Question

llion). 2. A stream of water vapor flowing at a rate of 350 mol/h is brought from so to 100°C and 1 atm. (a) Estimate the required cooling rate (kW) three wa steam tables. (ii) using heat capacity data in Table B.2, and (iii) usings data in Table B.8. (b) Which of the answers in part (a) is most accurate, and whe is the physical significance of the difference between the values calculated w (i) and (ii)? ht from 590°C and 10 bar se (kW) three ways: i) from the 2. and (iii) using specific enthalpy accurate, and why? (e) What values calculated with methods

Answer 1

Solutions given that h: 3$0 mol/h > mass flow rate of water m= hMHO 350 mol x 18 = 6300 = 0.00175 kg/s state ! TE 590c Pilo bar Sat Bolo bar = 179,99 As given tempe rature (TE 590°) is more than the Saturation temperature. Csat @lo bar = 179,9°c) at given Pressure. slate I must be in Superheated State. : Slate 2 - e Ta looc :P2 = 1 atm sat @latm = 100c As saturation temperature at given Pressure is equal to the gluen temperature. we cannot exactly decide the cond? ton of the steam. It may be in saturated 19 quid conditor/ Saturated vapour condition I wet condition,

og From Steam tables = . using Properties of Superheated steam Ti = 590? if he 3676 kJ/kg. Pic lo bar. Ja using properlies of Saturated Steam Pressure talde T2 = 100°c Pas latm= 1501325 bar s heyə big @ 1.0635 baj 267665/ng => required cooling rate's ge: toan 0.00175 (2676 - 3676) .=FLr5 kW) (ii) From table B.2, (Heat' capacities) - s Look up the heat capacity coefficients : The coefficients of water vapour are az33.46x103 S 6 = 0,6880x1os C= 0.7604xio 8 I d =-3,543 xid? molat Specific heat is . cp= atbit (7²7 dt3 : = 33.46*1634 0.6880x105T+0.76014109723.593710'3 "KI . moloc

- Required cooling rate, B = Stepdt = (350 mo!? 5 [33.46x140.6880xióST+0.7604xiospa 10900 - 3,593x10=72f3] dt. [35ox mol 7 [ 33.46 +16 (100-590) + 0.6880x1õ(100+590) goose L 4.0.160 41% (100% spa)=3,53ko Co 3600 sec 2 3 :(2.097 ved] [-1.062] [-1.74 kw -1.74 kw and using the enthally function from Appendix Table .B. 8 = The specific enthalpy of the prilet Steam which is at 590°c is Houst kJ/mol The specific enthalpy of the outlet steam which is at lo8c is Hout= 2.54 kI/mol ::$= n( Hout - H&p) 1796 1,097 mol 099 mois - sec si mou

The answer in Part B is most accurate because Steam tables de compiled from experimental data. This method also considered the pressure changes in the System Part Cil is based on a mathematical model to determine the enthalpy, which uses in equation to approximate the behaviour of a system. This modet does not consider, thie effects of Pressure on the enthalpy of the system Part(iii also does not consider the effects of Pressure ou to enthalpy The difference between the values calculated in Part 8 and part ch is due to the enthally change from thie system Pressure gorning from 10 bal to Latt. The temperature. Change effects are accounted for in botti methods.