II/IV B.Tech (Regular) DEGREE EXAMINATION. (1X12 = 12 Marks) Answer ONE question from each unit.

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1 Page 1 of 8 Hall Ticket Number: 14CH 404 II/IV B.Tech (Regular) DEGREE EXAMINATION June, 2016 Chemical Engineering Fourth Semester Engineering Thermodynamics Time: Three Hours Maximum : 60 Marks Answer Question No.1 compulsorily. (1X12 = 12 Marks) Answer ONE question from each unit. (4X12=48 Marks) 1. Write briefly about the following (1X12=12 Marks) (a) First law of thermodynamics (b) State functions (c) Reversible process (d) Entropy (e) Ideal gas (f) Third law of thermodynamics (g) Significance of residual properties (h) Significance of Gibbs energy (i) Coefficient of performance (j) Heat pump (k) Applications of liquefied gases. (l) Ton of refrigeration. UNIT-I 2. (a) Explain the phase rule. (3M) (b) How many degrees of freedom have each of the following systems? (3 x 3M = 9M) (i) Liquid water in equilibrium with its vapor. (ii) Liquid water in equilibrium with a mixture of water vapor and nitrogen. (iii) A liquid solution of alcohol in water in equilibrium with its vapor. 3. Air at 1bar and K is compressed to 5 bar and K by two different mechanically reversible processes: (a) Cooling at constant pressure followed by heating at constant volume. (b) Heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and U and H of the air for each path. The following heat capacities for air may be assumed independent of temperature: C V =20.78 C P =29.10 J mol -1 K -1. Assume also for air that PV/T is constant, regardless of the changes it undergoes. At 298K and 1 bar the molar volume of air is m 3 mol -1. UNIT-II 4. Air is compressed from an initial condition of 1 bar and K to a final state of 5bars and K by two different mechanically reversible processes in a closed system: (a) Isothermal compression (b) Adiabatic compression followed by cooling at constant volume. (4M) (8M)

2 Page 2 of 8 Assume air to be an ideal gas with the constant heat capacities, C V =(5/2)R and C P =(7/2)R. The initial and final molar volumes of air are and m 3. Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process. 5. (a) A central power plant rated at kw, generates steam at 585K and discards heat to a river at 295K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power? (b) A 40 kg steel casting (C P = 0.5 kj Kg -1 K -1 ) at a temperature of K is quenched in 150 kg of oil (C P = 2.5 kj kg -1 K -1 ) at K. If there are no heat losses, what is the change in entropy of (i) the casting, (ii) the oil, and (c) both considered together? 6. Prove the following results: (a) dh = Cp dt + [V T( P] dp (b) du = Cv dt + [T( v P] dv UNIT-III 7. (a) Develop the Clapeyron equation for two phase systems. (b) Explain the duct flow of compressible fluids. UNIT-IV 8. (a) With a neat sketch, explain the Absorption Refrigeration. (b) A house has a winter heating requirement of 30kW and a summer cooling requirement of 60kW. Consider a heat pump installation to maintain the house temperature at K in winter and K in summer. This requires the circulation of refrigerant through interior exchanger coils at K in winter and K in summer. Underground coils provide the heat source in winter and the heat sink in summer. For a year round ground temperature of K the heat transfer characteristics of the coils necessitate refrigerant temperatures of K in winter and K in summer. What are the minimum power requirements for winter heating and summer cooling? 9. (a) Describe the various steps in vapor compression cycle. (b) Explain the Liquefaction processes. 14CH 404

3 Page 3 of 8 Hall Ticket Number: II/IV B.Tech (Regular) DEGREE EXAMINATION June, 2016 Chemical Engineering Fourth Semester Engineering Thermodynamics Time: Three Hours Maximum : 60 Marks Answer Question No.1 compulsorily. (1X12 = 12 Marks) Answer ONE question from each unit. (4X12=48 Marks) 1. Write briefly about the following (1X12=12 Marks) (a) First law of thermodynamics: Total energy of the universe is constant. (b) State functions: Property dependent on internal condition, not on past history; Ex:density (c) Reversible process: Process which can retain its original state by reversing the infinitisma driving force. (d) Entropy: Intrinsic / state property defined by integral of (dq/t) (e) Ideal gas: Gas characterised by absence of intermolecular force. (f) Third law of thermodynamics: Entropy is zero for pure crystalline substance absolute zero temperature. (g) Significance of residual properties: M R = M-M ig (h) Significance of Gibbs energy: It is complete property information (i) Coefficient of performance: Target heat load/ input energy load (j) Heat pump: Continuous / cyclic device to transfer (pump) heat to higher temp reservoir from lower temperature reservoir. (k) Applications of liquefied gases: Ice manufacture, liquifying gases. (l) Ton of refrigeration: Rate at heat is to be removed to form ton of ice in a day from saturated water at 0 C UNIT-I 2. (a) Explain the phase rule F = C-P+2; Gibb s phase rule gives minimum number of intensive variables (F) required to fix the intensive state of the closed system at equilibrium with C components / species in P number of phases. (3M) (b) How many degrees of freedom have each of the following systems? (3 x 3M = 9M) (i) Liquid water in equilibrium with its vapor: F= 1-2+2=1 (T or P) (ii) Liquid water in equilibrium with a mixture of water vapor and nitrogen: F = 2-2+2= 2 (iii) A liquid solution of alcohol in water in equilibrium with its vapor: F = 2-2+2=2 Air at 1bar and K is compressed to 5 bar and K by two different mechanically reversible processes: The following heat capacities for air may be assumed independent of temperature: C V =20.78 C P =29.10 J mol -1 K -1. Assume also for air that PV/T is constant, regardless of the changes it undergoes. At 298 K and 1 bar the molar volume of air is m 3 mol (a) Cooling at constant pressure followed by heating at constant volume. Let cooling be done to temperature T T = [ ( /5) x 10 5 ]/ [ x 10 5 / 298]= 298/5 = 59.6 K

4 Page 4 of 8 For cooling up to T del T = (298/5-298) K U = del T and H = del T; W = x 10 5 / 298 del T; Q = H; For heating up to final condition del T = -(298/5-298) K U = del T and H = del T; W = 0 ; Q = U; (b) Heating at constant volume followed by cooling at constant pressure. Let heating be done to temperature T T = [ x 10 5 / 298]/ [ ( /5) x 10 5 ] = 298x5 = 1490 K For heating up to T: del T = ( ) K U = del T and H = del T; W = 0 ; Q = U; For heating up to final condition del T = ( ) K U = del T and H = del T; W = x 10 5 / 298 del T ; Q = H; UNIT-II 4. Air is compressed from an initial condition of 1 bar and K to a final state of 5bars and K by two different mechanically reversible processes in a closed system: Assume air to be an ideal gas with the constant heat capacities, C V =(5/2)R and C P =(7/2)R. The initial and final molar volumes of air are and m 3. Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process. (a) Isothermal compression U = 0 and H = 0 ; W = x 298 ln(5/1) ; Q = W; (4M) (b) Adiabatic compression followed by cooling at constant volume. Volume at final condition = x /5x10 5 = V Adiabatic compression is done up to a pressure of P: P = 1x10 5 (V1/V) γ = 1x10 5 (5) 1.4 Pa Temperature attained in adiabatic compression = T = T1(P/P1) ( γ-1)/ γ = (5) K U = W and H = 3.5 R (T ) ; W = x (T )/( 1.4-1) ; Q = 0; For cooling at constant volume intermediate step del T = ( x1.9) U = 2.5 R del T and H = 3.5 R del T ; W = 0 ; Q = U; (8M) 5. (a) A central power plant rated at kw, generates steam at 585K and discards heat to a river at 295K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power? Q c = / 0.7/ (1-295/585) kw; 6. (b) A 40 kg steel casting (C P = 0.5 kj Kg -1 K -1 ) at a temperature of K is quenched in 150 kg of oil (C P = 2.5 kj kg -1 K -1 ) at K. If there are no heat losses, what is the change in entropy of (i) the casting, (ii) the oil, and (c) both considered together? Equilibrium temperature after quenching is T = [(mcpt) s +(mcpt) w ]/ [(mcp) s +(mcp) w ]= [40x0.5x x2.5x298.15]/ [40x x2.5] K change in entropy of (i) the casting = 40x0.5xln (T/723.15) (ii) the oil = 150x2.5xln (T/298.15), and (c) both considered together (i) + (ii)

5 7. Prove the following results: (a) dh = Cp dt + [V T( P] dp H = H(T,P); (b) du = Cv dt + [T( v P] dv U = U(T,P); UNIT-III Page 5 of 8 ; ; ; ; ; ; 8. (a) Develop the Clapeyron equation for two phase systems. (b) Explain the duct flow of compressible fluids. Duct Flow Thermodynamics does provide equations that interrelate the changes occurring in pressure, velocity, cross-sectional area, enthalpy, entropy, and specific volume of a flowing stream. Adiabatic, steady-state, one-dimensional flow of a compressible fluid in the absence of shaft work and of changes in potential energy is considered here. Applying I law to a flow process with above constraints / conditions dh ud u..(1) From the fundamental property relations dh TdS dp.(2) Applying continuity equation to compressible fluid flow d A 0 which implies +A 0.(3) A Further considering as function of S and P i.e.,, d 1 P dp 1 S S ds P But from Physics S P where c is velocity of sound in fluid. Therefore P S Further S P T P βt S P P Therefore βt ds P dp..(4) There are 6 differentials dp,dv, du, ds,dh and da, and four equations. Therefore it is preferred to eliminate dh and dv. Substituting from equation (3) in equation (4) + A A βt P dp - TdS dp A A ds, and now replacing βt ds P dp from equation (1) and (2) as - [ TdS dp]

Page 6 of 8 Therefore 1 dp β c P 1TdS da 0..(5) Substituting TdS udu in place of dp from equation (1) and equation (2) 1 TdS udu β c P Therefore. 1 TdS da 0 β c udu P 1 1 TdS 1 da 0.

6 Page 6 of 8 Therefore 1 dp β c P 1TdS da 0..(5) Substituting TdS udu in place of dp from equation (1) and equation (2) 1 TdS udu β c P Therefore. 1 TdS da 0 β c udu P 1 1 TdS 1 da 0..(6) Equation (5) and (6) can be written as 1 dp β 1T ds da dx c P dx dx 0 Where u du β x is length of the duct. For dx c P irreversible 1 or T ds frictional dx 1 da 1 flow.. S 0 dx 0 UNIT-IV 9. (a) With a neat sketch, explain the Absorption Refrigeration. ABSORPTION REFRIGERATION CYCLE Absorption refrigeration cycle is similar to the vapor compression refrigeration cycle only, except the way vapor compression is achieved. In this cycle vapor from evaporator is absorbed in a relatively nonvolatile solvent at evaporator pressure and at lower temperature level of heat engine. Heat of absorption is rejected to surroundings. This is heat rejected by the engine. The liquid solution from the absorber, which contains a relatively high concentration of refrigerant, passes to a pump, which raises the pressure of the liquid to that of the condenser. Heat from the higher temperature source at T H is transferred to the compressed liquid solution, raising its temperature and evaporating the refrigerant from the solvent. Vapor passes from the regenerator to the condenser, and solvent, which now contains a relatively low concentration of refrigerant, returns to the absorber. The heat exchanger conserves energy and also adjusts stream temperatures toward proper values. Low-pressure steam is the usual source of heat for the regenerator.

7 Page 7 of 8 The most commonly used absorption-refrigeration system operates with water as the refrigerant and a lithium bromide solution as the absorbent. This system is obviously limited to refrigeration temperatures above the freezing point of water. For lower temperatures ammonia can serve as refrigerant with water as the solvent. An alternative system uses methanol as refrigerant and polyglycolethers as absorbent. Consider refrigeration at a temperature level of TC= K with a heat source of condensing steam at atmospheric pressure TH = K. For a surroundings temperature of TS = K, the minimum possible value of Q H. Q C Actual value. will be about three times this. (b) A house has a winter heating requirement of 30kW and a summer cooling requirement of 60kW. Consider a heat pump installation to maintain the house temperature at K in winter and K in summer. This requires the circulation of refrigerant through interior exchanger coils at K in winter and K in summer. Underground coils provide the heat source in winter and the heat sink in summer. For a year round ground temperature of K the heat transfer characteristics of the coils necessitate refrigerant temperatures of K in winter and K in summer. What are the minimum power requirements for winter heating and summer cooling? Winter:... Summer:... Type equation here. 10. (a) Describe the various steps in vapor compression cycle. (b) Explain the Liquefaction processes. Liquefaction of Gases

8 Page 8 of 8 Liquefied gases are in common use for a variety of purposes. For example, liquid propane in cylinders serves as a domestic fuel, liquid oxygen is carried in rockets, natural gas is liquefied for ocean transport, and liquid nitrogen is used for low-temperature refrigeration. In addition, gas mixtures (e.g., air) are liquefied for separation into their component species by fractionation. Liquefaction results when a gas is cooled to a temperature in the two-phase region. This may be accomplished in several ways 1. By heat exchange at constant pressure. 2. By an expansion process from which work is obtained. 3. By a throttling process. The first method requires a heat sink at a temperature lower than that to which the gas is cooled, and is most commonly used to pre-cool a gas prior to its liquefaction by the other two methods. An external refrigerator is required for a gas temperature below that of the surroundings. The constant-pressure process (1) approaches the two-phase region (and liquefaction) most closely for a given drop in temperature. The throttling process (3) does not result in liquefaction unless the initial state is at a high enough pressure and low enough temperature for the constant-enthalpy process to cut into the two-phase region. This does not occur when the initial state is at A. If the initial state is at A', where the temperature is the same but the pressure is higher than at A, then isenthalpic expansion by process (3') does result in the formation of liquid. The change of state from A to A' is most easily accomplished by compression of the gas to the final pressure at B, followed by constant-pressure cooling to A'. Liquefaction by isentropic expansion along process (2) may be accomplished from lower pressures (for given temperature) than by throttling. The throttling process (3) is the one commonly employed in small-scale commercial liquefaction plants. The temperature of the gas must of course decrease during expansion. This is indeed what happens with most gases at usual conditions of temperature and pressure. The exceptions are hydrogen and helium, which increase in temperature upon throttling unless the initial temperature is below about 100 K for hydrogen and 20 K for helium. Liquefaction of these gases by throttling requires initial reduction of the temperature to lower values by method 1 or 2. for air6 shows that at a pressure of 100 atm the temperature must be less than 169 K for any liquefaction to occur along a path of constant enthalpy.---prepared By: Dr.N.R.Gopal

S.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks : 100

Total No. of Questions 12] [Total No. of Printed Pages 7 Seat No. [4162]-189 S.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks