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natali 33 [55]
4 years ago
9

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.
Physics
1 answer:
Alex777 [14]4 years ago
5 0

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

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