When fermentation units are operated with high aeration rates, significant amounts of water can be evaporated into the air passi
ng through the fermentation broth. Since fermentation can be adversely affected if water loss is significant, the air is humidified before being fed to the fermenter. Sterilized ambient air is combined with steam to form a saturated air–water mixture at 1 atm and 90°C. The mixture is cooled to the temperature of the fermenter (35°C), condensing some of the water, and the saturated air is fed to the bottom of the fermenter. For an air flow rate to the fermenter of 10 L/min at 35°C and 1 atm, estimate the rate at which steam must be added to the sterilized air and the rate (kg/min) at which condensate is collected upon cooling the air–steam mixture.
1 answer:
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Answer:
0.96kg/s
Explanation:
Hello! To solve this exercise we must use the first law of thermodynamics, which states that the sum of the energies that enter a system is the same amount that must go out. We must consider the following!
state 1 : is the first flow in the input of the chamber
h1=entalpy=335.02KJ/kg
m1=mass flow=0.56kg/s
state 2 : is the second flow in the input of the chamber
h2=entalpy=83.915KJ/kg
state 3:is the flow that comes out
h3=entalpy=175.90 kJ/kg
now use the continuity equation that states that the mass flow that enters is the same as the one that comes out
m1+m2=m3
now we use the first law of thermodynamics
m1h1+m2h2=m3h3
335.02m1+83.915m2=175.9m3
as the objective is to find the cold water mass flow(m2) we divide this equation by 175.9
1.9m1+0.477m2=m3
now we subtract the equations found in the equation of continuity and first law of thermodynamics
m1 + m2 = m3
-
1.9m1 + 0.477m2=m3
----------------------------------
-0.9m1+0.523m2=0
solving for m2
the mass flow rate of the cold-water is 0.96kg/s
Answer:
program:
#include<stdio.h>
#include<string.h>
int main()
{
char character;
int count =0;
int i;
while (1)
{
while((character=getchar())!='^z')
{
if (character=='\n')
{
printf("\/n");
}
else
printf("%c",character);
}
break;
}
}
Output:
Answer:
shear strength = 2682.31 Ib/ft^2
Explanation:
major principal stress = 100 Ib / in2
minor principal stress = 20 Ib/in2
Normal stress = 3000 Ib/ft2
<u>Determine the shear strength when direct shear test is performed </u>
To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a
for direct shear test
use Mohr Coulomb criteria relation between normal stress and shear stress
Shear strength when normal strength is 3000 Ib/ft = 2682.31 Ib/ft^2
attached below is the detailed solution
