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A series of two turbines and a heat exchanger are used to obtain shaft work from steam in a...

Question:

A series of two turbines and a heat exchanger are used to obtain shaft work from steam in a steady state process.The steam enters the first turbine at 10 bar and 250 {eq}^o{/eq}C, and exits the turbine at 3 bar. In the heat exchanger, the steam is heated back up to 250 {eq}^o{/eq}C, while the pressure remains at 3 bar. This steam enters the second turbine in which it is expanded to 1 bar. Each turbine has an efficiency of 80 %.

For each kilogram of steam entering the process, find the amount of work produced in each turbine, and the amount of heat added in the heat exchanger.

Work

When an amount of force is applied on the object to displace it for an amount that is known as the work when a person pushes the box, the work is done by him is the amount of force applied by him to the displacement length of the box.

Answer and Explanation: 1

Given Data

  • Enter pressure of the steam is: {eq}{P_1} = 10\;{\rm{bar}} = 0.1\;{\rm{MPa}} {/eq}
  • Enter temperature of the steam is: {eq}{T_1} = 250^\circ {\rm{C}} {/eq}.
  • Exit pressure of the steam is: {eq}{P_2} = 3\;{\rm{bar}} {/eq}.
  • Temperature again heated is: {eq}\dot T = 250^\circ {\rm{C}} {/eq}.
  • Expanded pressure of the steam is: {eq}{P_E} = 1\;{\rm{bar}} {/eq}.
  • Efficiency of the turbine is: {eq}\eta = 80\% {/eq}.


From the properties of the steam

Enthalpy of the superheated steam at temperature {eq}{T_1} {/eq} and pressure{eq}{P_1} {/eq} is: {eq}{h_1} = 2943.1\;{\rm{kJ/kg}} {/eq}

Temperature of the steam at the pressure{eq}{P_2} {/eq} is:{eq}{T_1} = 133.52^\circ {\rm{C}} {/eq}.

Enthalpy of the steam at the pressure{eq}{P_2} {/eq} is:{eq}{h_2} = 2724.9\;{\rm{kJ/kg}} {/eq}.

Enthalpy of the steam at temperature{eq}250^\circ {\rm{C}}{/eq} and pressure {eq}3\;{\rm{bar}}{/eq} is: {eq}{h_3} = 2967.9\;{\rm{kJ/kg}} {/eq}.

Enthalpy of the steam at the pressure of the {eq}1\;{\rm{bar}}{/eq} is: {eq}{h_4} = 2675.0\;{\rm{kJ/kg}} {/eq}.


Expression to calculate the enthalpy change due to the expansion is

{eq}\Delta h = {h_1} - {h_2} {/eq}

Substitute the value in above expression

{eq}\begin{align*} \Delta h &= 2943.1\;{\rm{kJ/kg}} - 2724.9\;{\rm{kJ/kg}}\\ \Delta h& = 218.2\;{\rm{kJ/kg}} \end{align*} {/eq}


Expression to calculate the work done is

{eq}W = \Delta h \times \eta {/eq}

Substitute the value in above expression

{eq}\begin{align*} W &= 218.2\;{\rm{kJ/kg}} \times 0.8\\ W &= 174.56\;{\rm{kJ/kg}} \end{align*} {/eq}

Thus the work done at the turbine is {eq}174.56\;{\rm{kJ/kg}} {/eq}.


Expression to calculate the change in enthalpy in the travelling across the system is

{eq}\delta h = {h_3} - \Delta h {/eq}

Substitute the value in above expression

{eq}\begin{align*} \delta h& = 2967.9\;{\rm{kJ/kg}} - 218.2\;{\rm{kJ/kg}}\\ \delta h &= 2749.7\;{\rm{kJ/kg}} \end{align*} {/eq}


Expression to calculate the enthalpy of the steam at the turbine is

{eq}{h_T} = \delta h - {h_4} {/eq}

Substitute the value in above expression

{eq}\begin{align*} {h_T}& = 2749.7\;{\rm{kJ/kg}} - 2675.0\;{\rm{kJ/kg}}\\ {h_T}& = 74.7\;{\rm{kJ/kg}} \end{align*} {/eq}


Expression to calculate the work done of the turbine is

{eq}W = {h_T} \times \eta {/eq}

Substitute the value in above expression

{eq}\begin{align*} W &= 74.7\;{\rm{kJ/kg}} \times 0.8\\ W &= 59.75\;{\rm{kJ/kg}} \end{align*} {/eq}

Thus the heat added in the heat exchanger is {eq}59.75\;{\rm{kJ/kg}} {/eq}.


Learn more about this topic:

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Work: Definition, Characteristics, and Examples

from

Chapter 8 / Lesson 3
88K

Pushing a wall all day may feel like work, but unless you get that wall moving you're not doing any work according to the rules of physics. In this video lesson, you'll learn how work is defined as well as how to calculate the amount of work done on an object.


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