Question

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)

Answer 1

1. $\begin{aligned}ΔU &= - 316.5 kJ\\ m &= 2.268 kg\\ R &= 430 J/kg.K\\ k &= 1.35\\ T_i &= 204.4°C = 477.4 K\\ \\ a)\ W &= ∫Pdv = 0\ (constant\ volume\ process)\\ W &= 0 \end{aligned}$

$\begin{aligned} b) ΔU &= Q - W\\ Q &= ΔU\\ Q &= - 316.5\ kJ \ (heat\ transfer\ from \ system)\end{aligned}$

c)\$∆H_v = q\\$

$∆H = -316.5 = -316.5\ kJ$

d) \The change in entropy,

Specific heat at constant volume,

$k = Cp/Cv, \ Cv+R = Cp\\ k = Cv+R/Cv\\ k×Cv = Cv + R\\ Cv(k - 1) = R\\ Cv = R / k-1\\ Cv = 430/0.35\\ Cv =1228.57 J/kg.K\\ Cv =1.22857 J/kg.K$

Final temperature,

$ΔU = m×Cv×(Tf-Ti)\\ T_f-T_i = ΔU/m×Cv\\ T_f = ΔU/m×Cv + T_i\\ T_f = -316.5/2.268×1.22857 + 204.4\\ T_f = 90.812°C = 363.812 K$

Entropy change,

$ΔS = 2.268×1.22857×\ln(363.812/477.4)\\ ΔS = -0.75711 kJ/K$

3.

$V_t= 14 × 10 ×3.5= 490\ m³\\ V_p = 0.2m³ × 50 = 10\ m³\\ V_a \textsf{ (Volume of air)} = 490 - 10 = 480\ m³$

$\begin{aligned}P= 50 × 150kCal/hr ×4.187kJ/kCal = 31402.5kJ/hr \end{aligned}$

$P_i = 101.3kPa\\ T_i = 16°C = 273 + 16 = 289K$

$T_f = ?\\ t = 10 ×60 = 600\ s$

$Q = Pt = 31402.5 × 600 = 18.84\ MJ$

Heat released after 10 minutes is therefore 18.84 MJ.

$PV = nRT$

$101300 × 480 = n × 8.314 ×289$

n = 20236.8 moles of air

but, 1 mole of air = 29

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

$mc_a ∆T = Q\\ 586000 × 0.718 × ( T_2 - 16) = 18.84\ MJ$

$T_2 = 60.78°C$