Explanation Show the T-s diagram as in Figure (1). Express enthalpy at state 4. h 4 = h g @ 1.4 MPa (I) Here, enthalpy of saturation vapor at pressure of 1 .4 MPa is h g @ 1.4 MPa . Express entropy at state 4. s 4 = s g @ 1.4 MPa (II) Here, entropy of saturation vapor at pressure of 1 .4 MPa is s g @ 1.4 MPa . Express quality at state 5 s . x 5 s = s 4 − s f @ 10 kPa s f g @ 10 kPa (III) Here, entropy of saturation liquid at pressure of 10 kPa is s f @ 10 kPa and entropy of evaporation at pressure of 10 kPa is s f g @ 10 kPa . Express enthalpy at state 5 s . h 5 s = h f @ 10 kPa + x 5 s h f g @ 10 kPa (IV) Here, enthalpy of saturation liquid at pressure of 10 kPa is h f @ 10 kPa and enthalpy of evaporation at pressure of 10 kPa is h f g @ 10 kPa . Express enthalpy at state 5 using isentropic efficiency of turbine. h 5 = h 4 − η T ( h 4 − h 5 s ) (V) Express work done on low pressure turbine. w turb,low = h 4 − h 5 (VI) Express mass flow rate of low pressure turbine. m ˙ lowturb = W ˙ turb,II w turb,low (VII) Here, rate of work produced by low pressure turbine is W ˙ turb,II . Express mass flow rate of steam at the inlet of the high-pressure turbine. m ˙ total = m ˙ lowturb + m ˙ steam,turb (VIII) Here, mass flow rate of steam at the turbine exit is m ˙ steam,turb . Express work done by high pressure turbine. w turb,high = W ˙ turb,I m ˙ total (IX) Here, rate of work done by high pressure turbine at exit is W ˙ turb,I . Express enthalpy at state 3. h 3 = w turb,high + h 4 (X) Express enthalpy at state 4 s . h 4 s = h 3 − ( h 3 − h 4 ) η T , high (XI) Here, thermal efficiency of high pressure turbine is η T , high . Express quality at state 4 s . x 4 s = h 4 s − h f @ 1.4 MPa h f g @ 1.4 MPa (XII) Here, enthalpy of saturation liquid at pressure of 1 .4 MPa is h f @ 1.4 MPa and enthalpy of evaporation at pressure of 1 .4 MPa is h f g @ 1.4 MPa . Express entropy at state 4 s . s 4 s = s f @ 1.4 MPa + x 4 s ( s f g @ 1.4 MPa ) (XIII) Here, entropy of saturation liquid at pressure of 1 .4 MPa is s f @ 1.4 MPa and entropy of evaporation at pressure of 1 .4 MPa is s f g @ 1.4 MPa . Due to throttling process, write the entropy at state 3 is equal to entropy at state 4 s . s 3 = s 4 s (XIV) Conclusion: Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 1 .4 MPa and 10 kPa . h g @ 1.4 MPa = 2788.9 kJ / kg s g @ 1.4 MPa = 6.4675 kJ / kg ⋅ K s f @ 10 kPa = 0.6492 kJ / kg ⋅ K s f g @ 10 kPa = 7.4996 kJ / kg ⋅ K h f @ 10 kPa = 191.81 kJ / kg h f g @ 10 kPa = 2392.1 kJ / kg h f @ 1.4 MPa = 829.96 kJ / kg h f g @ 1.4 MPa = 1958.9 kJ / kg s f @ 1.4 MPa = 2.2835 kJ / kg ⋅ K s f g @ 1.4 MPa = 4.1840 kJ / kg ⋅ K Substitute 2788.9 kJ / kg for h g @ 1.4 MPa in Equation (I). h 4 = 2788.9 kJ / kg Substitute 6.4675 kJ / kg ⋅ K for s g @ 1.4 MPa in Equation (II). s 4 = 6.4675 kJ / kg ⋅ K Substitute 6.4675 kJ / kg ⋅ K for s 4 , 0.6492 kJ / kg ⋅ K for s f @ 10 kPa and 7.4996 kJ / kg ⋅ K for s f g @ 10 kPa in Equation (III). x 5 s = 6.4675 kJ / kg ⋅ K − 0.6492 kJ / kg ⋅ K 7.4996 kJ / kg ⋅ K = 0.7758 Substitute 191.81 kJ / kg for h f @ 10 kPa , 0.7758 for x 5 s and 2392.1 kJ / kg for h f g @ 10 kPa in Equation (IV)

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 104RP

To determine

The temperature, pressure, and the flow rate of steam at the inlet of the high-pressure turbine.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.9, Problem 103RP

Chapter 10.9, Problem 105RP

Students have asked these similar questions

1- Determine the following: 1- RSHF? 2- C.C.C in tons-ref. 3- Mass of supply air? Fresh Spray chilled water S air 100% RH To 34 C db & 26 wbt S Operation fan room I Exhaust air Ti 22 C db & 50% RH

How do I solve this task A weight for a lift is suspended using an adapter. The counterweight is held up with 4 screws. The weight F is 3200kg. The screws have a strength class of 8.8. Safety factor 3 Which is the smallest bunch size that can be used?+_Sr/Fm =0,16Gr=0,71ơ=800·0.8=640 MPaAs=?Fmax= As·ơ·GrFs=ơs·AsFFm= Fs· GFSF =SF / FFm · FFm Fpreload =Fload / SF → Fload /3Fpreload per screw =Fload / SF → Fload /4As=Fpreload per screw /ơ·Gr → As= Fpreload per screw / 640· 0.71 The correct answer should be M12 with As=84.3mm²

... TELEGRAM ديسمبر ۲۰۲ عند الساعة سوأل الوجه البينة ۲۷ - Find the equivalent resistance between A and B bellows For the circuit shown. • All resistances in Ohms. 2 C 2 A 4 B www 4 E 5 www ww 8 bar K. Dr. Abduljabbo Hammade 27/12/2024

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 temperature, pressure, and the flow rate of steam at the inlet of the high-pressure turbine.

Solution Summary: The author analyzes the temperature, pressure, and flow rate of steam at the inlet of the high-pressure turbine.

Videos

The temperature, pressure, and the flow rate of steam at the inlet of the high-pressure turbine.

Want to see the full answer?

Chapter 10 Solutions