is $\left( {{A_1}} \right)$ is $0.013{\rm{ }}{{\rm{m}}^2}$ .