diff options
Diffstat (limited to '2417/CH6/EX6.41/Ex6_41.sce')
-rwxr-xr-x | 2417/CH6/EX6.41/Ex6_41.sce | 23 |
1 files changed, 23 insertions, 0 deletions
diff --git a/2417/CH6/EX6.41/Ex6_41.sce b/2417/CH6/EX6.41/Ex6_41.sce new file mode 100755 index 000000000..0b89e2b0a --- /dev/null +++ b/2417/CH6/EX6.41/Ex6_41.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.41\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.41 (page no. 304)
+// Solution
+
+//For Methane(CH4,MW=16)
+p=500; //evaluate specific volume at p pressure //Unit:psia
+pc=674; //critical temperature //Unit:psia
+T=50+460; //evaluate specific volume at T temperature //Unit:R
+Tc=343; //critical temperature //Unit:R
+R=1545/16; //gas constant R = 1545/Molecular Weight //ft*lbf/lbm*R
+pr=p/pc; //reduced pressure //unit:psia
+Tr=T/Tc; //reduced temperature //unit:R
+//Reading figure 6.28 at these values gives
+Z=0.93; //compressibility factor
+//Z=(p*v)/(R*T)
+v=Z*((R*T)/(p*144)); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+printf("Using the value of Z=0.93,the specific volume is %f ft^3/lbm\n",v);
+//For ideal gas,
+v=(R*T)/(p*144); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+printf("For the ideal gas,the specific volume is %f ft^3/lbm\n",v);
|