Explanation The given cycle is the simple Rankine cycle with water as the working fluid and operated at the boiler and condenser pressures of 5 MPa and 10 kPa respectively. The simple Rankine cycle involves following processes and following conditions: Processes: 1. Process 1 to 2 – Isentropic compression in a Pump 2. Process 2 to 3 – Heat addition at constant pressure in a Boiler 3. Process 3 to 4 – Isentropic expansion in a Turbine 4. Process 4 to 1 – Heat rejection at constant pressure in a Condenser Conditions: The pressure is constant for the process 2 to 3 and the process 4 to 1. P 2 = P 3 and P 4 = P 1 The entropy is constant for the process 1 to 2 and the process 3 to 4. s 1 = s 2 and s 3 = s 4 Here, water is the working fluid and neglect the pump work. Hence, the net power output of the plant is equal to the turbine output. Write the formula for net power output of the plant. W ˙ net,out = W ˙ T, out = m ˙ ( h 3 − h 4 ) (I) Here, the mass flow rate is m ˙ , the enthalpy is h and the subscripts 3 and 4 indicates the process states and condition of steam. Write the formula for mass flow rate for the given plant. m ˙ = Q ˙ in q in (II) Write the formula for heat input of the plant ( q in ) . q in = h 3 − h 2 (III) Write the formula for quality of water at the exit of the turbine (state 4) is expressed as follows. x 4 = s 4 − s f s f g (IV) The enthalpy and entropy at state 4 is expressed as follows. h 4 = h f , 4 + x 4 h f g , 4 (V) Here, the enthalpy is h , the entropy is s , the quality of the water is x , the suffix f indicates the fluid condition, the suffix f g indicates the change of vaporization phase; the suffix 4 indicates the state 4. At state 1: The water exits the condenser as a saturated liquid at the pressure of 10 kPa . Hence, the enthalpy at state 1 is as follows. h 1 = h f @ 10 kPa Refer Table A-5, “Saturated water-Pressure table”. The enthalpy ( h 1 ) at state 1 corresponding to the pressure of 10 kPa is 191.81 kJ/kg . At state 3: (Turbine inlet) Refer Table A-6, “Superheated water”. The enthalpy ( h 3 ) and entropy ( s 3 ) at state 3 corresponding to the pressure of 5 MPa and the temperature of 600 ° C is 3666.9 kJ/kg and 7.2605 kJ/kg ⋅ K respectively. At state 4: (Condenser inlet) Refer Table A-5, “Saturated water-Pressure table”. Obtain the following properties corresponding to the pressure of 10 kPa . h f , 4 = 191.81 kJ/kg h f g , 4 = 2392.1 kJ/kg s f , 4 = 0.6492 kJ/kg ⋅ K s f g , 4 = 7.4996 kJ/kg ⋅ K Conclusion: When the pump work is neglected, the enthalpy at state 1 and state 2 is equal

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

9th Edition

ISBN: 9781260048667

Author: Yunus A. Cengel Dr.; Michael A. Boles

Publisher: McGraw-Hill Education

See similar textbooks

Videos

Question

Chapter 10.9, Problem 127FEP

To determine

The power output of this plant from the following:

(a) 118 kW

(b) 140 kW

(c) 177 kW

(d) 286 kW

(e) 450 kW

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.9, Problem 126FEP

Chapter 10.9, Problem 128FEP

Students have asked these similar questions

36 2) Use the method of MEMBERS to determine the true magnitude and direction of the forces in members1 and 2 of the frame shown below in Fig 3.2. 300lbs/ft member-1 member-2 30° Fig 3.2. https://brightspace.cuny.edu/d21/le/content/433117/viewContent/29873977/View

Can you solve this for me?

5670 mm The apartment in the ground floor of three floors building in Fig. in Baghdad city. The details of walls, roof, windows and door are shown. The window is a double glazing and air space thickness is 1.3cm Poorly Fitted-with Storm Sash with wood strip and storm window of 0.6 cm glass thickness. The thickness of door is 2.5 cm. The door is Poor Installation. There are two peoples in each room. The height of room is 280 cm. assume the indoor design conditions are 25°C DBT and 50 RH, and moisture content of 8 gw/kga. The moisture content of outdoor is 10.5 gw/kga. Calculate heat gain for living room : الشقة في الطابق الأرضي من مبنى ثلاثة طوابق في مدينة بغداد يظهر في مخطط الشقة تفاصيل الجدران والسقف والنوافذ والباب. النافذة عبارة عن زجاج مزدوج وسمك الفراغ الهوائي 1.3 سم ضعيف الاحكام مع ساتر حماية مع إطار خشبي والنافذة بسماكة زجاج 0.6 سم سماكة الباب 2.5 سم. الباب هو تركيب ضعيف هناك شخصان في كل غرفة. ارتفاع الغرفة 280 سم. افترض أن ظروف التصميم الداخلي هي DBT25 و R50 ، ومحتوى الرطوبة 8…

Chapter 10 Solutions

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

Ch. 10.9 - Why is the Carnot cycle not a realistic model for...Ch. 10.9 - Why is excessive moisture in steam undesirable in...Ch. 10.9 - A steady-flow Carnot cycle uses water as the...Ch. 10.9 - A steady-flow Carnot cycle uses water as the...Ch. 10.9 - Consider a steady-flow Carnot cycle with water as...Ch. 10.9 - Water enters the boiler of a steady-flow Carnot...Ch. 10.9 - What four processes make up the simple ideal...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...

Knowledge Booster

Learn more about

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

The power output of this plant from the following: (a) 118 kW (b) 140 kW (c) 177 kW (d) 286 kW (e) 450 kW

Solution Summary: The author explains the simple Rankine cycle with water as the working fluid and operated at the boiler and condenser pressures of 5MPa and 10

Videos

The power output of this plant from the following: (a) 118 kW (b) 140 kW (c) 177 kW (d) 286 kW (e) 450 kW

Want to see the full answer?

Chapter 10 Solutions

The power output of this plant from the following:

(a) 118 kW

(b) 140 kW

(c) 177 kW

(d) 286 kW

(e) 450 kW