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

Answer:

a) $(dP)_{v} = 9.692 kPa$

b) $(dP)_{T} = -9.692 kPa$

c) dP = 0 Pa

Explanation:

The specifies equation is :

$dz = (\frac{\delta z}{\delta x}) _{y} dx + (\frac{\delta z}{\delta y}) _{x} dy$

Note that:

$dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV$

1% increase in temperature at specific volume:

$dT = \frac{0.01}{1} *350\\dT = 3.5 K$

a) Change in pressure of helium at constant volume:

$(dP)_{v} = \frac{R}{v} dT$

R = 2.0769 kJ/kg-K

dT = 3.5 K

v = 0.75 m³/kg

$(dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa$

b)

dv = (1%/100%) *0.75

dv = 0.0075 m³/kg

Change in pressure of helium at constant temperature:

$(dP)_{T} = \frac{-RT}{v^{2} } dv$

R = 2.0769 kJ/kg-K

T = 350 K

v = 0.75 m³/kg

dv = 0.0075 m³/kg

$(dP)_{T} = \frac{-(2.0769*350)}{0.75^{2} } *0.0075\\(dP)_{T} = -9.692 kPa$

c) The change in pressure of helium :

$dP = (dP)_{v} + (dP)_{T}$

dP = 9.692 - 9.692

dP = 0