You have been asked to design a heat exchanger to cool a fluid from 80°c to 50°c. the fluid has a mass flow rate of 2kg/s and a specific heat of 3500 j/(kg*k). you have chilled water as your heat transfer media, with an incoming temperature of 15c and a flow rate of 2.5kg/s. the overall heat transfer coefficient is 2000 w/(m2*k). a) calculate the area required for a crossflow, single pass heat exchanger, where both fluids are unmixed. b) calculate the area required for a concentric tube, parallel flow heat exchanger

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

i think its 5049.

i hope this !

you didnt put any of the following