1. Basic state values of matter

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1 1. Basic state values of matter Example 1.1 The pressure inside a boiler is p p = Pa and p v = Pa inside a condenser. Calculate the absolute pressure inside the boiler and condenser if the barometric pressure p b = MPa. Find the vacuum in the condenser in %! p p = 11.5 MPa: p v = 94.4 kpa; p b = MPa; p a boil =? ; p a cond =? a) The absolute pressure inside the boiler p a boil = p b + p p = = MPa The absolute pressure in the condenser p a cond = p b - p v = = MPa b) The vacuum inside the condenser in % : 1

2 x [%] = x.100 = 96.3 % Example 1.2 A pressure vessel of a volume of 0.1 m 3 contains 1.25 kg of oxygen. What is the density and specific volume? V = 0.1 m 3 ; m = kg 0 2; ρ =? ; v =? air density specific volume of oxygen 2. THE IDEAL-GAS EQUATION OF STATE f(p, v, T)=0 Example 2.1 Calculate the densitz and specific volume of carbon dioxide CO 2 under normal physical conditions. Solution : ρ =? ; v =? Normal conditions are: pressure p = MPa, temperature t = 0 C. Using the equation of state p v = r T ; The gas constant for CO 2 r = 188,97 J/(kg.K); absolute temperature T = t = K Then Density Example 2.2 A pressure vessel contains nitrogen at a temperature of t = 20 C and pressure p = 2.2 MPa. The maximum allowed overpressure is 6 MPa. What is the maximum temperature that the nitrogen can be heated to if the barometric pressure is p b = 0,1 MPa? t = 20 C ; p = 2.2 MPa ; p pmax = 6 MPa ; p b = 0.1 MPa; t 2 =? The isochoric process p 1 = 2.2 MPa, T 1 = t = = 293 K p 2 = p a max = p pmax + p b = = 6.1 MPa 2

3 t 2 = T = = 539 C 3. Ideal-gas mixtures Example kg of dry air comprises 23.2 mass % oxygen and 76.8 % nitrogen. Find the volume fractions of air (i.e. the individual volumes), the gas constant, the mean virtual molar mass and the partial pressures of oxygen and nitrogen if the air pressure is p = MPa. w O2 = 0.232; w N2 = 0.768; p = MPa ; x O2 =?; x N2 =?; r =?; M =?; p O2 =?; p N2 =? Volume composition of air The volume fraction for air The volume fraction for nitrogen The gas constant for air The mean virtual molar mass for air or The partiál pressure of oxygen is p o2 = x O2 p = = MPa The partiál pressure of nitrogen is p N2 = x N2 p = = MPa 4. I. LAWS OF THERMODYNAMICS Example 4.1 A steel component weighing 0.2 kg heated inside a furnace is then inserted inside a calorimeter containing 0.5 kg of water at a temperature of t = 20 C. After stabilisation, the temperature inside the calometer is 75 C. Calculate the temperature of the component before it was inserted into the calorimeter. The specific heat capacitz of steel is c = kj/(kg.k). 3

4 m Fe = 0.2 kg; m H2O = 0.5 kg; t 1H2O =20 C ; t 2H2O = 75 C; c Fe = kj/(kg.k) The heat transferred from the steel to the water is Q H2O =m H2O. c H2O (t 2H2O - t 1H2O ) = (75-20) = J From the heat balance Q Fe = m Fe. c Fe (t Fe - t 2H2O ) = Q H2O we can calculate 5. Reversible thermodynamic processes of ideal gasses Example 5.1 A closed container contains 0.6 m 3 of air with a pressure of p 1 = Pa and temperature t 1 = 20 C. What is the pressure and temperature after losing kj of heat (i.e. cooling)? V= 0.6 m 3 ; p 1 = Pa ; t 1 = 20 C ; Q 12 = kj; p 2 =?; t 2 =? In an isochoric process for an isochoric process ; where p 2 and T 2 are the sought values. Here, it is necessary to use the First Law of Thermodynamics Q = m.. Δu = m. c v (t 2 - t 1 ), kde is the mass of air Pressure Example 5.2 How much heat must be received by 0.5 kg of oxygen at a pressure of p 1 = Pa and temperature t 1 = 35 C in order to carry out work A 12 = J under a constant pressure, and what will the final volume and temperature be? m = 0.5 kg; p 1 = Pa; t 1 = 35 C; A 12 = J; V 2 =? ; t 2 =? The work of the isobaric process (p 2 = p 1 = p 1,2 ) 4

5 , where In ab isobaric process ; The heat received Q 12 = m.c p (t 2 - t 1 ) = (238-35) = J = kj Example kj of heat is rejected in an isothermic compression of 0.3 m 3 of air with a pressure of p 1 = 10 6 Pa and temperature t = 300 C. Calculate the final volume and pressure! Solution : V 1 = 0.3 m 3 ; p 1 = 10 6 Pa; t = 300 C; Q 12 = -490 kj; V 2 =? ; p 2 =? From i sothermic work we get The pressure ratio The final pressure The final volume (from the isothermic equation p 1 V 1 = p 2 V 2 ) Example kg of air with an initial temperature of t 1 = 30 C and pressure p 1 = Pa is adiabatically compressed to p 2 = l0 6 Pa. Deterermine the final volume, the final temperature and volume work done on this system (κ = l.4 ). Solution : t 1 = 30 C ; p 1 = Pa ; p 2 = Pa ; v 2 =? ; t 2 =? ; a 12 =? From the temperature and pressure ratio in an adiabatic process we get the final temeprature t 2 = T = = C 5

6 The final volume (from the equation of state) The work done on the system Example kg of air is polytropically compressed from p 1 = Pa and t 1 = 18 C to p 2 = Pa and t 2 = 125 C. What is the polytropic, final volume the work done on the system the heat lost and the change in internal energy.. Solution : m= 1.5 kg; p 1 = Pa; t 1 = 18 C ; p 2 = Pa ; t 2 = 125 C ; n =?; V 2 =?; A 12 =?; Q 12 =?; ΔU =? The temperature-pressure ratio in a polytropic process is and the logarithm is The polytropic exponent n = The final volume The work in a polytropic process which is work consumed The heat lost The change in internal energy (from the First Law of Thermodynamics) ΔU = Q 12 - A 12 = ( ) = J Example 5.6 Air at a temperature of 127 C is compressed isothermically to a quarter of the initial volume and then adiabatically expands to its original pressure. Calculate the temperature of the air after the adiabatic expansion! t 1,2 = 127 C; v 2 = v 1 /4 ; p 3 = p 1 ; t 3 =? In an isothermic process from 1 to 2: p 1 / p 2 = v 2 / v 1 In an adiabatic process from 2-3: tj. The temperature 6

7 t 3 = T = = - 4 C 6. II. The law of thermodynamics, entropy, T-s diagram Example kg of oxygen at a pressure p 1 = Pa and temperature t 1 = 127 C expands isobarically to double its initial volume, then is compressed isothermically to p 2 = Pa. Calculate the change in entropy of the oxygen! p 1 = Pa ; t 1 = 127 C; v 2 = 2 v 1 ; p 3 = p 2 = 39,2.105 Pa; Δs =? The change in entropy of an isobaric process Δs 12 = c p 1n (T 2 /T 1 ) = c p 1n (v 2 /v 1 ) = c p 1n (2.v 1 /v 1 ) = 0, n 2 = 0, ,6931 = 0, J/(kg.K) The change in entropy of an isothermic process Δs (23) = q 2,3 / T 2,3, kde T 2,3 = T 1.( v 2 /v 1 ) = 400 (2.v 1 /v 1 ) = 800 K The total change in entropy Δs (13) = Δs (12) + Δs (23) = ( ).10 3 = J/(kg.K) 7

8 Example kg of air is compressed polytropically from pressure p 1 = Pa and tempreature t 1 = 18 C to pressure p 2 = Pa and temperature t 2.= 125 C. Find the entropy change! solution: m = 1.5 kg; p 1 = Pa ; t 1 = 18 C ; p 2 = 9, Pa ; t 2 = 125 C Δs =? The change in entropy in a polytropic process ΔS = m Δs = 1.5 ( ) = J/K (Note: The exponent of this polytropic process was determined in Exercise 5.47) Example 6.3 Inside a recuperation exchanger with a gas turbine the air is heated by the exhaust gases from temperature t 1 = 140 C to t 2 = 270 C. The gases are cooled from temperature t 3 = 340 C to temperature t 4 = 210 C. The gas is considered ideal having the properties of air. The temperature of the ambient t o = 20 C. There is no heat loss. Calculate the exergy loss! t 1 = 140 C ; t 2 = 270 C ; t 3 = 340 C ; t 4 = 210 C ; t 0 = 20 C; Δe =? Energy loss Δe = T 0 Δs, where Δs is the change in entropy Δs = Δs 1 + Δs 2 Δs = ( ).10 3 = J/(kg.K) 8

9 The exergy loss Δe = = J/kg = 10.4 kj/kg 7. Carnot cycle Example kg of air is executing the Carnot cycle between the temperatures t H = 327 C and t C = 27 C. The highest temperature is Pa; the lowest is Pa. Calculate the work, thermal efficiancy, the received and rejected heat and the exergic efficiency if the ambient temperatrure is t o = 20 C! Soluition : t H = t max = 327 C; t C = t min = 27 C, p max = Pa, p min = 1, Pa ; a o =? ; η t =?; q H =? /q C / =? ; η E =? The pressure in point 2 of the cycle The pressure in point 4 of the cycle The heat received The heat rejected The thermal efficiency of the cycle The exergetic efficiency is given by the ratio of the actually exerted and maximum work η E = a/a max ; a max = q H - T o Δs = q H T o (q H /T 1 ) ; a 0 - q H - q C 9

10 8. Cycle of IC-engine Example 8.1 The working gas in internal combustion engines with a combined input of heat is air. If the pressure p 1 = Pa, temperature t 1 = 30 C, compression ratio ε = 7, pressure ratio Ψ = 2, cutoff ratio ϕ = 1.2, calculate the state values in the characteristic points of the cycle, the input heat, the work done by the system and the thermal efficiency of the cycle. The specific heat capacity is constant. p 1 = Pa; t 1 = 30 C; ε = 7; Ψ = 2; ϕ = 1,2 ; q 1 =? ; a C =? : η t =? Quantities Point p [MPa] v [m 3 /kg] T [K] x x x quantities given, the rest calculated Point 1 Point 2 v 2 = v 1 / ε = / 7= m 3 /kg 10

11 Pressure p 2 = p 1. ε κ = ,4 = Pa Temperature Point 3 Pressure p 3 = p 2. Ψ = = Pa specific volume v 3 = v 2 = m 3 /kg Temperature Point 4 Pressure p 4 = p 3 = Pa specific volume v 4 = v 3 ϕ = 0,127. 1,2 = 0,1525 m 3 /kg Temperature T 4 = ϕ.t 3 = = 1580 K Point 5 Pressure specific volume v 5 = v 1 = m 3 /kg Temperature Heat received q H = c v ( T 3 - T 2 )+ c p ( T 4 - T 3 ) = 0, ( )+1005.( )= J/kg Heat lost Cycle work Thermal efficiency of the c ycle 9. Compressor Example 9.1 A twostage compressor sucks in air at a temperature of t 1 = 20 C and a pressure of MPa and compresses it to a pressure of p 2 = 6 MPa. Calculate the power of the engine necessary for driving the compressor and the amount of cooling water for both stages of the compressor and also for the intercooler if the ratio of the output and input pressures is the same for both stages and the mechanical 11

12 efficiency of each stage is 0.7. The temperature of the water increases by 15 K. Compression in both stages is polytropic with an exponent n = 1.3. The power of the compressor is V = 0.14 m 3 /s. Solution : t 1 = 20 C; p 1 = MPa, p 2 = 6 MPa, η k = 0,7, Δt H2O = 15 K; n =1.3; V = 0.14 m 3 /s ; P NT =?, P VT =?, m NT =?, m VT =?, m MCH =? The pressure ratio in both stages is 12

13 The temperature t 1 = t 1 = 20 C The volume V 1 = V 1 = V = 0.14 m 3 /s The power necessary for compression of the air in the low-pressure stage is and for the high-pressure stage is P VT = P NT = W The power of the engine for both stages is The temperature after compression in both stages is The heat loss from both stages during compression is The amount of cooling water for the low-pressure (high-pressure) stage of the compressor is The heat loss from the intercooler is The amount of cooling water for the intercooler is The total amount of cooling water that is necessary is m H2O = m H2O NT + m H2O VT + m H2O MCH = = kg/s 10. Gas flow Example

14 A pressure vessel contains oxygen at a pressure of p 1 = 5 MPa. The gas leaves through a nozzle into the ambient with a pressure of p 2 = 4 MPa. The initial temperature of the oxygen is 100 C. Calculate the theoreatical speed upon exit and the mass flow rate (kg/s), if the area of the end of the nozzle is S = m 2. Find the speed and mass flow rate of the oxygen upon exit in to the ambient which has a pressure of p 2 = MPa. p 1 = 5 MPa; p 2 = 4 MPa, t 1 = 100 C, S = m, 2 p 2 = MPa, w 2 =?, m =? w 2 =?, m =? a) The pressure ratio Mass flow b) The pressure ratio Mass flow 11.Cycle of gas turbines Example 11.1 Calculate the thermal efficiency of a cycle of a gas turbine with a compression ration of 10, which has a heat input at a constant pressure, κ = 1.4. ε = 10; κ = 1.4 ; η t =? The thermal efficiency of a turbine with heat input with p=const. η t = 1-( 1/ ε κ-1 ) = 1 - (1/10 1,4-1 ) = 1 - ( 1/ 2,51 ) = = REAL GASES AND VAPOURS 14

15 Example 12.1 Calculate the state of water vapour with a pressure of p = 0.59 MPa and a temperature t = 190 C. Solution : p = MPa ; t = 190 C; the state of the water vapour =? From the table for saturated water H 2 0 it is possible to determine the temperature of the saturated vapour with a pressure of 0.59 MPa, t 23 = C < t = 190 C. This is a case of superheated vapour and, using the table for superheated water vapour (i.e. the i-s diagram), it is possible to find the specific volume v, the enthalpy i and entropy s of the vapour for the pressure p and temperature t. The superheating of the water vapour is Δt = t - t 23 = = K Example 12.2 What is the heat received by 1 kg of water vapour with a constant pressure of p = 1.47 MPa if the vapour quality increases from x 1 = 0.8 to x 2 = 0.96? p = 1.47 MPa ; x 1 = 0.8 ; x 2 = 0.96 ; q 12 =? ( dp = 0 ) The heat received q 12 = i 2 - i 1 is equal to the enthalpy difference. a) The calculation using the i-s diagram for water vapour: 15

16 For pressure p and quality x 1 from the i-s diagram: i 1 = 2390 kj/kg. For p and x 2 : i 2 = 2702 kj/kg. The heat received q = i 2 -i 1 = = 312 kj/kg. b) The calculation using the tables for saturated water H 2 O: Enthalpy i 1 = i 1 + x 1 (i 1- i 1) i 2 = i 2 + x 2 (i 2 - i 2 ) for p = const. i 1 = i 2 = i', i 1 = i 2=i" The heat received: q 12 = (x 2 - x 1 )( i - i ) =(x 2 -x 1 ).l 23 from the table for pressure p l 23 = kj/kg q 12 = ( ) = = kj/kg Example 12.3 Calculate the heat received by 6 kg of water vapour with a volume of 0.6 m 3 at a temperature of p 1 = 0.6 MPa, so that the pressure increases to p 2 = 1 MPa at a constant volume. Find also the final quality of vapour. Solution : m = 6 kg ; V = 0.6 m 3 ; p 1 = 0.6 MPa ; p 2 = 1 MPa ; q 12 =? ; x 2 =? a) Using the tables for water vapour, the specific volume of the vapour is: From the condition v 1 = v 2 = v and considering the fact that in the initial and final states (1 and 2) the vapour is wet v = v 1 + x 1 (v 1 - v 1) 16

17 v = v 2 + x 2 ( v 2 - v 2) we get the vapour quality x 1 = ( v - v 1) / ( v 1- v 1) ; x 2 = ( v - v 2) / ( v 2 - v 2) From the table for saturated water we substitute for the values of specific volumes of saturated liquid and vapour ; The heat received at a constant volume of 1 kg of vapour is q 12 = u 2 - u 1 = (i 2 - p 2 v 2 )-(i 1 - p 1 v 1 ) = ( i 2 - i 1 ) - v.( p 2 - p 1 ) The enthalpy is i 1 = i 1 + x ,1 = = kj/kg i 2 = i 2 + x ,2 = = kj/kg q 12 = ( ) (1-0.6) = = J/kg The total heat received is Q 12 = m. q 12 = = kj b) The calculation using the i-s diagram for water vapour is: The enthalpy in point 1 for pressure p 1 and the specific volume is v is i 1 =1325 kj/kg. The enthalpy in point 2 for pressure p 2 and the specific volume v is i 2 = 1795 kj/kg The vapour quality in point 2 is x 2 = 0.51 The heat received is q 12 = (i 2 - i 1 ) - v.( p 2 -p 1 ) = ( ) (1-0.4) = = J/kg. Example kg of water vapour at a pressure of p 1 = 3 MPa and temperature t 1,2 = 300 C is compressed to one fifth of the original volume while 17

18 maintaining the temperature. Calculate the final state of the vapour, the work done on the szstem (i.e. consumed) and the heat lost. Illustrate the process in a p-v, T-s and i-s diagram for water vapour. p 1 = 3 MPa; t 1 = t 1 = t = 300 C ; v 2 = v 1 /5; state 2 =? ; a 12 =? ; q 12 =? 18

19 a) The calculation using tables for p 1, t - state 1 v 1 = m 3 /kg i 1 = 2988 kj/kg s 1 = 6.53 kj/(kg.k) The specific volume v 2 = v 1 /5 = /5 = m 3 /kg Since the volume of saturated vapour t = 300 C is m 3 /kg, the final state 2 falls into the area of wet vapour, p 2 = MPa. v 2 = v 2 + x 2 ( v 2 - v 2 ) the quality in point 2 The enthalpy in point 2 i 2 = i 2 + x 2 l 23,2 = = kj/kg The e ntropy in point 2 s 2 = s 2 + x 2 ( s 2 - s 2) = ( ) = kj/(kg.k) The difference in internal energies Δu = u 2 - u 1 = (i 2 - i 1 ) -(p 2 v 2 - p l v 1 ) = ( ) ( ).10 6 = J/kg The lost heat q 12 = T(s 2 - s 1 ) = 573 ( )= -847 kj/kg The consumed work a 12 = q 12 - Δu = ( ) = kj/kg b) The calculation using the i-s diagram for water vapour: for p 1 and t we get v 1 = 0.08 m 3 /kg, i 1 = 2990 kj/kg, s 1 = 6.53 kj/(kg.k), v 2 = v 1 /5 = 0,08/5 = m 3 /kg For temperature t and v 2 we obtain x 2 = 0.73, p 2 = 8.6 MPa from the i-s diagram i 2 = 2375 kj/kg, s 2 = 5.05 kj/(kg.k) The difference in internal energies Δu = u 2 - u 1 = (i 2 - i 1 ) - (p 2 v 2 - p 1 v 1 ) = ( ) ( ).10 6 = = J/kg 19

20 The lost heat q 12 = T(s 2 - s 1 ) = 573 ( ) = = kj/kg The consumed work a 12 = q 12 - Δu = (-512.6) = kj/kg Example kg of water vapour at a pressure of p 1 = 0.3 MPa and temperature t 1 = 300 C expands adiabatically to pressure p 2 = 0.05 MPa. Calculate the final parameters of the vapour, the volume work and the enthalpy change. Plot a T-s and i-s diagram. p 1 = 0.3 MPa ; t 1 = 300 C ; p 2 = 0.05 MPa ; state 2=? ; a 12 =? ; Δi =? Using the tables for water vapour: For p 1, t 1 we find i 1 = 2988 kj/kg ; v 1 = m 3 /kg;, s 1 = 6.53 kj/(kg.k) based on the condition for an adiabatic process s 2 = s 1 and regarding the fact that s 2 < s 2 at a pressure of p 2 (this is actually a case of wet vapour) 20

21 The specific volume v 2 = v 2+ x 2 (v 2 - v 2) = ( ) = m 3 /kg Enthalpy i 2 = i 2 + x 2 l 23,2 = = kj/kg (Note: l 23,2 = 2304 kj/kg) The difference in enthalpies Δi = i 2 - i 1 = = kj/kg The difference in internal energies Δu = u 2 - u 1 = Δi - (p 2 v 2 - p l v 1 ) = ( ).10 6 = = J/kg The volume work a 12 = -Δu = 613,63 kj/kg Example 12.6 Calculate the temperature at which ice melts under a pair of skates with blades 0.25 m long and 1 mm thick (the weight of the skater is m = 70 kg) if 1/10 of the surface area of the blades is touching the ice. The specific heat for melting ice is l 12 = 334 kj/kg, the density of the ice is ρ L = kg/m 3, the density of water is ρ v = kg/m 3. L = 0.25 m; δ = 1 mm ; m = 70 kg ; K = 0.1 ; l 12 = 334 kj/kg; ρ L = kg/m 3 ; ρ V = kg/m 3 ; t 12 =? The temperature is calculated using the Clausius-Clapeyron equation Let T 12 = 273 K (i.e. the temperature of melting ice at barometric pressure p) and dt related to it. The elementary change dt and dpare replaced by ΔT = T 12(p+Δp) - T 12,p = > T 12(p+Δp) = T 12,p + ΔT = = K, t 2 = C 13. VAPOUR FLOW Example 13.1 Calculate the output speed of water vapour and the diameter of the nozzle of a Lavalovy jet, if the output quantity of the water vapour is m = 216 kg/h at a pressure p 1 = MPa and the temperature t 1 = 220 C. The output preessure is p 2 = MPa. m = 216 kg/h; p 1 = MPa ; t 1 = 220 C ; p 2 = MPa,w 2 =? ; d 2 =? (using the i-s diagram for water vapour) 21

22 For pressure p 1 and temperature t 1 we calculate the enthalpz i 1 = 2890 kj/kg Entalpie v bodě 2 i 2 = 2420 kj/kg The output speed of water vapour The diameter of the output noyyle of the jet 14. Cycle of thermal power station Example 14.1 The cycle of a thermal power station works with additional heating of water vapour. Pressure p 1 = 12 MPa, temperature t 1 = 530 C, pressure p 2 = 3500 Pa. The heating is carried out at a pressure of p 1 = 2.3 MPa to a temperature of t 1 = 480 C. Calculate the increase in the quality of the vapour upon exit from the turbine and the increase in thermal efficiency of the cycle with additional heating compared to that without. p 1 = 12 MPa; t 1 = 530 C; p 2 = 3500 Pa, p 1 = 2.3 MPa; t 1 = 480 C; Δx =?; Δη t =? 22

23 From the i-s diagram for water vapour for states 1, 2, l, 2 and 2" we calculate i 1 = 3426 kj/kg ; i 2 = 2956 kj/kg i 1 ' = 3420 k J/kg; i 2 ' = 2180 kj/kg; i 2 " = 1972 kj/kg and the quality of the vapourx 2 = 0,85 ; x 2 = 0,763. Using the tables for water vapour we obtain i 3 = kj/kg The thermal efficiency of the cycle without additional heating The thermal efficiency of the cycle with additional heating The increase in the thermal efficiency Δη t = η; t2 - η t1 = = The increase in the quality of the vapourδx = x 2 - x 2 = 0,85-0,763 = 0,087 23

24 15. Refrigerating Example 15.1 An ammonia refrigerating station with a power output of Q = 110 kw works at an evaporation temperature of t 1 = -15 C. Saturated vapour (see figure below) leaves the evaporator. The condensation temperature t 3 = 30 C, then follows undercooling of the condensate to temperature t 3 = 25 C. Calculate the theoretical level of refrigeration of the cycle, the amount of refrigerant NH 3 in circulation and the theoretical power output of the motor necessary for driving the compressor of the station. Q = 110 kw; t 1 = -15 C; t 3 = 30 C, t 3 ' = 25 C ; q vth =? ; m NH3 =?, P =? From the i-log p NH 3 diagram - for the given temperatures - we obtain the enthalpies i 1 = 1662 kj/kg, i 2 = 1894 kj/kg ; i 3 = 536 kj/kg The theoretical level of cooling q vth = i 1 - i 4 = i 1 - i 3 ' = = 1126 kj/kg The amount of refrigerant in circulation m NH3 = Q / q vth = 110/1126 = kg/s The theoretical power output for driving the compressore is P = m( i 2 - i 1 ) = ( ) = kw 16. MOIST (ATMOSPHERIC) AIR Example kg moist air with a temperature of t 1 = 40 C and relative humidity ϕ 1 = 0.2 with a pressure of p = MPa is moistened with 5 kg of water with a temperature t w = 20 C. Calculate the state of the air after moistening. 24

25 Solution : m = 1000 kg ; t 1 = 40 C ; ϕ 1 = 0.2 ; p =0.098 MPa, m w = 5 kg, t w = 20 C ; the state after moistening =? The specific humidity can be determined using the i-x diagram for state 1: x 1 = kg/kg s.v. The amount of dry air The increase in specific humidity Δx = x 2 - x 1 = m w / m v = 5 / 991 = kg/kg s.v. x 2 = x 1 + Δx = = kg/kg s.v. During moistening, the slope of the straight line is equal to the enthalpy of the moistening medium (i.e. water) i w = 83.7 kj/kg. For i w and x 2 we get for state 2: i 2 = 65 kj/kg s.v. ; t 2 =27.9 C 17. Heat transfer Example 17.1 Calculate the thickness of a concrete wall with a heat conductivity of λ = 1.28 W/(m.K), through which heat flows with a magnitude q = 349 W/m 2. The temperatures on the two sides of the wall are t 1 = 20 C, t 2 = -20 C. Solution : λ = 1.28 W/(m.K); q = 349 W/m 2 ; t 1 = 20 C ; t 2 = -20 C ; δ =? From the heat flow: 25

26 Example 17.2 The lining inside a boiler comprises two layers (δ 1 = 0.35 m, λ 1 = 1.4 W/(m.K), δ 2 = 0.25 m, λ 2 = 0.55 W/(m.K)). The temperature of the inside surface is t 1 = 900 C, the temperature of the outside surface is t 3 = 90 C. Calculate the heat lost and the temperature between the layers. δ 1 = 0.35 m, λ 1 = 1.4W/(m.K), δ 2 = 0.25 m, λ 2 = 0.55 W/(m.K), t 1 = 900 C, t 3 = 90 C ; q =?, t 2 =? The heat flow through the double wall is From the heat flow through the first layer we get Example 17.3 Calculate the heat load q [W/m] and the heat diffusion coefficient k [W/(m.K)], temperatures t S1, t S2 of the wall of the pipe of the heat exchanger. The dimensions of the pipe are: d 1 = 0.1m, d 2 = m. The mean temperatures of the media are t i = 350 C, t e = 1500 C. The heat transfer coefficients on the inside surface of the pipe are α 1 = 300, α 2 = 110 W/(m 2.K). The heat conductivity coefficient of the material of the pipe is λ = 45 W/(m.K). d 1 = 0.1 m ; d 2 = m ; t i = 350 C ; t e = 1500 C, α 1 = 300 W/(m2.K), α 2 = 110 W/(m 2.K) ; λ = 45 W/(m.K) ; q =? ; k =? ; t S1 =? t S2 =? The heat flow through the double cylindrical wall is From the heat flow from the ambient 1 to the wall of the exchanger q = α 1 2 π r 1 (t 1 - t i ) we get The temperature is Example 17.4 Horizontal piping of a diameter of d = m is heated by a natural flow of air. The mean temperature of the surface of the piping is t s = 4 C, the temperature of the surrounding air is t o = 36 C. Calculate the heat transfer coefficient from the air to the wall of the piping. Solution : d = m ; t s = 4 C, t o = 36 C, α =? For the heat transfer by natural convection the Nu number is calculated as Nu = C ( Gr Pr ) n 26

27 The "best-guess" (i.e. mean) temperature is t = ( t s + t o )/ 2 = ( ) / 2 = 20 C Δt = t s - t o = 36-4 = 32 C The physical parameters of air at this temperature are λ = W/(m.K) υ = m 2 /s Pr = β = 1/293 1/K Grashof's number is Gr Pr = = This value determines the transition mode, for which C = 0.54, n = 1/4 Then The heat transfe coefficient is Příklad 17.5 Insulated piping covered with a layer of paint passes through a room whose walls have a temperature of 27 C. The diamater of the piping is d = 0.1 m, the length of the piping is l = 8 m. Calculate the heat transferred by radiation if the surface temperature is 7 C. How does this heat change if the piping passes through a 0.2 x 0.2 m casing whose internal surface is covered by aluminium paint. t 1 = 27 C ; d = 0,1 m ; l = 8 m, t 2 = 7 C ; Q =? ; Q =? - for the 0.2 x 0.2 m casing The heat radiated is (for ε 1 = 0.39 ) Q = ε 1 c o S 1 [(T 1 /100) 4 - (T 2 /100) 4 ] = π.0.1. [ ( 300 / 100 ) 4 -( 280/100 ) 4 ] = 112 W Q = ε n c o S l [ (T 1,/100) 4 - (T' 2 /100) 4 ] = π [(300 / 100) 4 - (280/100) 4 ] = 430 W where for ε 1 =0.92 ; ε 2 = 0.39 Example 17.6 Inside a heat exchanger, black oil is cooled from t 1 = 300 C to t 2 = 200 C and crude oil is simultaneously heated from t 1 = 25 C to t 2 = 175 C. Calculate the mean temperature gradient inside this exchanger in the case of a) parrallel flow, b)counterflow. What is the difference between the surface areas of the exchanger in the cases of a) and b) if, in both cases, the heat transfers and the heat transfer coefficients are the same. t 1 = 300 C; t 2 = 200 C; t 1 = 25 C; t' 2 = 175 C ; Δt s =?; Δt p =?; S s /S p =? a ) The parallel flow exchanger The mean logarithmic temperature gradient is 27

28 b) The counterflow exchanger The mean logarithmic temperature gradient is The ratio of the surface areas in the case of parallel flow and counterflow with the same heat transfer and heat transfer coefficients is S s / S p = Δtp /Δts = / =

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20 m neon m propane. g 20. Problems with solutions:

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