Answer:
a. T1 = 176.810k or -96.19⁰c
B. 0.05832kg
C. 93.92kj
D. -3.934kj
Step-by-step explanation:
These informations have been provided for tank
V1 = 0.07m³
M1 = 0.02kg
P1 = 105kpa
For pipe line:
P1 = 1000kpa
Tp = 50⁰c = 323k
P2 = 1000koa
T2 = 430k
If you check the table a-2(appendix)
Helium has
Cv = 3.1156kj/kgK
Cp = 5.1947kj/kgK
r = 1.667
R = 2.0785kj/kjk
A. Initial temperature
Using the ideal gas equation:
P1v1 = m1RT1
T1 is unknown
T1 = 105x0.07/2.0785x0.02
T1 = 176.810K
T1 = 176.810k
= -273+176.810
T1 = -96.19⁰C
B. Mass added to tank
M2 = mp+m1
Mp = m2-m1
Mp = (P1v1/RT2)-(p1v1/RT2)
= (1000*0.07)/(2.0785*430) - (105*0.07)/(2.0785*176.81)
Mp = (70/893.755)-(7.35/367.49958)
Mp = 0.07832-0.02
Mp = 0.05832kg
C. Change in internal energy:
U2-U1 = (m1+mp)cvT2-(m1CvT1)
∆U = (0.02+0.05832)(3.116)(430)-(0.02)(3.116)(176.81)
∆U = 93.92kj
D. Heat loss from tank
∆u = mphp + Q
hp = cpTP
93.92 = 0.05832(5.1947)(32.3)+ Q
Q = -3.934kj