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# Chemical Engineering 102 Problem Set #4 Due: Tues. 02/02 Note:) 50 mol/s of air flows at steady state in a heat exchanger. The temperature changes...

Please solve for #5.

Vapor toluene enters a compressor at 1 bar and 300ºC, and leaves at 15 bar. If the compressor is considered adiabatic, a.) what is the exit temperature (ºC) if the gas is treated as an ideal gas at all times? b.) what is the exit temperature (ºC) if the gas is not considered ideal at all times?

Chemical Engineering 102 Due: Tues. 02/02 Problem Set #4 Note: Treat all gases as ideal on this assignment unless stated otherwise 1.) 50 mol/s of air flows at steady state in a heat exchanger. The temperature changes from 10ºC to 250ºC while the pressure remains constant at 1 bar. Find: a.) <C p > H ; b.) <C p > S ; c.) the total amount of heat transferred, Q t ; d.) the change in entropy, S t . 2.) 21.6 kmol/hr of superheated benzene at 600 K and 9 bar is cooled to 300 K in a steady-flow, isobaric heat exchanger. Estimate the exchanger duty (heat, Q) in kW. Let T sat = 449.86 K at 9 bar. 3.) 120 kW of heat is removed from a heat exchanger with 25 mol/s methanol flowing through. The fluid enters started at 400ºC. What is the final temperature, if the pressure is held at 1 bar? You must iterate until your temperature is changing by less than 2% per iteration. 4.) Problem 4.21, parts b, f, m, and w 5.) Vapor toluene enters a compressor at 1 bar and 300ºC, and leaves at 15 bar. If the compressor is considered adiabatic, a.) what is the exit temperature (ºC) if the gas is treated as an ideal gas at all times? b.) what is the exit temperature (ºC) if the gas is not considered ideal at all times? c.) what is the % error in Kelvin temperatures if the ideal gas assumption is made? d.) what is the work requirement for the compressor in part (a.) in kJ/mol? Approximate answers (within ± 25%) 1.) <Cp> ~ 30 J/molK for each, c.) 400 kW; 2.) -500 kW; 3.) 350ºC 5.) b.) 380ºC; c.) small
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Q1:- For a adiabatic process we know that PV cons tan t
a) Now, for an ideal gas we know that
PV =nRT
Thus, using above two relation we get PT cons tan t
Now, we have P1 = 1bar and T2=300 oC =...

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