A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.
Answers
Answer:
V₂=1.76 m³
P=222.03 KPa
Explanation:
Given that
For tank 1
V₁=1 m³
T₁= 10°C = 283 K
P₁=350 KPa
For tank 2
m₂=3 kg
T₂=35°C = 308 K
P₂=150 KPa
We know that for air
P V = m R T
P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass
for tank 2
P₂ V₂ = m₂ R T₂
By putting the values
150 x V₂ = 3 x 0.287 x 308
V₂=1.76 m³
Final mass = m₁+m₂
m =m₁+m₂
The final volume V= V₂+V₁
V= 1.76 + 1 m³
V= 2.76 m³
The final temperature T= 19.5°C
T= 292.5 K
m =m₁+m₂
m =4.3 + 3 = 7.3 kg
Now at final state
P V = m R T
P x 2.76 = 7.3 x 0.287 x 292.5
P=222.03 KPa
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Sorry man i dont know the answer to this one
Answers
Answer: Tl = - 13.3°C
the lowest outdoor temperature is - 13.3°C
Explanation:
Given that;
Temperature of Th = 21°C = 21 + 273 = 294 K
the rate at which heat lost is Qh = 5400 kJ/h°C
the power input to heat pump Wnet = 6 kw
The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;
COPhp = Th/(Th - Tl)
COPhp = Qh/Wnet
Qh/Wnet = Th/(Th -Tl)
the amount of heat loss is expressed as
Qh = 5400/3600(294 - Tl)
the temperature of sink
( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)
now solving the equation
Tl = 259.7 - 273
Tl = - 13.3°C
so the lowest outdoor temperature is - 13.3°C
Answers
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
//declare arrays
string ItemName[100];
double ItemCost[100]={0.0};
//declare variable
string name="";
double Total=0.0;
int NumItems=0;
cout<<"Enter number of grocery item you will be entering";
cin>>NumItems
cout<<"\n";
if (NumItems>100)
{
cout<<"Enter grocery items less than 100"
cin>>NumItems
cout<<"\n";
}
for (int i=1;i>=NumItems;i++)
{
cout<<"Please enter the item name in one word only, Example: icecream\n";
cin>>ItemName[i]
cout<<"Please enter the cost as a decimal number, Example: 2.05\n\n";
cin>>ItemCost[i];
}
cout<<"Items"<<" "<<""$""<<"Cost"<<endl
for (int i=1;i>=NumItems;i++)
{
cout<<ItemName[i]<<" "<<"$"<<ItemCost[i]<<endl
}
for (int i=1;i>=NumItems;i++)
{
Total=Total+ItemCost[i];
}
cout<<"Total:$"<<Total<<endl;
system("PAUSE");
return 0;
}
Answers
Explanation:
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Answers
Answer:
Explanation:
mole fraction of propane after passing through the separator is
mole fractions of ethane and methane
after passing through separator are:
Change in partial pressures then can be written as:
where
and
are mole fractions after and before passing through the separator
Hence,