Heat capacity coeffficents
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_inτ) 0.00138593 (τ‐1)/(T_inτ) 0.0012 (τ‐1)/(T_in* 0.001044284
Equation 4.7 terms Equation 4.7 terms Equation 4.7 terms AT_in(τ‐1) = 1283.4 AT_in(τ‐1) = 1732.2 AT_in(τ‐1) = 953.5 BT_in^2(τ^2‐1)/2 = 11984.7 BT_in^2(τ^2‐1)/2 = 27114.3 BT_in^2(τ^2‐1)/2 = 8365.6 CT_in^3(τ^3‐1)/3 = ‐3646.17 CT_in^3(τ^3‐1)/3 = ‐8800.63 CT_in^3(τ^3‐1)/3 = ‐2325.36 D(τ‐1)/(T_inτ) = 0 D(τ‐1)/(T_inτ) = 0 D(τ‐1)/(T_inτ) = 0 sum of terms= 9,622 sum of terms= 20,046 sum of terms= 6,994 Rsum of terms, J = 80,000 Rsum of terms, J = 166,667 Rsum of terms, J = 58,148 nRsum of terms, J = 800,000 nRsum of terms, J = 2,500,000 nR*sum of terms, J = 1,055,040,483
f(T) = Q ‐ nRsum of terms = 0 f(T) = Q ‐ nRsum of terms = 0.03 f(T) = Q ‐ nRsum 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
