diff options
Diffstat (limited to '3841/CH5')
| -rw-r--r-- | 3841/CH5/EX5.1/Ex5_1.sce | 13 | ||||
| -rw-r--r-- | 3841/CH5/EX5.2/Ex5_2.sce | 11 | ||||
| -rw-r--r-- | 3841/CH5/EX5.3/Ex5_3.sce | 15 | ||||
| -rw-r--r-- | 3841/CH5/EX5.4/Ex5_4.sce | 11 | ||||
| -rw-r--r-- | 3841/CH5/EX5.5/Ex5_5.sce | 15 | ||||
| -rw-r--r-- | 3841/CH5/EX5.6/Ex5_6.sce | 10 | ||||
| -rw-r--r-- | 3841/CH5/EX5.7/Ex5_7.sce | 12 | ||||
| -rw-r--r-- | 3841/CH5/EX5.8/Ex5_8.sce | 8 | ||||
| -rw-r--r-- | 3841/CH5/EX5.9/Ex5_9.sce | 11 |
9 files changed, 106 insertions, 0 deletions
diff --git a/3841/CH5/EX5.1/Ex5_1.sce b/3841/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..9655c7aa0 --- /dev/null +++ b/3841/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,13 @@ +clear +//given +//find the heat required +t2=185 +t1=95 +W1=42 +cp=0.5 +g=0.92 +w1=8.31*g +W=W1*w1 +Q=W*cp*(t2-t1) +printf("\n W") +printf("\n heat required is %.2f ", Q) diff --git a/3841/CH5/EX5.2/Ex5_2.sce b/3841/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..4e84e4078 --- /dev/null +++ b/3841/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,11 @@ +clear +//given +//find the back pressure and heat required +// +g=9. +w1=0.0361 +Bp=g*w1 +//as 1 psi =2.036 +Bp1=Bp*2.0326 +printf("\n \n back pressure %.3f ",Bp) +printf("\n \n heat required is %.2f ",Bp1) diff --git a/3841/CH5/EX5.3/Ex5_3.sce b/3841/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..a088f27db --- /dev/null +++ b/3841/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,15 @@ +clear +//given +//find the total weight of a given temperature and velocity +// +v=9500. +p=5. +t=180. +V=v/(12.**3) +//normal barometric pressure 14.7 +bP=p+14.7 +bP1=144.*bP +//640 is total temperature conveting into k +W=(bP1*V)/(53.3*640.) + +printf("\n \n total weight %.2f ",W) diff --git a/3841/CH5/EX5.4/Ex5_4.sce b/3841/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..2f81c3643 --- /dev/null +++ b/3841/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,11 @@ +clear +//given +// +//find the rate of flow outlet +t=600. +c=1200. +t2=400. +AT=t+460. +AT1=t2+460. +Rfo=c*(AT1/AT) +printf("\n \n Rate of flow outlet %.2f ",Rfo) diff --git a/3841/CH5/EX5.5/Ex5_5.sce b/3841/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..b189b0e06 --- /dev/null +++ b/3841/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,15 @@ +clear +//given +// +//find the final pressure gage and convert absloute temperature to normal temprature +a=210. +t=160. +t2=60. +//absloute temperature to convert is 460 +AT=160.+460. +AT1=60.+460. +IP=210.+14.7 +FP=IP*(520./620.) +printf("\n FP") +FPg=(FP-14.7) +printf("\n \n final pressue gage is %.2f ",FPg) diff --git a/3841/CH5/EX5.6/Ex5_6.sce b/3841/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..1bd22b8b4 --- /dev/null +++ b/3841/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,10 @@ +clear +//given +// +//find the final pressure gage +a=240. +b=15. +IP=14.7 +FP=IP*(a/b) +Fpg=FP-IP +printf("\n \n final pressure gage is %.2f ",Fpg) diff --git a/3841/CH5/EX5.7/Ex5_7.sce b/3841/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..e80a608a3 --- /dev/null +++ b/3841/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,12 @@ +clear +//given +//find out weight of hydrogen +// +//1lb hydrogen +8 lb oxygen gives +T=1.+8. +//this problem deals with 24 lb oxygen so +t=24./8. +//multiplying all equation by 3 +TT=3+24 + +printf("\n \n 3 lb of hydrogen requires %.2f lb",TT) diff --git a/3841/CH5/EX5.8/Ex5_8.sce b/3841/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..83892357f --- /dev/null +++ b/3841/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,8 @@ +clear +//given +o=0.14 +h=0.86 +O=120*o +H=120*h +O2=134.4+275.5 +printf("\n \n Total O2 uniting with oil %.2f ",O2) diff --git a/3841/CH5/EX5.9/Ex5_9.sce b/3841/CH5/EX5.9/Ex5_9.sce new file mode 100644 index 000000000..f5419fd5c --- /dev/null +++ b/3841/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,11 @@ +clear +// +//find the total weight and airfuel ratio +//given data +O2=409.9 +lb=0.231 +w=409.9 +W=w/lb +AFR=W/120. +printf("\n \n total weight %.2f ",W) +printf("\n \n air fuel ratio %.2f ",AFR) |