Thermodynamics - Mass on Vertical Piston

[Sirsh]

I keep getting this question wrong, and I am struggling to understand why this is.

Homework Statement

Homework Equations

F = p*A
Ap = π*r²

The Attempt at a Solution

Denoting p_atm as atmospheric force, and p₀ as pressure inside the cylinder.

Firstly, a force balance on the system. Force UP = Force DOWN.
p₀*Ap = m*g + p_atm*Ap

p_atm = 101.4 kPa , and at 104.4°C p₀ is equal to a pressure between 110 kPa (at 102.3°C) and 120 kPa (at 104.8°C).

By interpolation p0 is equal to 118.4 kPa.

Rearranging the force balance, m = ((p0-patm)*Ap)/g = ((118.4-101.4)*(π*(0.042)²)/9.81 = 0.00960 kg.

This seems like a VERY unrealistic situation considering this very small mass is creating 17 kN/m² of force..

Any help would be appreciated.

 

[Sirsh]

I think I realized my mistake, I need to times my mass by 1000 due to using kN/m² = kPa I should be finding mass with N/m².

Edit still wrong...

 

[Chestermiller]

The area of a circle is πD²/4, not πD².

Chet

 

[Sirsh]

The area of a circle is πD²/4, not πD².

Chet

Sorry Chester, I changed the 84mm diameter into 0.042m radius when I put it into the formula.

It seems my working is correct, except the quiz I'm doing is playing up. Which is really confident battering :/

 

[Chestermiller]

Sorry Chester, I changed the 84mm diameter into 0.042m radius when I put it into the formula.

It seems my working is correct, except the quiz I'm doing is playing up. Which is really confident battering :/

Sorry, my mistake. I rechecked your calculation, and it looks correct. But the requirement of 3 significant figures is an issue. If you interpolate in your pressure table, you get 118.4. But if I interpolate in my pressure table, I get 117.9; and if I interpolate in my temperature table, I get 118.0. (I did these interpolations using semi-log interpolation). So the pressure difference can be anywhere from 17 to 16.5.

Chet

 

[Sirsh]

Sorry, my mistake. I rechecked your calculation, and it looks correct. But the requirement of 3 significant figures is an issue. If you interpolate in your pressure table, you get 118.4. But if I interpolate in my pressure table, I get 117.9; and if I interpolate in my temperature table, I get 118.0. (I did these interpolations using semi-log interpolation). So the pressure difference can be anywhere from 17 to 16.5.

Chet

As always, appreciate your help Chester. In the end it worked out to be correct!