Determine the percent error in using the ideal gas model to determine the specific volume of (a) water vapor at 4000 lbf/in. 2, 10008F. (b) water vapor at 5 lbf/in. 2, 2508F. (c) ammonia at 40 lbf/in. 2, 608F. (d) air at 1 atm, 5608R. (e) Refrigerant 134a at 300 lbf/in. 2, 1808F.

Question

Engineering

Jefferson Barnes

27 October, 03:49

Determine the percent error in using the ideal gas model to determine the specific volume of (a) water vapor at 4000 lbf/in. 2, 10008F. (b) water vapor at 5 lbf/in. 2, 2508F. (c) ammonia at 40 lbf/in. 2, 608F. (d) air at 1 atm, 5608R. (e) Refrigerant 134a at 300 lbf/in. 2, 1808F.

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Ducky · Answer 1

a) 24.01%

b) 0.35%

c) 3.43%

d) 0%

e) 37.31%

Explanation:

A) water vapor=> H2O

From table @4000 lbf/in^2, 1000°F v = 0.1752 ft^3/ib

m = 18.02 ib/ibmol

P = 4000 lbf/in^2

From the ideal gas model

PV/m = mRT/m

V = RT/Pm

Where R = 1945 (ft. ibf) / (ibmol.°R)

T = 1000°F = 1459.67°R

Therefore

V = (1945 * 1459.67) : (4000 * 18.02) = 2255190.19 : 72080 = 0.21727 ft^3/ib

Error = (V - v) / v * 100% = (0.21727 - 0.1752) : 0.1752 = 0.2401 * 100% = 24.01%

B) water vapor=> H2O

From table @ 5 lbf/in^2, 250°F v = 84.21 ft^3/ib

m = 18.02 ib/ibmol

P = 5 lbf/in^2

From the ideal gas model

PV/m = mRT/m

V = RT/Pm

Where R = 1945 (ft. ibf) / (ibmol.°R)

T = 250°F = 709.67°R

Therefore

V = (1945 * 709.67) : (5 * 18.02) = 84.51 ft^3/ib

Error = (V - v) / v * 100% = (84.51 - 84.21) : 84.21 = 0.003538 * 100% = 0.35%

C) ammonia = >

From table @40 lbf/in^2, 60°F v = 7.9134 ft^3/ib

m = 17.03 ib/ibmol

P = 40 lbf/in^2

From the ideal gas model

PV/m = mRT/m

V = RT/Pm

Where R = 1945 (ft. ibf) / (ibmol.°R)

T = 60°F = 519.67°R

Therefore

V = (1945 * 519.67) : (40 * 17.03) = 8.189 ft^3/ib

Error = (V - v) / v * 100% = (8.189 - 7.9134) : 7.9134 = 0.03432 * 100% = 3.43%

D) Air at 1 atm = >

No table so we assume air is ideal

m = 28.97 ib/ibmol

P = 14.71 lbf/in^2

From the ideal gas model

PV/m = mRT/m

V = RT/Pm

Where R = 1945 (ft. ibf) / (ibmol.°R)

T = 560°R

Therefore

V = (1945 * 560) : (14.71 * 28.97) = 14.099 ft^3/ib

Since Air is ideal error is zero = 0%

E) refrigerant = >

From table @300 lbf/in^2, 180°F v = 0.1633 ft^3/ib

m = 102.03 ib/ibmol

P = 300 lbf/in^2

From the ideal gas model

PV/m = mRT/m

V = RT/Pm

Where R = 1945 (ft. ibf) / (ibmol.°R)

T = 180°F = 639.67°R

Therefore

V = (1945 * 639.67) : (300 * 102.03) = 0.22422 ft^3/ib

Error = (V - v) / v * 100% = (0.22422 - 0.1633) : 0.1633 = 0.37306 * 100% = 37.31%