How to calculate "joint pressure tank"

[selamatg]

Dear All,

I'm looking formula to calculate "joint pressure tank" as picture shown in the attachment.

I try searching on the internet have not met the formula.

I Assumed as below:

Volume Tank A (VTA)
= 4 x 2 x 2 x 22 : 7
= 50.28 inch3

Volume Tank B (VTB)
= 4 x 1 x 1 x 22 : 7
= 12.57

Volume pressure tank A (VPA)
= 50.28 x 100 PSI
= 5,028

Volume pressure tank B (VPB)
= 12.57 x 50 PSI
= 628.6

To calculate joint pressure (not include pipe)
= (VPA+VPB) / (VTA+VTB)
= (5,028 + 628.6 ) / (50.28 + 12.57)
= 5,656.6 / 62.85
= 90 PSI

Please correct me if above calculation are wrong.

Best regards,

Selamatg

[selamatg]

Hi all,

Regarding the drawing, there is cylinder tank with 4" diameter x 4" Length (100 PSI) and 2" diameter x 4" lenghth (50 PSI), how many PSI if both tank joined.

Thanks in advance


Selamatg

[Cloxxki]

Hi all,

Regarding the drawing, there is cylinder tank with 4" diameter x 4" Length (100 PSI) and 2" diameter x 4" lenghth (50 PSI), how many PSI if both tank joined.

Thanks in advance


Selamatg

I'll throw in my idea first, based solely on intuition (no formal education on the subject, at least that I remember)

Since the cross-section of the cilinders is identical, we can scratch X and Y from the formula's, which-ever those may be. Makes it rather simple, I would guess.

100 psi x 4 = 400 units of stuff
50 psi x 2 = 100 units of stuff
6 volume units, total
500 units of stuff, total
500/6= 83.33 psi when joined.

If I got this wrong, I'll learn from the correct answer.
In school, I usually got away with my method of making it up as I went along. Even if the classes and text books taught the correct formulas, I'd have a perfectly wiped memory of that going into a test anyway. I would have worded it mote convincingly if it were a test, though. I'm rusty.

Thanks,

J

[sucahyo]

Vote to the first result. 90PS.

VTA*PA + VTB*PB = VTJ*PJ
PJ = (VTA*PA + VTB*PB) / VTJ
PJ = (VTA*PA + VTB*PB) / (VTA + VTB)

[Harvey]

You sent me searching for an easier way but it always boils down to using the Ideal Gas Law in one form or another.
pV = nRT

Absolute Pressure x Volume = number of moles x Gas Constant x Absolute Temperature

You could write this as n = pV / RT

n[tankA] = p[tankA] x V[tankA] / RT

n[tankB] = p[tankB] x V[tankB] / RT

n[joined] = n[tankA] + n[tankB]
V[joined] = V[tankA] + V[tankB]

p[joined] x V[joined] = n[joined]RT

p[joined] = n[joined]RT / V[joined]


Note the formula units are SI units. So you would need to convert your PSI to Pascals and Volume into Cubic Meters.

You may also find this site helpful: Fluid Concepts

1 PSI = 6.894×10³ Pascal
1 cubic inch = 16.387 064 × 10^−6 Cubic Meters

p[tankA] = 689400 Pascals
v[tankA] = 0.00082394157792 Cubic Meters
p[tankB] = 344700 Pascals
v[tankB] = 0.00020598539448 Cubic Meters
R = 8.314472
T = 300 Kelvin (Just to plug in)

n[tanka] = (689400 x 0.00082394157792) / 8.314472 x 300 = 0.2278 mole
n[tankb] = (344700 x 0.00020598539448) / 8.314472 x 300 = 0.0285 mole
n[joined] = 0.2563 mole
v[joined] = 0.0010299269724 m³
p[joined] = (0.2563 x 8.314472 x 300) / 0.0010299269724 = 620723.380 Pascals or 90.038 PSI

So it would seem that everyone has nearly the same result for 90 PSI