Question

Consider helium at 350 K and 0.75 m3/kg. Using Eq. 12-3, determine the change in pressure corresponding to an increase of (a) 1 percent in temperature at constant specific volume, (b) 1 percent in specific volume at constant temperature, and (c) 1 percent in both the temperature and specific volume.

Answer 1

a)
`(dP)_(v) = 9.692 kPa`

b)
`(dP)_(T) = -9.692 kPa`

c) dP = 0 Pa

Step-by-step explanation:

The specifies equation is :

`dz = ((\delta z)/(\delta x)) _(y) dx + ((\delta z)/(\delta y)) _(x) dy`

Note that:

`dP = (R)/(v) dT - (RT)/(v^(2) ) dV`

1% increase in temperature at specific volume:

`dT = (0.01)/(1) *350\\dT = 3.5 K`

a) Change in pressure of helium at constant volume:

`(dP)_(v) = (R)/(v) dT`

R = 2.0769 kJ/kg-K

dT = 3.5 K

v = 0.75 m³/kg

`(dP)_(v) = (2.0769)/(0.75) * 3.5\\(dP)_(v) = 9.692 kPa`

b)

dv = (1%/100%) *0.75

dv = 0.0075 m³/kg

Change in pressure of helium at constant temperature:

`(dP)_(T) = (-RT)/(v^(2) ) dv`

R = 2.0769 kJ/kg-K

T = 350 K

v = 0.75 m³/kg

dv = 0.0075 m³/kg

`(dP)_(T) = (-(2.0769*350))/(0.75^(2) ) *0.0075\\(dP)_(T) = -9.692 kPa`

c) The change in pressure of helium :

`dP = (dP)_(v) + (dP)_(T)`

dP = 9.692 - 9.692

dP = 0