A steam power plant operates on the ideal Rankine cycle. Steam enters the turbine at the following conditionsPressure: 3 MpaTemperature: 350°CThe steam further condenses in the condenser at 75kPa pressure. Calculate the thermal efficacy of the cycle. Q1. A steam power plant operates on the ideal Rankine cycle. Steam entersthe turbine at the following conditionsPressure: 3 MpaTemperature: 350°CThe steam further condenses in the condenser at 75kPa pressure. Calculatethe thermal efficacy of the cycle.

Answered: Q1. A steam power plant operates on the…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

See similar textbooks

Related questions

Question

A steam power plant operates on the ideal Rankine cycle. Steam enters the turbine at the following conditions Pressure: 3 Mpa Temperature: 350°C The steam further condenses in the condenser at 75kPa pressure. Calculate the thermal efficacy of the cycle.

Q1. A steam power plant operates on the ideal Rankine cycle. Steam enters
the turbine at the following conditions
Pressure: 3 Mpa
Temperature: 350°C
The steam further condenses in the condenser at 75kPa pressure. Calculate
the thermal efficacy of the cycle.

Expert Solution

Step 1

Rankine cycle

Through Rankine cycle model, the performance of a steam power plant is measured. It is an ideal heat engine thermodynamic cycle through which the heat is converted in to work during change of phase.

There are four steps of the Rankine cycle. The states are identified by numbers.

Process 1–2: Isentropic compression.

Process 2–3: Constant pressure heat addition in boiler

Process 3–4: Isentropic expansion

Process 4–1: constant pressure condensation

Step 2

For condenser, state 1,

Given:

$P_{1} = 75 kPa, (saturated liquid)$

The value of enthalpy, specific volume and specific internal energy for this value of pressure can be noted down by steam table such as:

$h_{1} = h_{f} at 75 kpa= 384.44 \frac{kJ}{kg} v_{1} = v_{f} at 75 kpa = 0.001037 \frac{m^{3}}{kg} T_{1} {=T}_{saturation} at 75 kpa = 91.76 ° C$

State 2, steam,

Given:

$P_{2} = 3 Mpa$

Step 3

Step 1-2 is isentropic compression, so $S_{1} = S_{2}$ ,

Work can be calculated as;

$\begin{array}{rcl} w_{p u m p, i n} & = & v_{1} (P_{2} - P_{1}) \\ = & 0.001037 \frac{m^{3}}{kg} \times (3000 - 75) kpa \\ = & 3.03 \frac{kJ}{kg} \end{array}$

And enthalpy will be,

$\begin{array}{rcl} h_{2} & = & h_{1} + w_{p u m p, i n} \\ = & (384.44 + 3.03) \frac{kJ}{kg} \\ = & 387.47 \frac{kJ}{kg} \end{array}$

Step by step

Solved in 5 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemical-engineering and related others by exploring similar questions and additional content below.

Recommended textbooks for you

Introduction to Chemical Engineering Thermodynami…

Chemical Engineering

ISBN:

9781259696527

Author:

J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:

McGraw-Hill Education