## what is the final temperature  for the flowing

Lecture # 18 Handout   by J.W. Van Zee 19‐Sep‐17
Problem 4.2.  Statement: FIND: For steady flow through a heat exchanger at approximately atmospheric pressure, what tis the final temperature
for the flowing: a. When 800 kJ of heat is added to 10 moles of ethylene initially at 200C Ethylene = C2H4 b. When 2,500 kJ is added to 15 moles of 1 butene initially at 260C. 1‐butene =  c. When 10^6  Btu is added to 40 lb mole of ethylene initially at 500 F. Ethylene = C2H4
Known: Q, kJ=  varies by part Properties: schematic: T_0, C varies by part Heat capacity coeffficents:
T_0, K varies by part Ethylene 1‐Butene T,K = varies by part T=? n, moles=  varies by part A 1.424 1.967
B 1.44E‐02 3.16E‐02 Assumptoins: none C ‐4.39E‐06 ‐9.87E‐06
D 0 0 Q varies by part Analysis:
Use equation 4.7 or equaiton 4.8. Here we use equation 4.7 to intergrate the Cp/R from  T0  = 200C to T . This results from an energy balance on the heat exchanger for a flow system Part c.  500 F= 533.13 K which yields     0=Q + ΔH and  ΔH = H_in ‐ H _out and  where becuae of the integral: dH =  ‐Cp*dT. Part c.  1 btu =  1055 J
Part c.  1 lb mole = 453.6 g moles Part a.  n, mole =   10 Part b.  n, mole =   15 Part c.  n, mole =   18,144
Q, kJ = 800 Q, kJ = 2,500 Q, kJ = 1,055,040                T_in,K 473.15 solving for T_out,K T_in,K 533.15 solving for T_out,K T_in,K 533.13 solving for T_out,K T_out,K 1374.4476 T_out,K 1413.8 T_out,K 1202.746053 R, J/mol‐K = 8.3143 R, J/mol‐K = 8.3143 R, J/mol‐K  8.3143 τ = T_out/T_in = 2.90488767 τ = T_out/T_in 2.6517 τ = T_out/T 2.256008953 τ ‐ 1= 1.90488767 τ ‐ 1= 1.6517 τ ‐ 1= 1.256008953 τ ^2‐ 1= 7.43837237 τ ^2‐ 1= 6.0316 τ ^2‐ 1= 4.089576397 τ^3 ‐ 1 = 23.5125238 τ^3 ‐ 1 = 17.6457 τ^3 ‐ 1 = 10.48212992 (τ‐1)/(T_in*τ) 0.00138593 (τ‐1)/(T_in*τ) 0.0012 (τ‐1)/(T_in* 0.001044284
Equation 4.7 terms Equation 4.7 terms Equation 4.7 terms A*T_in*(τ‐1) = 1283.4 A*T_in*(τ‐1) = 1732.2 A*T_in*(τ‐1) = 953.5 B*T_in^2*(τ^2‐1)/2 = 11984.7 B*T_in^2*(τ^2‐1)/2 = 27114.3 B*T_in^2*(τ^2‐1)/2 = 8365.6 C*T_in^3*(τ^3‐1)/3 = ‐3646.17 C*T_in^3*(τ^3‐1)/3 = ‐8800.63 C*T_in^3*(τ^3‐1)/3 = ‐2325.36 D*(τ‐1)/(T_in*τ) = 0 D*(τ‐1)/(T_in*τ) = 0 D*(τ‐1)/(T_in*τ) = 0 sum of terms= 9,622              sum of terms= 20,046              sum of terms= 6,994                       R*sum of terms, J = 80,000            R*sum of terms, J = 166,667            R*sum of terms, J = 58,148                     n*R*sum of terms, J = 800,000          n*R*sum of terms, J = 2,500,000        n*R*sum of terms, J = 1,055,040,483
f(T) = Q  ‐ n*R*sum of terms =  0                      f(T) = Q  ‐ n*R*sum of terms =  0.03                  f(T) = Q  ‐ n*R*sum of terms =  (81)                            use solve to find this value of T_out in K= 1374 use solve to find this value of T_out in K= 1414 use solve to find this value of T_out in K= 1203

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