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SOLUTIONS to Problem Set I - EEE 2010

# 1219 at 35 and this enthalpy we find temperature to

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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 = 1-x1; 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 H-x 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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