Question

from introduction of chemical engineering thermdynamics 7th edition.

chapter 9 question 11 part(c).

can you show me how did you read the graph and get H4=117.5 and S4=0.262.

i need you to show the graph and how to read it to get thoses amswers.

ARE YOU GOING TO SOLVE IT ??????

you guys know no thing. you never solve any problem right and you always late.

0 Step 6 of 9 Done (c) Draw the flow diagram for the process as follows: Qc (4) (3) Heat exchanger (2) (1) Temperature at point-4 is 70F Pressure of the system is 14.667 psia. From graph G.2 the enthalpy of the fluid at point-4 (H) is 17,5 Btulb Entropy of the fluid at point-4 (S.) is 0.262 Btu/lb-R Calculate the flow rate of refrigerant as follows: н,-н, 5 Btu/s 117.5 Btu/lb-37.978 Btu/lb 0,0629 lb-/s 0.0629 lb./ Therefore, the flow rate of refrigerant is 9.11. A refrigerator with tetrafluoroethane (Table 9.1, Fig. G.2) as refrigerant operates with an evaporation temperature of -15F) and a condensation temperature of 80F) Saturated liquid refrigerant from the condenser flows through an expansion valve into the evaporator, from which it emerges as saturated vapor Obras protegidas por direitos de a Problems 335 (a) For a cooling rate of 5(Btu)(s)-, what is the circulation rate of the refrigerant? (b) By how much would the circulationn rate be reduced if the throttle valve were replaced by a turbine in which the refrigerant expands isentropically? (c) Suppose the cycle of (a) is modified by the inclusion of a countercurrent heat exchanger between the condenser and the throttle valve in which heat is trans ferred to vapor returning from the evaporator. If liquid from the condenser enters the exchanger at 80(°F) and if vapor from the evaporator enters the exchanger at -15CF) and leaves at 70 F), what is the circulation rate of the refrigerant? (d) For each of (a), (b), and (c), determine the coefficient of performance for isentropic compression of the vapor.

Answer 1

4-11 CSateutcon): HE vaporator 4A 2,150F Clskan klae evaporalon paine T: - 15°E at P - 14.667 psia aud Mo salurated hgaud aud Sai vapor So 100 799 Btu Hrig= 7-5DS Btu Llom Avap = H2 = om Siq= 0.01733 Btu 0-227 14 Bty Svapz Lom Raukine qeid al paint hes saturatad T 80 F S4=0.07892 Btu lom Rantine He= 37.978 Btu is Afaigesaiuin cuicuiaton sate 4on seuthalpte apansin paaces aud HIHA sec Hom Patt d.026 Sec tao 9937 928 Hz-H Toa0.039om Sec Ans (a Scanned with CamSdanne

DATE AGE NO haing iseutapie expansien he arymess sactuon is S Slig Svap- Slig O07892-0.01733 0.2214-D.01333 O.2935 = Hrig+1(Hvap- Hag). H, -2.D5+0.2935( 100.999-25D5) HiE 34.89.2 Boty aredation sate saduced u) 0.0359ln Sec ennount 100.999-348 92 H2-41 Ans (b exapasahs cachage the satusaladl vapoh rm tine iA uperlcated in a taeat e 70CE ai a paissuse di 14.667 P'sia the paepertues from straue table auu HoA= F S Blu S2A =0.262 Llom Atom lautis H2A-H4 539.98 0.0629 22S ANs (C

9serdaopee compaesuon d sat vapor at Sa= Svap and P 101-37 (Psia) Ha = 118.3 Bty Fos @ HEH4 100 099-39.28 He-H4 118.3-32498 T00-399 Hs-H coP 35896 Ho-Hi 100.299-34.892 IL Ha-H 18.3-100 399 cDP 3. 96 89 COPE SA O.D62 korny la entsepy is constaut Bty Ehe (Ha-HA) me Ha= 188 lom -(138-112.5)XO.062.9 1-289 Sec 3.8991 CDP@ w 1289 Ans