Solve the question asked in the first picture and use the second picture for inf

Question

Solve the question asked in the first picture and use the second picture for information if needed. Please explain and show all steps for 5 stars!

Revisit problem 6 from Chapter 10 of Solen and Harb on page 186. Assume that the total molar flow rate of the reactor feed stream is 350 kgmol/h. Assume that cooling water is available at 85degreeF and has a specified maximum temperature of 120degreeF. Find the area for a cooling jacket on the reactor, using the heat exchanger design equation. Assume that the overall heat transfer coefficient is U0 = 20 Btu h-1 ft-2degreeF-1. Assume that the hot process stream enters and leaves the reactor at 500degreeF. In a steady-state process, the stream enters a reactor where 2/3 of the species A is converted to species B. The heat of reaction at the reactor temperature equals -918 kJ/kgmol of A reacted. How much heat (per kgmol of entering stream) must be transferred from the reactor to maintain a steady temperature in the reactor? (As indicated in the chapter, kinetic and potential energy effects can be neglected.) While there is some shaft work from the stirrer in the reactor, such work is usually negligible and can be considered so in this case.

Explanation / Answer

enthaly of vaurization is 42668 BTU/kgmol

hence in exchanger

max temp differnece of water is 120-85 =35

42668 *350= 20 *A*35

hence area A = 21334 ft^2

.........

as temperature of steam at exit and inlet is same (steam only changing phase) hence co current and counter current design will be same..

here as we have to find energy per kgmol of steam we will choose basis of 1 mol of steam

1 mole of steam contains 0.27 species

and 2/3 of species A is converted to species B

hence 0.27*2/3 =0.18 moles of A is converted to B

heat of reaction is per mole is 918 KJ

hence for 0.18 mole, heat of reaction is -918*0.18 = -165.24 KJ

hence 165.24 KJ heat should be extracted in order to maintain same temperature

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