CHEMICAL ENERGY BALANCE - EXAMPLE 11.3 : For liquid benzene, the CP constants are : a = 129440, b = - 169.5, c = 0.64781. Reference temperature is 298 K. The temperature of benzene is 60 degree Celsius. Calculate the enthalpy of benzene by using the formula H = a (DT) + (b/2) (T^2 - TREF^2) + (c/3) (T^3 - TREF^3) where ^ is power, DT is temperature difference with TREF = 298 K. H is in J / kmol. DT = T - TREF.
CHEMICAL ENERGY BALANCE - ANSWER 11.3 : T = 60 + 273 = 333 K. H = 129440 (35) + (-169.5 / 2) (333^2 - 298^2) + (0.64781 / 3) (333^3 - 298^3) = 4917920 J / kmol. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).
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