A tank with it's insulation not fully in tact has a volume of 0.07 cubic meters. Initially the tank contains 0.02 kg of helium at 105 kPa. The tank is filled with helium from a supply line at 1000 kPa and 50 Celsius. The filling process ends when the pressure of helium in the tank reaches the supply line pressure, at which time the temperature of the helium in the tank is 430 K. Determine the following: a. Initial temperature of the helium in the tank.
b. The amount of mass added to the tank.
c. The change in energy inside the tank.
d. The heat loss from the tank a.

a. T1 = 176.810k or -96.19⁰c

B. 0.05832kg

C. 93.92kj

D. -3.934kj

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

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