Thermodynamics Chapter 3 Solution Sta Maria

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1. An automobile tire is inflated to 32psig pressure at 50 degree F. After being driven, thetemperature rises to 75 degree F. Determine the final gage pressure assuming the volumeremains constant. (Electrical Engineering Board Exam Problem) Solution: Since volume is constant, use Charles’ Law on constant volume processes. P1/T1=P2/T2 Note: P and T should be absolute. (32+14.7)/ (50+460) =P2/ (75+460) P2=48.99 psia Converting back to gage pressure, P2g=48.99-14.7 P2g=34.29 psig

2. If 100 ft^3 of atmospheric air at zero Farenheit temperature are compressed to a volume of 1 ft^3 at a temperature of 200 deg Farenheit, what will be the pressure of the air in psi?

answer: 2109 psia 1V1/T1 = P2V2/T2 since given yung mga ffg: P1 = 14.7 psia since atmospheric pressure of air ung cnbi V1 = 100 ft^3 T1 = 0 + 460 = 460 kelvin para maging absolute temperature V2 = 1 ft^3 T2 = 200 + 460 = 660 kelvin absolute temp. na yan then,

P2 = (P1V1T2)/(T1V2) P2 = (14.7 psia x100 ft^3 x 660 k)/(460 k x 1 ft^3) P2 = 2109.13 psia 3.

Here is the method. Convert temperatures (T1=85degF and T2=70degF) to centigrade, then add 273 to convert to kelvin. Use the kelvin temperatures throughout. Since PV = nRT PV/(nT) = R Providing we are consistent with units and have T in kelvin: P1V1/(n1T1) = P2V2/(n2T2) where P is pressure in psia V is volume in ft^3 n is weight (proportional to mass and moles) in lb T is in kelvin The volume is fixed - it remains 10ft^3 throughout. For the discharge process: P1V1/(n1T1) = P2V2/(n2T2) 500x10/(25xT1) = 300x10/(n2xT2) 5/(25xT1) = 3/(n2xT2) This allows the weight of gas, n2, to be found. For the final stage: P1V1/(n1T1) = P2V2/(n2T2) 300x10/(n2xT2) = P2x10/(n2xT1) 300/T2 = P2/T1 This allowed P2 to be found. 3.

Four hundred cubic centimeters of gas at a pressure of 740 mmHg absolute andtemperature of 18 degree C undergoes a process until the

pressure and temperaturebecomes 760 mmHg absolute and 0 degrees C respectively. Calculate the final volume of the gas. (Electrical Engineering Board Exam Problem) Since all the parameters (temperature, pressure and volume) changes, use combined gas law PV/T=constant P1V1/T1=P2V2/T2 (740)(400)/(18+273)=(760)V2/(0+273) V2=365.3 cc

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6 6 m^3 tank contains helium at 400K and is evacuated from atmospheric pressure to a pressure of 740 mm Hg vacuum. Determine (A) mass of helium remaining in the tank, (

mass of helium pump out, � the temperature of the

helium falls to 10 degrees Centigrade. What is the pressure in kPa? Answer: A 0.01925 kg B 0.7123 kg C 1.886 kPa mass 1 = P1V1/R1T1 P(abs) = P(atm) - P(vac) = 20 mm Hg = 2666 pascal R= 2077.67 pa. m^3/kg. K mass of helium remaining in the tank original mass of helium: m= PV/RT = 101325.024*6/2077.67*400 = .7315 kg A. m1= P(abs) * V / R * T = 2666 * 6 / 2077.67* 400 = .01925 kg Mass of helium pump out: B. m2 = mass original - m1 = .71237 kg C. R and volume is still the same after the temperature is changed. T3 = 10 C + 273 = 283 K

p3 p3 p3 p3

= = = =

nRT3/V .01925 kg x (2077.67 pa. m^3/kg. K) x (283 K) / (6m^3) 1886.43779 Pa 1.886 kPa

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Lift Load = Buoyancy Force = (Weight of air inside balloon if it were filled with outside dry air) - (Weight of actual gas inside balloon) Weight of gas inside balloon = (density of gas) (acceleration of gravity at earth's surface) (balloon volume)

Density of gas = (absolute pressure)(Molecular Weight of Gas) / [(Specific Gas Constant)(absolute temperature)] Balloon Volume = (4/3) π (balloon radius)^3 Use wikipedia to look up the constants and conversion factors. Let me know if you have any problems. Chip 9. 10. A gas initially at 15 psia and 2 ft³ undergoes a process to 90 psia and 0.60 ft³, during which the enthalpy increases by 15.5 Btu; Cv = 2.44 Btu/ lb . °R. Determine (a) Change in U; (b) Cp and (c) R Given: P1 = 15 psia V1 = 2 ft³ Cv = 2.44 Btu/ lb . °R P2 = 90 psia V2 = 0.60 ft³ Change in H = 15.5 Btu Solution: a. Change in H = Change in U + P2V2 – P1V1 15.5 Btu =Change in U +[ { 90 lb/in² x 0.60 ft³ - 15 lb/in² x 2 ft³}{ 12² in²/1² ft² x 1 Btu/778 lbf – ft}] Change in U = 11.05 Btu b. k = Change in H/Change in U = 15.5/11.05 = 1.40 Cp = kCv = 1.40 ( 2.44 Btu/lb . °R) = 3.42 Btu/lb . °R c. Cp = Cv + R R = Cp – Cv = 3.42 – 2.44 = 0.98 Btu/lb . °R x (778 lbf – ft/1 Btu) = 762.4 lbf – ft/lbm - °R

11. For a certain gas R=0.277 kJ/ kg.K and k=1.384 (a) What are the value of Cp and Cv ? (b) What mass of this gas would occupy a volume of 0.425 cu m at 517.11 kPa and 26.7degree Celsius (c) If 31.65 kJ are transferred to this gas at constant volume in (b) , what are the resulting temperature and pressure? Part (a): R = Cp - Cv and k = Cp/Cv You just have to solve two equations and two unknowns: 0.277 = Cp - Cv 1.384 = Cp/Cv => Cp = 1.384*Cv 0.277 = 1.384Cv - Cv = 0.384Cv Cv = 0.277/0.384 = 0.7214 = 0.721 kJ/kgK Cp = 1.384*0.7214 = 0.9984 = 0.998 kJ/kgK Part (b): Use ideal gas law PV =mRT m = PV/RT = (517.11 kPa)*(0.425 m^3) / (0.277 kJ/kgK)*(273.15+26.7) = 2.646 kg = 2.65 kg Hint: T must be in Kelvins. Take the centigrade temperature and add 273.15 to it. Part (c) Q =m*Cv*deltaT = 31.65 kJ = (2.646 kg)*(0.7214 kJ/kgK)*(delta T) delta T = 16.58 degrees.

This means the temperature increased by 16.58 degrees. T(final) = 26.7 degrees C + 16.58 degrees = 43.28 degrees C = 43.3 degrees C (When you are dealing with deltaTs, the difference is the same for Kelvin or Centigrade.) For the final pressure, use the Ideal Gas Law Again P =mRT/V = (2.646 kg)*(0.277 kJ/kgK)*(273.15+43.28 degrees)/(0.425 m^3) = 545.7 = 546 kPA