Air at 80 kPa and 400 K enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the airstream is decreased from 230 to 30 m/s as it passes through the diffuser. Find (a) the exit temperature of the air and (b) the exit area of the diffuser.

a)  425.6 K = 152.6 degree Celsius

b)  678.6 square centimeter

Explanation:

Initial enthalpy = Final enthalpy

m(h1 + v1^2/2) = m(h2 + v2^2/2)

(h1 + v1^2/2) = (h2 + v2^2/2)

h 2 = (h1 + v1^2/2) - (v2^2/2)

h 2 = 400.98 + (230^2 – 30^2)/2 * 10^-3

= 426.98 KJ/Kg

T2 = T1’ + (T2’ -T1’)(h2-h1’)/( h2’-h1’)

T2 = 420 + (430-420)(431.43-421.26)(426.98-421.26)/(431.43 – 421.26)

T2 = 425.6 K = 152.6 degree Celsius

Area = mRT2/P2V2

Area = (600/60*60*1) *0.287*425.6/(100*30) *10^4

Area = 678.6 square centimeter

i would say the answer is food.

hope this !

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz