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Unformatted text preview: 2 = 108 BTU/lbm (pure H2O)
H1 = 20 BTU/lbm; (pure H2SO4)
HE = H – (x1⋅H1 + x2⋅H2); HE = −133 BTU/lbm Ans.
k. Solution to problem 12.47(SVNA) Mix m1 lbm NaOH with m2 lbm 10% soln. @ 68 °F.
BASIS: m2 = 1 lbm ;
x3 = 0.35;
x2 = 0.1 m1 = 1 lbm (guess);
m3 = m1 + m2
Given m1 + m2 = m3; m1 + x2m2 = x3m3
Find m1, m3 m1 = 0.385 lbm;
m3 = 1.385 lbm
From Example 12.8 and Fig. 12.19
H1 = 478.7 BTU/lbm;
H2 = 43 BTU/lbm
H3 = (m1H1 + m2H2) / m3;
H3 = 164 BTU/lbm
From fig. 12.19 at 35% and this enthalpy, we find temperature to be about 205 ° F
l. Solution to problem 12.52(SVNA) Initial solution (1) at 80 °F; Fig. 12.19:
m1 = 1 lbm; x1 = 0.40;
H1 = 77 BTU/lbm
Saturated steam at 35(psia); Table F.4:
m2 = (x1.m1/x3) – m1
H2=1161.1 BTU/lbm; x3=0.38;
m3 = m1+m2;
m3 = 1.053lbm;
m2 = 0.053lbm
H3 = (m1H1 + m2H2) / m3
H3 = 131.2 BTU /lbm
We see from Fig. 12.19 that for this enthalpy at 38% the temperature is about 155
° F.
m. Solution to problem 12.53 (SVNA) Read values for H, H1, & H2 from Fig. 12.17 at 100 °F:
H = 56 BTU/lbm;
H1 = 8 BTU/lbm;
H2 = 68 BTU/lbm
x1 = 0.35;
x2 = 1x1;
ΔH = H x1⋅H1 − x2⋅H2
ΔH = 103 BTU/lbm
n. Solution to problem 12.57(SVNA) Graphical solution: If the mixing is adiabatic and water is added to bring the
temperature to 140°F, then the point on the Hx diagram of Fig. 12.17
representing the final solution is the intersection of the 140°F isotherm with a
straight line between points representing the 75 wt % solution at 140°F and pure
water at 40°F. This intersection gives x3, the wt % of the final solution at 140 °F:
x3 = 42%;
m1 = 1 lb
By a mass balance:
x3 = (0.75 m1) / (m1 + m2);
m2 = (0.75 m1/x3) – m1
m2 = 0.786lbm EQUATIONS OF STATE
A. GASES
a. Solution
Tank = 1m3; P = 19.5bar = 1.95 MPa
T = 225 °C. Steam table reports values at 2 and 1.8 MPa, therefore using interpolation
v 195MPa = 0.10701m3/kg
mass = V/v = 1/0.10701 = 9.345 kg
Z = PV/RT = Vactual/ Videal = 0.909
Water or steam is not an ideal gas
Using IG EOS v = (R/M). T/P
R/M = 461.5 J/kg.K;
v = 0.117732m3/kg using I.G. Law
Therefore, mass = 1/0.117732 = 8.4938 kg
The answer is less by ~ 1kg ! Not Good !
b. Pressure cooker Design
IMP: Tsat increases with P, therefore, we can cook food quicker at higher altitudes
7 kg H2O, T=250°F;
Psat = 29.82psi; vl = 0.017001 ft3/lbm; vg = 13.826 ft3/lbm
Volume of liquid water, Vwater = 0.26236 ft3
Volume of the cooker = 2×Vwater = 906.72in3
Since the diameter of the stove top heating element is 8in. let’s take the base of the
cooker to be 10 in.
Therefore, height = Volume/base area = 11.54in.
Shape of cooker: 11.54in. high and 10in. diameter (Acceptable shape!)
c. Methyl Chloride
B = 245.5 cm3/mol;
C = 25.20cm6/mol
Assuming ideal gas W = RT ln(P1/P2) = 12.432 kJ/mol
Z = 1+ B/V = PV/RT
Or (PV2/RT) – V – B = 0; Quadratic in V
Reversible Process, therefore W =
So we need to find V1 and V2
2 solutions of the quadratic; V1 = 30713.44cm3/mol and V2 = 310.24cm3/mol
Since ideal gas law gives approximately V1ig = 31023...
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 Spring '08
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