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MALAYAN COLLEGES LAGUNA

1PROCESSES OF IDEAL GASESISOMETRIC (Isochoric) PROCESS - Constant Volume ProcessIsometric process is a reversible constant volume process. P T 2 T2 2 V=C P2 1 1 T1 P1 Q V S1 S2 S V1 = V2

a.) Relationship between P and T.T1/T2 = P1/P2b.) Non-flow work.Wn =

2c.) The change of internal energyU = mcv(T2 T1) d.) The heat transferred Q = mcv(T2 T1) e.) The change of enthalpy H = mcp(T2 - T1) f.) Irreversible non-flow constant volume processQ = U + Wn where: For reversible non-flow, Wn = 0 For irreversible nonflow, Wn 0 Wn = non-flow workPROBLEMSA reversible, non-flow, constant volume process decreases the internal energy by 316.5kJ for 2.268 kg of a gas for which R=430J/kg-K and k=1.35. For the process, determine a) the work, b) the heat, and c) the change of enthalpy if the initial temperature is 204.4oC.Twenty kilojoules of heat is added at constant volume to 2.5 kg ideal gas (k=1.25, R=320 J/kg-K). The initial temperature of the gas is 32C. Compute for a) Wnf, b) T2, c) the change of internal energy, d.) the change of enthalpy. (0, 311K, 20KJ, 25KJ)3. A gas whose composition is not known has 42.2kJ of paddle work input at constant volume of 566L. Initially, P1=138kPa, t1=26.7oC, finally t2=82.2oC, What are (a.) the change of internal energy, and (b.) the heat transferred if k=1.21?4.The mass of a gas is 1.8 kg, for which R = 0.38 kJ/kg.K and k = 1.23 undergoes a constant volume process where P1 = 650kPa and t1 = 70oC. The final pressure is P2 = 1950 kPa. The process resulted to additional 111 kJ of heat when the gas is stirred internally. Calculate (a )final temperature, T2, (b) the work, and (c) the change of internal energy.

ISOBARIC PROCESS Constant Pressure ProcessIsobaric process is a reversible constant pressure process of a substance. P 1 2 T 2 T2 P = C P 1 T1 Q V1 V2 V S1 S2 S a) Relationship between V and T T1/T2 = V1/V2 b) Nonflow work Wn = = P(V2 V1) = mR(T2 T1) c) Change of internal energy U = mcv (T2 T1)

d) Heat transferred Q = mcp (T2 T1) e) Change of enthalpy H = mcp (T2 T1)PROBLEMS:A certain gas with R = 320 J/kg-K and k = 1.3 undergoes a constant pressure process where the initial temperature is 30oC. If 150 kJ are added to 4 kg of this gas, determine; a) T2, b) change of enthalpy, c) change of internal energy, d) work for non-flow process.An ideal gas (R=2,075 J/kg-K & k=1.659) goes through a reversible constant pressure process where 525KJ are added to 2.3kg of the gas. The initial temperature of the gas is 310K. Compute for a) T2, b) Wnf, c) U, d) H (353.7K, 208.7KJ)3. If 120 kJ are added to 3.0 kg of nitrogen gas at constant pressure when the initial temperature is 32oC, find (a) T2, (b) H, (c) U, and (d) work for a non-flow process. (R=0.2969 KJ/kg-K, k=1.399)4.Hydrogen gas expands from 5 cu ft and 80oF to 15 cu ft while the pressure remains constant at 15.5 psia. Compute (a) T2, (b) H, (c) U and (d) S. (e) For a reversible non-flow process, what is the work?

ISOTHERMAL PROCESS Constant Temp. ProcessAn isothermal process is reversible constant temperature process of asubstance. T P 1 1 2 T PV= C 2 S1 S2 S V1 V2 V (a.) Relationship between P and V P1V1 = P2V2 (b.) Nonflow work Wn = P1V1lnV2/V1 = mRT1lnV2/V1 (c.) The change of internal energy U = OQ

(d.) The heat transferred Q = Wn = P1V1lnV2/V1 = mRT1lnP1/P2 (e.) The change of enthalpy H = 0

PROBLEMS1. During a reversible process there are abstracted 317 kJ/s from 1.134 kg/s of a certain gas while the temperature remains constant at 2.7oC. For this gas, Cp = 2.232 and Cv = 1.713 kJ/kg-K. The initial pressure is 586 kPa. For a non-flow (PE = 0, KE = 0) process, determine (a) V1,V2 and P2, (b) the work and Q, ( c ) H. 2. A 3.6 kg mass of air goes through an isothermal process with a temperature of 305K. The initial and final pressures of the air are 550KPa and 135KPa, respectively. Compute for a) Wnf, b) U and H c.) the heat. (442.64KJ, 0) 3. During an isothermal process at 88oF, the pressure on 8 lb of air drops from 80 psia to 5 psig. For an internally reversible process, determine (a) the pdV or the work of a nonflow process, (b) Q,, (c) U and H.4. Air flows steadily through an engine at constant temperature, 400 K. Find the work per kilogram if the exit pressure is one-third the inlet pressure and the inlet-pressure is 207 kPa. Assume that the kinetic and potential energy variation is negligible.

ISENTROPIC PROCESSAn isentropic process is a reversible adiabatic process. Adiabatic simply means no heat. A reversible adiabatic is one of constant entropy. P T 11 PVK= C 2 2 V1 V2 V S1. Relationship among P, V and T.(a) Relationship between P and V P1V1k = P2V2k = C(b) Relation between T and V From P1 V1k = P2 V2k and P1V1/T1 = P2V2/T2, we have T2/T1 = [V1/V2]k-1(c ) Relation between T and P T2/T1 = [p2/p1](k-1)/k

2. Nonflow work.From PVk = C, P = CV-kWn = PdV = C-kdV = C V-kdVIntegrating and simplifyingWn = P2V2 P1V1 / (1- k) = mR(T2 T1) / (1 k)3. The change of internal energy.U = mCv (T2 T1)4. The heat transferred.Q = 05. The change of enthalpy.H = mCp (T2 T1)6. The change of entropy.S = 0 PROBLEMSAn ideal gas having a mass of 2 kg. at 465K and 415 KPa expands in a reversible adiabatic process (isentropic). The gas constant R is 242 J/kg-K and k = 1.4. Determine (a) Cp and Cv, (b) the work, (c) the U, and (d) Q.In an isentropic process, the volume of an ideal gas (Cv= 0.65 KJ/kg-K and k = 1.29) initially 2m3 was expanded to 3.4m3. The pressure and the temperature of the gas before the process was 190KPa & 110C, respectively. Find (a) U; (b) H; (c) Wn & (d) QIn an isentropic process, the volume of 500 gm. of an ideal gas (Cp = 0.52 KJ/kg-K, Cv = 0.32 KJ/kg-K) was increased from 20L to 100L. If the initial pressure of the gas was 760KPa, compute for (a) P2; (b)T2; (c) U & (d) Wn.From a state defined by 300 psia, 100 cu ft and 240oF, Helium undergoes and isentropic process to 0.3 psig. Find (a) V2 and t2, (b) U and H, (c) pdV,, (d) Q and S and (e) the work if the process is non-flow.The internal energy of a certain ideal gas is given by the expression u=850+0.529PV Btu per lb, where p is in psia. Determine the exponent k in PVk=C for this gas undergoing an isentropic process. (Ans. k=1.53)

POLYTROPIC PROCESSA polytropic process is an internally reversible process during whichPVn = C and P1V1n = P2V2n = P1V1n PT 1 1 PVn = C PVn = C 2 Q 2

V S where n is any polytropic exponent or constant. 1. Relationship among p, V, and T(a) Relation between P and VP1V1n = P2V2n(b) Relation between T and VT2/T1 = [V1/V2]n-1(c) Relation between T and pT2/T1 = [p2/p1](n-1)/n

2. Nonflow workWn = pdV = P2V2 P1V1 / 1-n = mR(T2-T1) / 1-n3. The change of internal energy.U = mCv(T2 T1)4. The heat transferred.Q = U + Wn = mcv (T2 T1) + mR (T2 T1)/(1 - n) = m [(cv ncv + R)/(n 1)] (T2 T1) = m [(cp ncv)/(1 n)] (T2 T1) = mcv [(k n)/(1 n)] (T2 T1) Q = mCn(T2 T1) Cn = Cv [(k n)/(1 n)], the polytropic specific heat5. The change of enthalpyH = mCp (T2 T1)PROBLEMSAn ideal gas (R = 0.19 KJ/kg-K, Cp = 0.8 KJ/kg-K) with a mass of 4.5 kgs. undergoes a polytropic process that changed its state from 135 KPa and 280K to 825 KPa and 445K. Calculate (a) n; (b) U; (c) H; (d) Wn; and (e) Q. (Ans: 1.342, 453KJ, 594KJ, -412.5KJ, 41.5KJ) 2. Four and a half kg. of an ideal gas ( R=210J/kg-K and Cp = 1.04KJ/kg-K ) initially at 138 KPa and 278K goes through a polytropic process so that its pressure and temperature becomes 827KPa and 445K. Calculate (a) n; (b) U; (c) H; (d) Wn and (e) Q

3. During a polytropic process, 10 lb of an ideal gas, whose R = 40 ft.lb-R and Cp = 0.25 Btu/lb-R, changes state from 20 psia and 40oF to 120 psia and 340oF. Determine (a) n; (b) U; (c) H; (d) Q, (e) pdV. (1BTU = 778 ftlbf)

4. In a polytropic process with n = 1.4, 4.3 m3/s of an ideal gas ( Cv = 0.66 KJ/kg-K, k = 1.2 and = 1.169 kg/m3) has an initial and final temp. of 60C and 220C, respectively. Calculate (a) Work non-flow per unit time; and (b) Heat flow rate

5. Compress 4 kg/s of CO2 gas polytropically (PV1.2 = C) from P1 = 103.4 kPa, t1 = 60oC to t2 = 227oC. Assuming ideal gas action, find P2, W, and Q for a non-flow process.

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