Answer to Question #173960 in Mechanical Engineering for Randal Rodriguez

Question #173960

Note: Draw the P-V, T-S diagrams for all problems

1. A reversible nonflow constant volume process increases the internal energy by 316.5 kJ for 2.268 kg m of a gas for which R = 430 J/kg m -K and k = 1.35. For the process, determine a) Work, b) heat

transferred c) change in enthalpy d) change in entropy if the initial temperature is 204.4 o C..

2. A certain ideal gas has a constant R = 38.9 ft-lf f /lb m -R with k = 1.4. a) If 3 lb m of this gas undergoes a reversible nonflow constant volume process from 120 psia,740 o F to 140 o F find P 2 , ∆U, ∆H, Q, Work. b) If the process in “a” had been steady flow find work and ∆S.

3. A group of 50 persons attended a secret meeting in a room 14 m x 10 m x 3.5 m high. The room is

completely sealed off and insulated. Each person gives off 150 kCal/hr of heat and occupies a volume

of 0.2 m 3 . The room has an initial pressure 101.3 kPaa and temperature of 16 o C. Calculate the room

temperature after 10 min. NOTE: (4.187 kJ = 1 kCal)



1
Expert's answer
2021-03-23T08:02:13-0400

1. "\\begin{aligned}\u0394U &= - 316.5 kJ\\\\\nm &= 2.268 kg\\\\\nR &= 430 J\/kg.K\\\\\nk &= 1.35\\\\\nT_i &= 204.4\u00b0C = 477.4 K\\\\\n\\\\\na)\\ W &= \u222bPdv = 0\\ (constant\\ volume\\ process)\\\\\nW &= 0\n\\end{aligned}"


"\\begin{aligned} b)\n\u0394U &= Q - W\\\\\nQ &= \u0394U\\\\\nQ &= - 316.5\\ kJ \\ (heat\\ transfer\\ from \\ system)\\end{aligned}"


"c)\\""\u2206H_v = q\\\\"

"\u2206H = -316.5 = -316.5\\ kJ"


"d) \\"The change in entropy,

Specific heat at constant volume,

"k = Cp\/Cv, \\ Cv+R = Cp\\\\\nk = Cv+R\/Cv\\\\\nk\u00d7Cv = Cv + R\\\\\nCv(k - 1) = R\\\\\nCv = R \/ k-1\\\\\nCv = 430\/0.35\\\\\nCv =1228.57 J\/kg.K\\\\\nCv =1.22857 J\/kg.K"


Final temperature,

"\u0394U = m\u00d7Cv\u00d7(Tf-Ti)\\\\\nT_f-T_i = \u0394U\/m\u00d7Cv\\\\\nT_f = \u0394U\/m\u00d7Cv + T_i\\\\\nT_f = -316.5\/2.268\u00d71.22857 + 204.4\\\\\nT_f = 90.812\u00b0C = 363.812 K"


Entropy change,

"\u0394S = 2.268\u00d71.22857\u00d7\\ln(363.812\/477.4)\\\\\n\u0394S = -0.75711 kJ\/K"



3.

"V_t= 14 \u00d7 10 \u00d73.5= 490\\ m\u00b3\\\\\nV_p = 0.2m\u00b3 \u00d7 50 = 10\\ m\u00b3\\\\\nV_a \\textsf{ (Volume of air)} = 490 - 10 = 480\\ m\u00b3"

"\\begin{aligned}P= 50 \u00d7 150kCal\/hr \u00d74.187kJ\/kCal = 31402.5kJ\/hr \\end{aligned}"

"P_i = 101.3kPa\\\\\nT_i = 16\u00b0C = 273 + 16 = 289K"

"T_f = ?\\\\\nt = 10 \u00d760 = 600\\ s"


"Q = Pt = 31402.5 \u00d7 600 = 18.84\\ MJ"

Heat released after 10 minutes is therefore 18.84 MJ.


"PV = nRT"

"101300 \u00d7 480 = n \u00d7 8.314 \u00d7289"

n = 20236.8 moles of air

but, 1 mole of air = 29

"\\therefore" the mass of air in the room = 586kg


"mc_a \u2206T = Q\\\\\n586000 \u00d7 0.718 \u00d7 ( T_2 - 16) = 18.84\\ MJ"

"T_2 = 60.78\u00b0C"

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