1mot alr is heated reversibly at constant pressure from Isito 300 c, and then cooled reversibly atconstant volume backto the initral temperature, the initial prevure is L03bar. Calculate thenet heat flow and the overall change of entropy Sketch theen T-s diagram. Ans. L lo1.51g , 0.2461glk]process

Given:V1=1 m3T1=15°C =288 KT2=300°C =573 KP1=1.03 bar

1mot air is haated reversibly at constant pressure from 15c to 300 č, and then cooled reversibly at constant volume back to the initral temperature, the initial presure is n03bar. Calculate the net heat flow and the overall change of entropy. Sketch the process on T-s diagram . Ans. L lo1-5 1y, 0.246 1 lK]

1mot air is haated reversibly at constant pressure from 15c to 300 č, and then cooled reversibly at constant volume back to the initral temperature, the initial presure is n03bar. Calculate the net heat flow and the overall change of entropy. Sketch the process on T-s diagram . Ans. L lo1-5 1y, 0.246 1 lK]

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

Chapter46: Room Air Conditioners

Section: Chapter Questions

Problem 15RQ

See similar textbooks

Similar questions

p89#10. a polytropic process of air from 150psia,300degree farenheit and 1ft^3 occurs to p2=20psia in accoridance with pV^1.3=C. Dtermine (a)t2 and V2,(b) change in internal enegry,change in enthalpy and change in entropy,(c)integral of pdV and -integral of Vdp.(d) compute the heat from the polytopic specific heat and check by the equation Q=change in internal energy + pdV. (e) find the nonflow work and (f) the steady flow work for change in kineteic energy=0.
Please show the complete solution (Formula, Given and Conversion of Units) The gain of entropy during isothermal non- flow process of 5 lb of air at 60°F is 0.462 Btu/°R. Find the V1/V2.
What are the specific volume, specific internal energy, specific enthalpy andspecific entropy of steam at 1.50 Mpa and temperature is 180 oC?
P88#1. a perfect gas has a value of R=58.8 ft.lbf/lbm degree Rankine and K=1.26. if 20 Btu are added to 5lbm of this gas at constant volume when the initial temperature is 90degree farenheit, find (a)T2, (b) enthalphy, (c)entropy, (d)internal energy and (e) work for nonflow process.
Compute the change in specific entropy when helium, initially at a temper-ature of 273 K and a pressure of 100 kPa, undergoes the following processes: a.An isochoric process with final pressure 250 kPa b.By computing the change in specific entropy, determine whether it is possible foran adiabatic process to bring the gas to a final temperature of 200 K and a densityof 0.3 kg/m3. If not, why not? And if so, how could it be done?
Steam at 3.1 MPa and with 74.28 °C SH rejects 1973.36 kJ per kg of steam heat at constantpressure, determine the (a) final temperature and (b) change in specific entropy.
p88#5. while the pressure remains constant at 689.5 kPa the volume of a system of air changes from 0.567m^3. what are the (a)internal energy,(b) enthalphy ,(c) Q, (d) entropy, (e) if the process is nonflow and internally reversible,what is work?
steam at pressure of 0.70 mpa has initial internal of 2850kj/kg. what are the temperature, enthalpy and entropy
Steam at a pressure of 0.90 Mpa has entropy of 4.9678 kJ/kg (K). What are the specific volume, internal energy, and enthalpy?
Saturated liquid at 240 C flows through a pipeline. What is its entropy in kJ/kg-k? choose the correct answer
Steam at a temperature of 250°C has a specific volume of 0.15436m3Kg. What are the pressure, specific internal energy, enthalpy,and entropy?
Please Solve only if you are 100% Sure , Don't make any kind of Mistake. Show all the work ,if possible attach Steam Table. 1)what is the state of steam at 623 K and 90 bar pressure? Determine the specfic entropy of this steam