Answer:
Heat loss=85.9W/m^2
ΔT1(Steel)=0.04C
ΔT2(Brick1)=110.13C
ΔT3(Mwood)=343.6C
ΔT1(Brick2)=154.18C
Explanation:
raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.
To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.
I attached the procedure
Answer:
2074.2 KW
Explanation:
<u>Determine power developed at steady state </u>
First step : Determine mass flow rate ( m )
m / Mmax = ( AV )₁ P₁ / RT₁ -------------------- ( 1 )
<em> where : ( AV )₁ = 8.2 kg/s, P₁ = 0.35 * 10^6 N/m^2, R = 8.314 N.M / kmol , </em>
<em> T₁ = 720 K . </em>
insert values into equation 1
m = 0.1871 kmol/s ( mix )
Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )
W( power developed at steady state )
W = m [ Yco2 ( h1 - h2 )co2
Attached below is the remaining part of the detailed solution
Answer:
okay it is faster and easier
Answer:
From the question, we have two variables
1. userNum1
2. userNum2
And we are to print "userNum1 is negative" if userNum1 is less than 0.
Then Assign userNum2 with 2 if userNum2 is greater than 10.
Otherwise, print "userNum2 is less or equal 10.".
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include<iostream>
using namespace std;
int main ()
{
// Declare variables
int userNum1, userNum2;
// Accept input for these variables
cin>>userNum1, userNum2;
// Condition 1
if(userNum1 < 0)
{
cout<<"userNum1 is negative"<<'\n';
}
// Condition 2
if(userNum2 > 10)
{
userNum2 = 2;
}
// If condition is less than 10
else
{
cout<<"userNum2 is less or equal to 10"<<\n;
}
return 0;
}
// End of Program.
Answer:
23.34 seconds
Explanation:
Flow rate = 1500
Arrival = 800 vehicle per hour
Cycle c = 60 seconds
Dissipation time = 10 seconds
Arrival time = 800/3600 = 0.2222
Rate of departure = 1500/3600 = 0.4167
Traffic density p = 0.2222/0.4167 = 0.5332
Real time = r
r + to + 10 = c
to = c-r-10 ----1
t0 = p*r/1-p ----2
Equate both 1 and 2
C-r-10 = p*r/1-p
60-r-10 = 0.5332r/1-0.5332
50-r = 0.5332r/0.4668
50-r = 1.1422r
50 = 1.1422r + r
50 = 2.1422r
r = 50/2.1422
r = 23.34 seconds