Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
A reheat-regenerative rankine cycle needs to produce 150MW of net power. Boiler pressure is at 15MPa. Reheat occurs at 10MPa. extractions occur at 5MPa and 2MPa for open feedwater heaters. These extractions occur after the expansion after reheating. Steam leaves the boiler and reheater at 600C. Condenser pressure is at 10kPa. SOLVE FOR: a.) Determine the amount of steam needed passing through the boiler to produce the required net power (kg/s) b.) Determine the amount of heat added to the system (kJ/s) c.) Determine the thermal efficiency of this cycle (%)

A reheat-regenerative rankine cycle needs to produce 150MW of net power. Boiler pressure is at 15MPa. Reheat occurs at 10MPa. extractions occur at 5MPa and 2MPa for open feedwater heaters. These extractions occur after the expansion after reheating. Steam leaves the boiler and reheater at 600C. Condenser pressure is at 10kPa. SOLVE FOR: a.) Determine the amount of steam needed passing through the boiler to produce the required net power (kg/s) b.) Determine the amount of heat added to the system (kJ/s) c.) Determine the thermal efficiency of this cycle (%)

Refrigeration and Air Conditioning Technology (MindTap Course List)
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter40: Typical Operating Conditions
Section: Chapter Questions
Problem 1RQ: The standard design conditions for air-conditioning systems established by the Air-Conditioning,...
See similar textbooks
icon
Related questions
Question

A reheat-regenerative rankine cycle needs to produce 150MW of net power. Boiler pressure is at 15MPa. Reheat occurs at 10MPa. extractions occur at 5MPa and 2MPa for open feedwater heaters. These extractions occur after the expansion after reheating. Steam leaves the boiler and reheater at 600C. Condenser pressure is at 10kPa.

SOLVE FOR:

a.) Determine the amount of steam needed passing through the boiler to produce the required net power (kg/s)

b.) Determine the amount of heat added to the system (kJ/s)

c.) Determine the thermal efficiency of this cycle (%)

PLS PLS USE THE ATTACHED IMAGE OF THE STEAM TABLE FOR THE VALUES FOR ENTHALPHYS

ANSWERS ROUND OFF TO SIG SIGNIFICANT FIGURES

TABLE 2: SATURATED WATER - PRESSURE TABLE Specific Volume, m3/kg Internal Energy, KJ/kg Enthalpy, KJ/kg Entropy, KJ/(kg-K) Press. Sat. Temp. oc Sat. Vapor Sat. Liquid Sat. Vapor Sat. Liquid Sat. Liquid Sat. Vapor Sat. Liquid Evap. Evap. Evap. Sat. Vapor Кра 10 45.81 0.001010 14.67 191.82 2246.1 2437.9 191.83 2392.9 2584.7 0.6493 7.5009 8.1502 Specific Volume, m3/kg Internal Energy, KJ/kg Enthalpy, KJ/kg Press. Entropy, KJ/(kg-K) Мра Sat. Temp. Sat. Liquid Sat. Vapor Sat. Liquid Evap. Sat. Vapor Sat. Liquid Evap. Sat. Vapor Sat. Liquid Evap. Sat. Vapor Psat t oc 2.0 212.42 0.001177 0.09963 906.44 1693.9 2600.3 908.79 1890.7 2799.5 2.4474 3.8935 6.3409 5.0 263.99 0.001286 0.03944 1147.81 1449.3 2597.1 1154.23 1640.1 2794.3 2.9202 3.0532 5.9734 CamScanner 0000
Transcribed Image Text:TABLE 2: SATURATED WATER - PRESSURE TABLE Specific Volume, m3/kg Internal Energy, KJ/kg Enthalpy, KJ/kg Entropy, KJ/(kg-K) Press. Sat. Temp. oc Sat. Vapor Sat. Liquid Sat. Vapor Sat. Liquid Sat. Liquid Sat. Vapor Sat. Liquid Evap. Evap. Evap. Sat. Vapor Кра 10 45.81 0.001010 14.67 191.82 2246.1 2437.9 191.83 2392.9 2584.7 0.6493 7.5009 8.1502 Specific Volume, m3/kg Internal Energy, KJ/kg Enthalpy, KJ/kg Press. Entropy, KJ/(kg-K) Мра Sat. Temp. Sat. Liquid Sat. Vapor Sat. Liquid Evap. Sat. Vapor Sat. Liquid Evap. Sat. Vapor Sat. Liquid Evap. Sat. Vapor Psat t oc 2.0 212.42 0.001177 0.09963 906.44 1693.9 2600.3 908.79 1890.7 2799.5 2.4474 3.8935 6.3409 5.0 263.99 0.001286 0.03944 1147.81 1449.3 2597.1 1154.23 1640.1 2794.3 2.9202 3.0532 5.9734 CamScanner 0000
TABLE 3-3: SUPERHEATED WATER T V u h KJ/kg P = 10.0 Mpa (311.06 OC) OC m3/kg KJ/(kg-K) m3/kg KJ/kg KJ/(kg-K) KJ/kg P = 15.0 Mpa (342.24 OC) KJ/kg Sat. Vap: 0.010337 2455.5 2610.5 5.3098 600 0.02491 3208.6 3582.3 6.6776 0.03837 3241.7 3625.3 6.9029 CamScanner
Transcribed Image Text:TABLE 3-3: SUPERHEATED WATER T V u h KJ/kg P = 10.0 Mpa (311.06 OC) OC m3/kg KJ/(kg-K) m3/kg KJ/kg KJ/(kg-K) KJ/kg P = 15.0 Mpa (342.24 OC) KJ/kg Sat. Vap: 0.010337 2455.5 2610.5 5.3098 600 0.02491 3208.6 3582.3 6.6776 0.03837 3241.7 3625.3 6.9029 CamScanner
Expert Solution
steps

Step by step

Solved in 4 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Power Plant Engineering
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.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Refrigeration and Air Conditioning Technology (Mi…
Refrigeration and Air Conditioning Technology (Mi…
Mechanical Engineering
ISBN:
9781305578296
Author:
John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS