Thermodynamics I Lecture 1 PDF

Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Chapter one 1-1 Introduction Thermodynamics: the science that deals with heat and work and those properties of matter that relate to heat and work. Its study of t...

Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Chapter one 1-1 Introduction Thermodynamics: the science that deals with heat and work and those properties of matter that relate to heat and work. Its study of the relationship between work, heat, and energy. Deals with the conversion of energy from one form to another. Deals with the interaction of a system and it surroundings..‫ ھﻮ اﻟﻌﻠﻢ اﻟﺬي ﯾﺘﻌﺎﻣﻞ ﻣﻊ اﻟﺤﺮارة واﻟﺸﻐﻞ وﺧﺼﺎﺋﺺ اﻟﻤﺎدة اﻟﺘﻲ ﺗﺘﻌﻠﻖ ﺑﺎﻟﺤﺮارة واﻟﺸﻐﻞ‬: ‫اﻟﺪﯾﻨﺎﻣﯿﻜﺎ اﻟﺤﺮارﯾﺔ‬.‫ دراﺳﺘﮫ ﻟﻠﻌﻼﻗﺔ ﺑﯿﻦ اﻟﺸﻐﻞ واﻟﺤﺮارة واﻟﻄﺎﻗﺔ‬.‫ ﯾﺘﻌﺎﻣﻞ ﻣﻊ ﺗﺤﻮﯾﻞ اﻟﻄﺎﻗﺔ ﻣﻦ ﺷﻜﻞ إﻟﻰ آﺧﺮ‬.‫ ﯾﺘﻌﺎﻣﻞ ﻣﻊ ﺗﻔﺎﻋﻞ اﻟﻨﻈﺎم وﻣﺤﯿﻄﮫ‬ Engineers concerned with power generating machinery should have a working knowledge of all matters dealing with the conversion of heat energy into work or power. The laws based on experimental results obtained from the study of gases and vapours are useful in the design of boilers, steam engines, steam turbines, internal combustion engines, gas turbines, refrigerating machines and air compressors. In the present days of industrialization, demand for energy is increasing rapidly. It is, therefore, necessary to design and operate thermal plants and machines at their highest level of performance for efficient utilization of fuels and natural resources available. ‫ﯾﺠﺐ أن ﯾﻜﻮن ﻟﺪى اﻟﻤﮭﻨﺪﺳﯿﻦ اﻟﻤﮭﺘﻤﯿﻦ ﺑﺂﻻت ﺗﻮﻟﯿﺪ اﻟﻄﺎﻗﺔ ﻣﻌﺮﻓﺔ ﻋﻤﻠﯿﺔ ﺑﺠﻤﯿﻊ اﻷﻣﻮر اﻟﻤﺘﻌﻠﻘﺔ ﺑﺘﺤﻮﯾﻞ‬ ‫ اﻟﻘﻮاﻧﯿﻦ اﻟﻤﺴﺘﻨﺪة إﻟﻰ اﻟﻨﺘﺎﺋﺞ اﻟﺘﺠﺮﯾﺒﯿﺔ اﻟﺘﻲ ﺗﻢ اﻟﺤﺼﻮل ﻋﻠﯿﮭﺎ ﻣﻦ دراﺳﺔ‬.‫اﻟﻄﺎﻗﺔ اﻟﺤﺮارﯾﺔ إﻟﻰ ﻋﻤﻞ أو طﺎﻗﺔ‬ ‫اﻟﻐﺎزات واﻷﺑﺨﺮة ﻣﻔﯿﺪة ﻓﻲ ﺗﺼﻤﯿﻢ اﻟﻐﻼﯾﺎت واﻟﻤﺤﺮﻛﺎت اﻟﺒﺨﺎرﯾﺔ واﻟﺘﻮرﺑﯿﻨﺎت اﻟﺒﺨﺎرﯾﺔ وﻣﺤﺮﻛﺎت اﻻﺣﺘﺮاق‬ ‫ ﯾﺘﺰاﯾﺪ اﻟﻄﻠﺐ ﻋﻠﻰ‬، ‫ ﻓﻲ أﯾﺎم اﻟﺘﺼﻨﯿﻊ اﻟﺤﺎﻟﯿﺔ‬.‫اﻟﺪاﺧﻠﻲ واﻟﺘﻮرﺑﯿﻨﺎت اﻟﻐﺎزﯾﺔ وآﻻت اﻟﺘﺒﺮﯾﺪ وﺿﻮاﻏﻂ اﻟﮭﻮاء‬ ‫ ﻣﻦ اﻟﻀﺮوري ﺗﺼﻤﯿﻢ وﺗﺸﻐﯿﻞ اﻟﻤﺤﻄﺎت واﻵﻻت اﻟﺤﺮارﯾﺔ ﺑﺄﻋﻠﻰ ﻣﺴﺘﻮى ﻣﻦ اﻷداء‬، ‫ ﻟﺬﻟﻚ‬.‫اﻟﻄﺎﻗﺔ ﺑﺴﺮﻋﺔ‬.‫ﻣﻦ أﺟﻞ اﻻﺳﺘﺨﺪام اﻟﻔﻌﺎل ﻟﻠﻮﻗﻮد واﻟﻤﻮارد اﻟﻄﺒﯿﻌﯿﺔ اﻟﻤﺘﺎﺣﺔ‬ 1 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-2 The thermodynamics laws are: 0th Law ⇒ Defines Temperature (T) 1st Law ⇒ Defines Energy (U) 2nd Law ⇒ Defines Entropy (S) 3rd Law ⇒ Gives Numerical Value to Entropy Zeroth law of thermodynamics: states that if two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. ،‫ﯾﻧص اﻟﻘﺎﻧون اﻟﺻﻔري ﻟﻠدﯾﻧﺎﻣﯾﻛﺎ اﻟﺣرارﯾﺔ أﻧﮫ إذا ﺗواﺟد ﺟﺳﻣﺎن ﻓﻲ ﺣﺎﻟﺔ ﺗوازن ﺣراري ﻣﻊ ﺟﺳم ﺛﺎﻟث‬ ‫ وﯾﺛﺑت اﻟﻘﺎﻧون اﻟﺻﻔري أن درﺟﺔ اﻟﺣرارة ھﻲ ﺧﺎﺻﯾﺔ أﺳﺎﺳﯾﺔ ﻟﻠﻣﺎدة‬،‫ﻓﮭﻣﺎ أﯾﺿًﺎ ﻓﻲ ﺣﺎﻟﺔ ﺗوازن ﻣﻊ ﺑﻌﺿﮭﻣﺎ‬.‫وھﻲ ﻗﺎﺑﻠﺔ ﻟﻠﻘﯾﺎس‬ T A = TB , TA=TC so TB=TC First law of thermodynamics: one of the most fundamental laws of nature is the conservation of energy principle. It simply states that during an interaction, energy can change from one form to another but the total amount of energy remains constant. ‫ إﻧﮫ ﯾﻨﺺ ﺑﺒﺴﺎطﺔ‬.‫ أﺣﺪ أھﻢ ﻗﻮاﻧﯿﻦ اﻟﻄﺒﯿﻌﺔ وھﻮ ﻣﺒﺪأ اﻟﺤﻔﺎظ ﻋﻠﻰ اﻟﻄﺎﻗﺔ‬:‫اﻟﻘﺎﻧﻮن اﻷول ﻟﻠﺪﯾﻨﺎﻣﯿﻜﺎ اﻟﺤﺮارﯾﺔ‬.‫ ﯾﻤﻜﻦ أن ﺗﺘﻐﯿﺮ اﻟﻄﺎﻗﺔ ﻣﻦ ﺷﻜﻞ إﻟﻰ آﺧﺮ ﻟﻜﻦ اﻟﻜﻤﯿﺔ اﻹﺟﻤﺎﻟﯿﺔ ﻟﻠﻄﺎﻗﺔ ﺗﻈﻞ ﺛﺎﺑﺘﺔ‬، ‫ﻋﻠﻰ أﻧﮫ أﺛﻨﺎء اﻟﺘﻔﺎﻋﻞ‬ 2 ‫‪Thermodynamics I‬‬ ‫‪lecture 1‬‬ ‫‪Dr. Yasmeen H.Abed‬‬ ‫‪Second law of thermodynamics: energy has quality as well as quantity, and actual‬‬ ‫‪processes occur in the direction of decreasing quality of energy. The second law of‬‬ ‫‪thermodynamics introduces a new property called entropy, S.‬‬ ‫‪Entropy is a measure of the randomness of the system or it is the measure of‬‬ ‫‪energy or chaos within an isolated system. It can be considered as a quantitative‬‬ ‫‪index that describes the quality of energy.‬‬ ‫اﻟﻘﺎﻧﻮن اﻟﺜﺎﻧﻲ ﻟﻠﺪﯾﻨﺎﻣﯿﻜﺎ اﻟﺤﺮارﯾﺔ ھﻮﻋﺪم إﻣﻜﺎﻧﯿﺔ اﻧﺘﻘﺎل اﻟﺤﺮارة ﻣﻦ ﺟﺴﻢ ﺑﺎرد اﻟﻰ ﺟﺴﻢ ﺳﺎﺧﻦ وﻟﻜﻦ اﻟﻌﻜﺲ‬ ‫ھﻮ اﻟﺼﺤﯿﺢ‪ ،‬ﻓﺎﻟﺤﺮارة ﺗﻨﺘﻘﻞ ﻣﻦ اﻟﺠﺴﻢ اﻟﺴﺎﺧﻦ اﻟﻰ اﻟﺠﺴﻢ اﻟﺒﺎرد‪.‬واﻟﻄﺎﻗﺔ اﻟﻤﻮﺟﻮدة ﻓﻲ ﻧﻈﺎم ﻣﻌﺰول ﺗﻨﺘﺸﺮ‬ ‫وﺗﺘﻮزع ﻓﯿﮫ ﺑﺎﻟﺘﺴﺎوي ﻣﻊ ﻣﺮور اﻟﺰﻣﻦ‪.‬‬ ‫وﻟﺬﻟﻚ اﻧﺘﺸﺎر اﻟﻄﺎﻗﺔ ﻓﻲ ﻧﻈﺎم ﯾﻌﻨﻲ ان ﺗﻤﯿﻞ اﻻﺧﺘﻼﻓﺎت ﻓﻲ ﺗﺮﻛﯿﺰ اﻟﻄﺎﻗﺔ ان ﺗﺨﺘﻔﻲ ﺑﻤﺮور اﻟﻮﻗﺖ ‪،‬ﻓﺘﺘﺴﺎوي‬ ‫درﺟﺔ اﻟﺤﺮارة‪ ،‬وﯾﺘﺴﺎوي اﻟﻀﻐﻂ ‪ ،‬وﺗﺘﺴﺎوي اﻟﻜﺜﺎﻓﺔ‪.‬وھﻜﺬا اﻻﻧﺘﺮوﺑﻲ أﺣﺪ ھﺬه اﻟﺨﺼﺎﺋﺺ ﯾﻤﻜﻦ أﺧﺬھﺎ ﻣﻘﯿﺎس‬ ‫ﻻﻧﺘﺸﺎر اﻟﻄﺎﻗﺔ أو اﻟﺤﺮارة‪.‬وﻟﺬﻟﻚ اﻟﻘﺎﻧﻮن اﻟﺜﺎﻧﻲ ﯾﺘﻌﻠﻖ ﺑﺎﻻﻧﺘﺮوﺑﻲ )اﻟﻔﻮﺿﻰ(‪.‬ﯾﺸﯿﺮ اﻟﻘﺎﻧﻮن اﻟﺜﺎﻧﻲ إﻟﻰ أن‬ ‫ﻋﻤﻠﯿﺎت اﻟﺪﯾﻨﺎﻣﯿﻜﺎ اﻟﺤﺮارﯾﺔ أي اﻟﻌﻤﻠﯿﺎت اﻟﺘﻲ ﺗﺘﻀﻤﻦ ﻧﻘﻞ أو ﺗﺤﻮﯾﻞ اﻟﻄﺎﻗﺔ اﻟﺤﺮارﯾﺔ‪ ،‬ﻏﯿﺮ ﻋﻜﺴﯿﺔ ﻷﻧﮭﺎ ﺟﻤﯿﻌًﺎ‬ ‫ﺗﺆدي إﻟﻰ زﯾﺎدة ﻓﻲ اﻹﻧﺘﺮوﺑﻲ ) زﯾﺎدة اﻟﻔﻮﺿﻰ(‪.‬‬ ‫‪Third law of thermodynamics: The 3rd law of thermodynamics will essentially‬‬ ‫‪allow us to make "sense" of entropies.‬‬ ‫ﯾﻨﺺ اﻟﻘﺎﻧﻮن اﻟﺜﺎﻟﺚ ﻋﻠﻰ أﻧﮫ ﻟﻼﻧﺘﺮﺑﻲ ﻣﻘﯿﺎس ﻣﺤﺴﻮس ﯾﻤﻜﻦ ﺣﺴﺎﺑﮫ ‪.‬ﻋﻨﺪﻣﺎ ﺗﻘﺘﺮب درﺟﺔ ﺣﺮارة اﻟﻨﻈﺎم ﻣﻦ‬ ‫اﻟﺼﻔﺮ اﻟﻤﻄﻠﻖ ‪ ،‬ﯾﺼﺒﺢ اﻹﻧﺘﺮوﺑﻲ ﺛﺎﺑﺘﺔ ‪ ،‬أو أن اﻟﺘﻐﯿﺮ ﻓﻲ اﻹﻧﺘﺮوﺑﯿﺎ ھﻮ ﺻﻔﺮ ‪.‬ﻋﻨﺪ اﺿﺎﻓﺔ اﻟﺤﺮارة ﺗﺰداد‬ ‫اﻻﻧﺘﺮوﺑﻲ ) ﺗﺰداد اﻟﻔﻮﺿﻰ( ‪.‬‬ ‫‪3‬‬ Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-3 Dimensions and units Any physical quantity can be characterized by dimensions. The arbitrary magnitudes assigned to the dimensions are called units. There are two types of dimensions, primary or fundamental and secondary or derived dimensions..‫ ﺗﺴﻤﻰ اﻟﻤﻘﺎدﯾﺮ اﻟﻤﺨﺼﺼﺔ ﻟﻸﺑﻌﺎد ﺑﺎﻟﻮﺣﺪات‬.‫ﯾﻤﻜﻦ ﺗﻤﯿﯿﺰ أي ﻛﻤﯿﺔ ﻣﺎدﯾﺔ ﺑﺎﻷﺑﻌﺎد‬.‫ واﻟﺜﺎﻧﻮﯾﺔ أو اﻟﻤﺸﺘﻘﺔ‬، ‫ اﻷﺑﻌﺎد اﻷوﻟﯿﺔ أو اﻷﺳﺎﺳﯿﺔ‬، ‫ھﻨﺎك ﻧﻮﻋﺎن ﻣﻦ اﻷﺑﻌﺎد‬ Primary dimensions are mass, m; length, L; time, θ; temperature, T Secondary dimensions are the ones that can be derived ‫ ﻣﺴﺘﻤﺪ‬from primary dimensions such as: velocity (m/s), pressure (Pa = kg/m.s2), Volume m3 , energy kg m2/s2 There are two unit systems currently available:  USCS (United States Customary System) or English (British) system.  SI (International System) English system has no apparent systematic numerical base, and various units in this system are related to each other rather arbitrarily (12 in =1 ft , 1 mile =5280 ft , etc.), which makes it confusing and difficult to learn. We, however, will use SI units exclusively in this course. The SI units are based on decimal relationship ‫ اﻟﻌﻼﻗﺔ اﻟﻌﺸﺮﯾﺔ‬between units. The prefixes used to express the multiples of the various units are listed in Table 1-1, Table 1-1 4 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 5 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-4 Unit Conversion Ex 1: convert 3 km/h to m/s Ex 2: convert 255 mL to L 1 km= 1000 m L=1000 m L 1 h= 60 min 1min=60 s mL=0.001 L 3 × × × =. 255𝑚𝐿 × = 0.255 𝐿 1000 𝑚 3000 𝑚 =3 = = 0.833 𝑚/𝑠 3600 𝑠 3600 𝑠 Ex 3: convert 0.028 km to cm Ex 4: convert 8432 L to ML km=1000 m ML=𝟏𝟎𝟔 L 1m=100 cm 8432 𝐿 × = 0.028 𝑘𝑚 × × = 2800 𝑐𝑚 = 8432 × 10 𝑀𝐿 = 0.008432 𝑀𝐿 Ex 5: convert 45 mile/h to m/s 1mi=5280 ft , 1m=3.281 ft , 1min=60 s , 1h=60min 𝑚𝑖 5280 𝑓𝑡 1𝑚 1ℎ 1 𝑚𝑖𝑛 237600 𝑚 𝑚 45 × × × × = = 20.1 ℎ 1 𝑚𝑖 3.281 𝑓𝑡 60𝑚𝑖𝑛 60 𝑠 13761 𝑠 𝑠 Ex 6: convert 9 ton/h to lbm/min 1 ton= 1000 kg 1h=60 min 1 kg= 2.205 lbm 9 𝑡𝑜𝑛 1000 𝑘𝑔 ℎ lbm lbm × × × = 68.027 ℎ 𝑡𝑜𝑛 60 𝑚𝑖𝑛 2.205 𝑘𝑔 𝑘𝑔 Important note: in engineering all equations must be dimensionally homogenous. This means that every term in an equation must have the same units. It can be used as a sanity check for your solution. ‫ ھﺬا ﯾﻌﻨﻲ أن ﻛﻞ ﻣﺼﻄﻠﺢ ﻓﻲ‬.‫ ﻓﻲ اﻟﮭﻨﺪﺳﺔ ﯾﺠﺐ أن ﺗﻜﻮن ﺟﻤﯿﻊ اﻟﻤﻌﺎدﻻت ﻣﺘﺠﺎﻧﺴﺔ اﻷﺑﻌﺎد‬: ‫ﻣﻼﺣﻈﺔ ﻣﮭﻤﺔ‬.‫ ﯾﻤﻜﻦ اﺳﺘﺨﺪام ﻛﻮﺳﯿﻠﺔ ﻟﻠﺘﺤﻘﻖ ﻣﻦ ﺳﻼﻣﺔ اﻟﺤﻞ اﻟﺨﺎص ﺑﻚ‬.‫اﻟﻤﻌﺎدﻟﺔ ﯾﺠﺐ أن ﯾﻜﻮن ﻟﮫ ﻧﻔﺲ اﻟﻮﺣﺪات‬ 6 ‫‪Thermodynamics I‬‬ ‫‪lecture 1‬‬ ‫‪Dr. Yasmeen H.Abed‬‬ ‫‪1-4 Definitions‬‬ ‫‪System: is a quantity of matter or a region in space‬‬ ‫‪chosen for study.‬‬ ‫‪Surroundings: Everything outside the system‬‬ ‫‪boundary.‬‬ ‫‪Boundary: The surface dividing the System from the‬‬ ‫‪Surroundings.‬‬ ‫اﻟﻨﻈﺎم‪ :‬ھﻮ ﻛﻤﯿﺔ ﻣﻦ اﻟﻤﺎدة أو ﻣﻨﻄﻘﺔ ﻓﻲ اﻟﻔﻀﺎء ﯾﺘﻢ اﺧﺘﯿﺎرھﺎ ﻟﻠﺪراﺳﺔ‪.‬‬ ‫اﻟﺒﯿﺌﺔ اﻟﻤﺤﯿﻄﺔ‪ :‬ﻛﻞ ﺷﻲء ﺧﺎرج ﺣﺪود اﻟﻨﻈﺎم‪.‬‬ ‫اﻟﺤﺪود‪ :‬اﻟﺴﻄﺢ اﻟﺬي ﯾﻔﺼﻞ اﻟﻨﻈﺎم ﻋﻦ اﻟﻤﺤﯿﻂ‪.‬‬ ‫‪1-Closed System (control mass ): fixed non-changing mass of fluid within the‬‬ ‫‪system, i.e., no mass transfer across the system boundary but can have energy‬‬ ‫‪exchange with the surroundings in the form of heat or work, can cross the‬‬ ‫‪boundary, and the volume of a closed system does not have to be fixed.. Example:‬‬ ‫‪piston-cylinder assembly‬‬ ‫اﻟﻨﻈﺎم اﻟﻤﻐﻠﻖ ) ﻛﺘﻠﺔ ﺛﺎﺑﺘﺔ( ‪ :‬ﻛﺘﻠﺔ ﺛﺎﺑﺘﺔ ﻏﯿﺮ ﻣﺘﻐﯿﺮة ﻣﻦ اﻟﺴﻮاﺋﻞ داﺧﻞ اﻟﻨﻈﺎم ‪ ،‬أي ﻻ ﯾﻮﺟﺪ اﻧﺘﻘﺎل ﻟﻠﻜﺘﻠﺔ ﻋﺒﺮ‬ ‫ﺣﺪود اﻟﻨﻈﺎم وﻟﻜﻦ ﯾﻤﻜﻦ أن ﯾﻜﻮن ﻟﮭﺎ ﺗﺒﺎدل ﻟﻠﻄﺎﻗﺔ ﻣﻊ اﻟﻤﻨﺎطﻖ اﻟﻤﺤﯿﻄﺔ ﻓﻲ ﺷﻜﻞ ﺣﺮارة أو ﺷﻐﻞ ‪ ،‬وﯾﻤﻜﻨﮭﺎ‬ ‫ﻋﺒﻮر اﻟﺤﺪود ‪ ،‬اﻟﺤﺠﻢ ﻓﻲ اﻟﻨﻈﺎم اﻟﻤﻐﻠﻖ ﻻ ﯾﺠﺐ أن ﯾﻜﻮن ﺛﺎﺑﺘﺎ‪.‬ﻣﺜﻞ اﺳﻄﻮاﻧﺔ اﻟﻤﻜﺒﺲ‪ ،‬ﻻ ﯾﻤﻜﻦ ﻟﻜﺘﻠﺔ اﻟﻐﺎز‬ ‫ﻋﺒﻮر ﺣﺪودھﺎ‪.‬أي ‪ ،‬ﻻ ﯾﻤﻜﻦ ﻷي ﻛﺘﻠﺔ أن ﺗﺪﺧﻞ أو ﺗﺘﺮك اﻟﻤﻜﺎن اﻟﻤﻐﻠﻖ ﻓﻲ اﻻﺳﻄﻮاﻧﺔ ‪ ،‬ﻟﻜﻦ ﺗﺘﺴﺮب اﻟﺤﺮارة‬ ‫اﻟﻰ اﻟﻤﺤﯿﻂ و ﯾﻤﻜﻦ اﻻﺳﺘﻔﺎدة ﻣﻦ ﺣﺮﻛﺔ اﻟﻤﻜﺒﺲ ﻛﺸﻐﻞ ﻟﻨﻈﺎم اﺧﺮ‪.‬‬ ‫‪7‬‬ ‫‪Thermodynamics I‬‬ ‫‪lecture 1‬‬ ‫‪Dr. Yasmeen H.Abed‬‬ ‫‪2-Open System (Control Volume): fixed volume in space, the mass and energy‬‬ ‫‪exchange permitted across the system boundary it usually encloses a device that‬‬ ‫‪involves mass flow such as a compressor, turbine, or nozzle.‬‬ ‫اﻟﻨﻈﺎم اﻟﻤﻔﺘﻮح )ﺣﺠﻢ ﺛﺎﺑﺖ(‪ :‬اﻟﺤﺠﻢ اﻟﺜﺎﺑﺖ ﻓﻲ اﻟﻨﻈﺎم ‪،‬ﻟﻜﻦ ﯾﺴﻤﺢ ﺑﺘﺒﺎدل اﻟﻜﺘﻠﺔ واﻟﻄﺎﻗﺔ ﻋﺒﺮ ﺣﺪود اﻟﻨﻈﺎم ‪،‬‬ ‫ﻋﺎدةً ﻣﺎ ﻧﺮاه ﻓﻲ ﺟﮭﺎز ﯾﺘﻀﻤﻦ ﺗﺪﻓﻘًﺎ ﻣﺴﺘﻤﺮ ﺑﺎﻟﻜﺘﻠﺔ ﻣﺜﻞ اﻟﻀﺎﻏﻂ ‪ ،‬اﻟﺘﻮرﺑﯿﻦ ‪ ،‬ﻓﻮھﮫ )اﻟﻨﻮزل( ‪ ،‬او ﺳﺨﺎن اﻟﻤﺎء‪.‬‬ ‫ﻓﻲ ﺳﺨﺎن اﻟﻤﺎء ﻛﺘﻠﺔ اﻟﻤﺎء ﺗﺪﺧﻞ و ﺗﺨﺮج ﻟﻜﻦ ﺣﺠﻢ اﻟﻤﻜﺎن ﺛﺎﺑﺖ ‪.‬ﻓﻲ اﻟﻔﻮھﮫ ﺣﺪود اﻟﻨﻈﺎم ﺗﻜﻮن ﺛﺎﺑﺘﺔ ﺣﻘﯿﻘﯿﺔ‬ ‫ﻣﺘﻤﺜﻠﺔ ﺑﺠﺪار اﻟﻔﻮھﮫ اﻣﺎ ﺣﺪود دﺧﻮل و ﺧﺮوج اﻟﻜﺘﻠﮫ ﻓﻲ ﺣﺪود ﺧﯿﺎﻟﯿﺔ‪.‬ﻓﻲ اﻟﻀﺎﻏﻂ ﯾﻜﻮن اﺣﺪ ﺣﺪود اﻟﻨﻈﺎم‬ ‫ﻣﺘﺤﺮﻛﺔ واﻻﺧﺮى ﺛﺎﺑﺘﮫ‪.‬‬ ‫‪3- Isolated System: Special case of closed system that does not interact at all with‬‬ ‫‪the surroundings, e.g., no heat transfer across system boundary.‬‬ ‫اﻟﻨﻈﺎم اﻟﻤﻌﺰول‪ :‬ﺣﺎﻟﺔ ﺧﺎﺻﺔ ﻣﻦ اﻟﻨﻈﺎم اﻟﻤﻐﻠﻖ ﻻ ﯾﺘﻔﺎﻋﻞ‬ ‫ﻋﻠﻰ اﻹطﻼق ﻣﻊ اﻟﺒﯿﺌﺔ اﻟﻤﺤﯿﻄﺔ ‪ ،‬ﻋﻠﻰ ﺳﺒﯿﻞ اﻟﻤﺜﺎل ‪ ،‬ﻋﺪم‬ ‫اﻧﺘﻘﺎل اﻟﺤﺮارة ﻋﺒﺮ ﺣﺪود اﻟﻨﻈﺎم‪.‬‬ ‫‪8‬‬ Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 4- Adiabatic system: A closed or open system that does not exchange energy with the surroundings by heat , but still work could be. A process can be adiabatic by some other ways, either the process is very fast, so that there is no time for heat transfer to take place, or the temperature difference between system and surrounding is zero, or the system is insulated in some way. ‫ ﺳواء اﻟﻧظﺎم ﻣﻐﻠق أو ﻣﻔﺗوح ﻻ ﯾﺗم ﺗﺑﺎدل اﻟطﺎﻗﺔ ﻣﻊ اﻟﻣﺣﯾط ﺑﺎﻟﺣرارة ﻻ ﻣن داﺧﻠﮫ وﻻ ﻣن‬: ‫اﻟﻧظﺎم اﻻدﯾﺑﺎﺗﻲ‬ ‫ و ﻋدم ﺗﺑﺎدل اﻟﺣرارة ھو اﻣﺎ ﺑﺳﺑب ﺣدوث اﻟﻌﻣﻠﯾﺔ ﺑﺳرﻋﺔ وﻻ ﯾوﺟد اﻟوﻗت‬.‫ ﻓﻘط ﺣدوث ﺷﻐل‬، ‫ﺧﺎرﺟﮫ‬ ‫ او ﯾﻛون اﻟﻧظﺎم ﻣﻌزول ﺑطرﯾﻘﺔ ﻣﺎ‬، ‫اﻟﻛﺎﻓﻲ ﻟﻠﺗﺑﺎدل او ﯾﻛون اﻟﻧظﺎم ﻣﺣﺎط ﺑوﺳط ذات درﺟﺔ ﺣرارة ﻣﺳﺎوﯾﺔ ﻟﮫ‬.‫ﻋن اﻟﺑﯾﺋﺔ اﻟﻣﺣﯾطﺔ ﺑﮫ‬ Important note: some thermodynamics relations that are applicable to closed and open systems are different. Thus, it is extremely important to recognize the type of system we have before start analyzing it..‫ ﺑﻌض اﻟﻌﻼﻗﺎت اﻟدﯾﻧﺎﻣﯾﻛﯾﺔ اﻟﺣرارﯾﺔ ﺗﻛون ﻣﺧﺗﻠﻔﺔ و ﺗﻧطﺑق ﻋﻠﻰ اﻷﻧظﻣﺔ اﻟﻣﻐﻠﻘﺔ واﻟﻣﻔﺗوﺣﺔ‬: ‫ﻣﻼﺣظﺔ ﻣﮭﻣﺔ‬.‫ ﻣن اﻟﻣﮭم ﻟﻠﻐﺎﯾﺔ اﻟﺗﻌرف ﻋﻠﻰ ﻧوع اﻟﻧظﺎم ﻟدﯾﻧﺎ ﻗﺑل اﻟﺑدء ﻓﻲ ﺗﺣﻠﯾﻠﮭﺎ‬، ‫وﺑﺎﻟﺗﺎﻟﻲ‬ Working substance: the substance contained within the boundaries of the system, and the various processes can be done on it, for example steam , air ….. Pure substance: the homogenous substance that has fixed chemical composition whether it is a single or a multi elements..‫ ﻣﺛل اﻟﺑﺧﺎر واﻟﮭواء‬، ‫ وﯾﻣﻛن إﺟراء اﻟﻌﻣﻠﯾﺎت اﻟﻣﺧﺗﻠﻔﺔ ﻋﻠﯾﮭﺎ‬، ‫ اﻟﻣﺎدة اﻟﻣوﺟودة داﺧل ﺣدود اﻟﻧظﺎم‬:‫ﻣﺎدة اﻟﻌﻣل‬.‫ اﻟﻣﺎدة اﻟﻣﺗﺟﺎﻧﺳﺔ ذات اﻟﺗرﻛﯾب اﻟﻛﯾﻣﯾﺎﺋﻲ اﻟﺛﺎﺑت ﺳواء ﻛﺎﻧت ﻣﻔردة أو ﻣﺗﻌددة اﻟﻌﻧﺎﺻر‬:‫اﻟﻣﺎدة اﻟﻧﻘﯾﺔ‬ Systems Surface free to do work 9 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-6 Properties of a System In classical thermodynamics, the substance is assumed to be a continuum, homogenous matter with no microscopic holes. This assumption holds as long as the volumes, and length scales are large with respect to the intermolecular spacing. System property: Any characteristic of a system substance used to describe the system condition and can be measured like pressure, temperature, volume is called a property. ‫ ﯾﺴﺘﻤﺮ‬.‫ ﯾُﻔﺘﺮض أن ﺗﻜﻮن اﻟﻤﺎدة ﻣﺘﺼﻠﺔ وﻣﺘﺠﺎﻧﺴﺔ ﺑﺪون ﺛﻘﻮب ﻣﺠﮭﺮﯾﺔ‬، ‫ﻓﻲ اﻟﺪﯾﻨﺎﻣﯿﻜﺎ اﻟﺤﺮارﯾﺔ اﻟﻜﻼﺳﯿﻜﯿﺔ‬.‫ھﺬا اﻻﻓﺘﺮاض طﺎﻟﻤﺎ ﻛﺎﻧﺖ اﻷﺣﺠﺎم وﻣﻘﺎﯾﯿﺲ اﻟﻄﻮل ﻛﺒﯿﺮة ﻓﯿﻤﺎ ﯾﺘﻌﻠﻖ ﺑﺎﻟﺘﺒﺎﻋﺪ ﺑﯿﻦ اﻟﺠﺰﯾﺌﺎت‬. (property ‫ ﺗﺴﻤﻰ أي ﺧﺎﺻﯿﺔ ﻟﻤﺎدة اﻟﻨﻈﺎم اﻟﻤﺴﺘﺨﺪﻣﺔ ﻟﻮﺻﻒ ﺣﺎﻟﺔ اﻟﻨﻈﺎم ﺑـ ) ﺧﺎﺻﯿﺔ‬: ‫ﺧﺎﺻﯿﺔ اﻟﻨﻈﺎم‬.‫وﯾﻤﻜﻦ ﻗﯿﺎﺳﮭﺎ ﻣﺜﻞ اﻟﻀﻐﻂ ودرﺟﺔ اﻟﺤﺮارة واﻟﺤﺠﻢ‬ There are two types of properties: 1. Intensive property : the property is independent of system mass (Value may not vary throughout the system), e.g., pressure, temperature ,density ‫ اﻟﻀﻐﻂ ودرﺟﺔ اﻟﺤﺮارة واﻟﻜﺜﺎﻓﺔ‬، ‫ ﻋﻠﻰ ﺳﺒﯿﻞ اﻟﻤﺜﺎل‬، ‫ﺧﻮاص ﻣﺴﺘﻘﻠﺔ ﻋﻦ ﻛﺘﻠﺔ اﻟﻨﻈﺎم‬ 2. Extensive property: the property value for the system is the sum of the values of the parts into which the system is divided (depends on the system size) e.g., U, V,S,H,W, Q. ،V ،U ،m ‫ھﻲ ﻣﺠﻤﻮع ﻗﯿﻢ اﻷﺟﺰاء اﻟﺘﻲ ﯾﻘﺴﻢ إﻟﯿﮭﺎ اﻟﻨﻈﺎم )ﯾﻌﺘﻤﺪ ﻋﻠﻰ ﺣﺠﻢ اﻟﻨﻈﺎم( ﻋﻠﻰ ﺳﺒﯿﻞ اﻟﻤﺜﺎل‬ Q ،W ،H ،S 10 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Specific property : which is defined as extensive property per unit mass. Like , v =V/m specific volume , u=U/m specific internal energy , s=S/m specific isotropy.. 1-7 Processes and Cycles System state : condition of the thermodynamic system as described by its properties (P1,T1,….) Phase : A quantity of matter which is homogenous in chemical composition can exist in three phases (solid, liquid ,gas) Process : Is the transformation of a system from one state to another state, for example heating ,cooling , compression , expansion…… Path : The series of states that a system passes throughout during a process..‫ ﺳﻠﺴﻠﺔ ﺗﻐﯿﺮ اﻟﺤﺎﻻت اﻟﺘﻲ ﯾﻤﺮ ﺑﮭﺎ اﻟﻨﻈﺎم ﺧﻼل اﻟﻌﻤﻠﯿﺔ‬:‫اﻟﻤﺴﺎر‬ Cycle : Is a sequence of processes that begins and ends at the same state..‫ ھﻲ ﺳﻠﺴﻠﺔ ﻣﻦ اﻟﻌﻤﻠﯿﺎت ﺗﺒﺪأ وﺗﻨﺘﮭﻲ ﻓﻲ ﻧﻔﺲ اﻟﺤﺎﻟﺔ‬:‫اﻟﺪورة‬ 11 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-8 Pressure Pressure is the force exerted by a fluid per unit area Force: A body when start moving, brought to rest or change direction it is said to exert a force on it. Newton's second law gives 𝐹 = 𝑚 ∗ 𝑎 where m=mass (kg) ∴𝐹 =𝑚∗ =𝑚 =𝑚 (𝑘𝑔 = 𝑁) Newton(N): is the force required to accelerate a unit mass at rate one meter per squared second Weight :is the attraction force between a body and earth. Weight is normally change with altitude and position. 𝑚 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚 ∗ 𝑔 𝑘𝑔. = 𝑁 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝜌 𝑉 ∗ 𝑔 𝑠 g: (gravitational acceleration) =9.81 m/s2 𝒎 𝑭 𝒌𝒈. 𝟐 𝑵 𝒔 Pressure = = ( = = 𝐏𝐚) 𝑨 𝒎𝟐 𝒎𝟐 𝑚 Work = 𝐹∗𝑥 ( 𝑘𝑔. ∗𝑚 = 𝑁∗𝑚 = 𝐽) 𝑠2 In fluids, gases and liquids, we speak of pressure; in solids this is stress. Prove 𝑃 =𝜌𝑔ℎ P= = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝐴 𝑚𝑔 𝜌 𝑉 𝑔 𝑃 = = 𝐴 𝐴 𝜌𝐴ℎ𝑔 = 𝐴 𝑃=𝜌𝑔ℎ 12 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Units: 1 Pa = 1 N/m2 standard atmosphere = 101,325 Pa 1 bar = 100,000 Pa = 100 kPa = 0.1 MPa a-Atmospheric pressure (Patm.) The local atmospheric pressure, Patm. measurement via a mercury Barometer. 𝑃 =𝜌 ∗ 𝑔 ∗ ℎ ( 760 mm) = 13600*9.81*0.76 =101325 Pa =101.325 kPa =1.01325 bar = 1 atm The density of mercury is 13.5 times the density of water Note: 1 standard atmosphere = 760 mm (29.9 in.) mercury = 10,260 mm water b-Gauge pressure ,( Pg. ): A pressure measured relative to the local atmospheric pressure, Patm. It is (+,-) sign Gauge pressure measurement via a manometer. The cross sectional area of the tube is not important. 13 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Bourdon Tube is a device that measures pressure using mechanical deformation. Pressure Transducers are devices that use piezoelectric to measure pressure. Pressure Transduce rs c-Absolute pressure , Pabs. : Pressure measured relative to a perfect vacuum.(+) 𝑃 =𝑃 ∓𝑃 In thermodynamics calculations, always use absolute pressure. Most pressure measuring devices are calibrated to read zero in the atmosphere. 14 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-9 Temperature and temperature scale Temperature :measure of hotness or coldness it not easy to give an exact definition for it level of temperature freezing cold, cold , warm, hot ,red-hot. Temperature scale based on some easily reference state such as freezing and boiling points of water. Ice point : is the temperature of mixture of ice and water that is in equilibrium with air saturated with vapour at a pressure of 1 atm. Steam point: is the temperature of a mixture of liquid water and water vapour in equilibrium at a pressure of 1 atm. Means of Temperature Measurement All temperature measurements are based on the change of certain property with temperature, such means can be listed here as ; :‫وﺳﺎﺋﻞ ﻗﯿﺎس درﺟﺔ اﻟﺤﺮارة‬ ‫ وﯾﻤﻜﻦ‬، ‫ﺗﺴﺘﻨﺪ ﺟﻤﯿﻊ ﻣﻘﺎﯾﯿﺲ درﺟﺔ اﻟﺤﺮارة ﻋﻠﻰ ﺗﻐﯿﯿﺮ ﺧﺎﺻﯿﺔ ﻣﻌﯿﻨﺔ ﻣﻦ ﺧﻮاص اﻟﻤﺎدة ﻣﻊ درﺟﺔ اﻟﺤﺮارة‬ ‫إدراج ھﺬه اﻟﻮﺳﺎﺋﻞ ﻋﻠﻰ اﻟﻨﺤﻮ اﻟﺘﺎﻟﻲ ؛‬  Volumetric expansion of a liquid (bulb thermometer)  Pressure exerted by gas (gas thermometer)  Vapour pressure of liquids(steam pressure thermometer)  Electric resistance of solids (resistance thermometer)  Thermoelectricity (thermocouple thermometer)  Thermal radiation (pyrometer) (‫اﻟﺘﻤﺪد اﻟﺤﺠﻤﻲ ﻟﻠﺴﺎﺋﻞ )ﻣﺤﺮار ﺑﺼﻠﻲ‬  (‫اﻟﻀﻐﻂ اﻟﻐﺎز )ﻣﻘﯿﺎس ﺣﺮارة اﻟﻐﺎز‬ (‫ﺿﻐﻂ ﺑﺨﺎر اﻟﺴﻮاﺋﻞ )ﻣﻘﯿﺎس ﺣﺮارة ﺿﻐﻂ اﻟﺒﺨﺎر‬ (‫اﻟﻤﻘﺎوﻣﺔ اﻟﻜﮭﺮﺑﺎﺋﯿﺔ ﻟﻠﻤﻮاد اﻟﺼﻠﺒﺔ )ﻣﺤﺮار اﻟﻤﻘﺎوﻣﺔ‬ (‫اﻟﻜﮭﺮﺑﺎء اﻟﺤﺮارﯾﺔ )ﻣﻘﯿﺎس ﺣﺮارة ﻣﺰدوج‬ (‫اﻹﺷﻌﺎع اﻟﺤﺮاري )اﻟﺒﯿﺮوﻣﺘﺮ‬ 15 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Temperature scales There are four major temperature scales that are used around the world – Fahrenheit and Celsius are frequently used in everyday, around the house measurements, while the absolute zero-based Kelvin and Rankine scales are more commonly used in industry and the sciences. A- The Celsius scale to measure temperatures. The Celsius scale is sometimes referred to as the centigrade scale, because it is based on a 100 degree division between the freezing and boiling points of water: water freezes at 0 degrees Celsius and boils at 100 degrees C. B- The Kelvin scale was adapted from the Celsius scale. Kelvin was designed in order to set the zero point of the temperature scale at absolute zero. To convert from Celsius to Kelvin by adding 273.15 to a Celsius temperature. Water freezes at 273.15 K, and boils at 373.15 K. Because of its direct relation to absolute zero, Kelvin temperature is widely used in scientific equations and calculations. For instance, the ideal gas law, uses Kelvin as its standard unit. T(K) = t(℃) + 273.15 , 0℃ = 273.15K , ∆T(K) = ∆t(℃) C- The Fahrenheit scale On this scale, the interval between the water freezes at 32 degrees Fahrenheit, and boils at 212 degrees F. It is divided to 180 degree. The Fahrenheit temperature scale includes negative temperatures, below 0 degrees F 9 t(℉) = t(℃) + 32 5 D- Rankine scale provides an absolute zero-based equivalent to the Fahrenheit scale. Essentially, it is for the Fahrenheit scale what Kelvin is for Celsius. Temperatures can be converted from Fahrenheit to Rankine by adding 459.67. Absolute zero is thus located at 0 degrees Rankine. Water freezes at 491.67 degrees R, and boils at 671.67 degrees R. T(R) = t(℉) + 459.67 T(R) = T(K) 16 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Microscopic point of view: Temperature is a measure of the internal molecular motion, e.g., average molecule kinetic energy At a temperature of –273.15C molecular motion ceases Temperature in units of degrees kelvin, K, is measured relative to this absolute zero temperature, so 0 K = -273C ‫ ﻣﺘﻮﺳﻂ‬، ‫ ﻋﻠﻰ ﺳﺒﯿﻞ اﻟﻤﺜﺎل‬، ‫ درﺟﺔ اﻟﺤﺮارة ھﻲ ﻣﻘﯿﺎس ﻟﻠﺤﺮﻛﺔ اﻟﺠﺰﯾﺌﯿﺔ اﻟﺪاﺧﻠﯿﺔ‬:‫وﺟﮭﺔ ﻧﻈﺮ ﺗﺤﺖ اﻟﻤﺠﮭﺮ‬ ‫ ﺗﺘﻮﻗﻒ اﻟﺤﺮﻛﺔ اﻟﺠﺰﯾﺌﯿﺔ ﻋﻦ درﺟﺔ‬، ‫ درﺟﺔ ﻣﺌﻮﯾﺔ‬- 273.15 ‫اﻟﻄﺎﻗﺔ اﻟﺤﺮﻛﯿﺔ ﻟﻠﺠﺰيء ﻋﻨﺪ درﺟﺔ ﺣﺮارة‬ ‫ ﻟﺬﻟﻚ‬، ‫ ﯾﺘﻢ ﻗﯿﺎﺳﮭﺎ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺪرﺟﺔ ﺣﺮارة اﻟﺼﻔﺮ اﻟﻤﻄﻠﻖ‬،K ، ‫اﻟﺤﺮارة ﺑﻮﺣﺪات درﺟﺔ ﻛﻠﻔﻦ‬ ‫ درﺟﺔ ﻣﺌﻮﯾﺔ‬273- = ‫ﺻﻔﺮ ﻛﻠﻔﻦ‬ 1-10 State and Equilibrium At a given state, all the properties of a system have fixed values. Thus, if the value of even One property changes, the state will change to different one. In an equilibrium state, there are no unbalanced potentials (or driving forces) within the system. condition of the thermodynamic system as described by its properties (P1,T1,….) A system is at steady-state if none of the system properties change with time A system is in equilibrium if when the system is isolated from its surroundings there are no changes in its properties Example: thermal equilibrium 17 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-11 Kilocalorie The measurement of heat is called calorimetry. The calorie, or gram calorie, is the quantity of heat required to raise the temperature of 1 gram of pure water 1°C. The kilocalorie, or kilogram calorie, is the quantity of heat required to raise the temperature of 1 kg of pure water 1°C; it is equal to 1,000 cal. 1 J= 0.239 Cal 1-12 Volume Volume: The space occupied by the substance, if increases than it expands, or decreases thus it is compressed. Unit (m3) and for liquid (L) , 1m3=1000L Specific volume: which is defined as volume per unit mass. That is, v, =V/m m3/kg Density : is defined as mass per unit volume 𝒎 𝒌𝒈 𝝆= 𝒎𝟑 𝑽 The reciprocal of density is the specific volume 𝟏 𝝆= 𝒗 1-13 The mass flow rate (for open system) (𝑚̇): is the mass flow rate of substance (fluid) which passes per unit time.its units in SI( kg/s) 𝑚̇ =𝜌 𝐶𝐴 Where 𝛒= the density kg/m3 C= the velocity m/s A= the cross section area m2 18 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed 1-14 The volumetric flow rate (for open system) 𝑉̇ : is the volume flow rate of substance (fluid) which passes per unit time. its unit in SI (m3/s) 𝑉̇ = 𝐶𝐴 Where A: cross sectional area of conduit m2 ‫ﻣﺳﺎﺣﺔ ﻣﻘطﻊ اﻻﻧﺑوب او اﻟﻘﻧﺎه‬ C: flow velocity inside the conduit m/s ‫ﺳرﻋﺔ اﻟﻣﺎﺋﻊ داﺧل اﻻﻧﺑوب او اﻟﻘﻧﺎه‬ ̇ / ̇ 𝑉̇ = 𝐶𝐴 = = 𝑣 𝑚̇ = = / 1-15 Examples : Example 1.1. On a piston of 10 cm diameter a force of 1000 N is uniformly applied. Find the pressure on the piston.( N/m2) & (kPa) Solution. Diameter of the piston d = 10 cm (= 0.1 m) Force applied on the piston, F = 1000 N ∴ Pressure on the piston, P = = = 127307 =. kN 127.307. = 127.307kPa 𝑚 Example 1.2. A tube contains an oil of specific gravity 0.9 to a depth of 120 cm. Find the gauge pressure at this depth (in kN/m2). Solution. Specific gravity of oil = 0.9 , 𝑆𝐺 = Depth of oil in the tube, h = 120 cm = (1.2 m) We know that 𝑃 = 𝜌 𝑔 ℎ = (0.9 𝜌 ) ∗ 9.81 ∗ 1.2 𝑁 = 0.9 × 1000 × 9.81 × 1.2 𝑚 𝑁 𝑘𝑁 = 10594.8 = 10.5948 𝑚 𝑚 19 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Example 1.3. The vacuum pressure reading of a tank (Pg) given 15 kPa and atmospheric pressure (Patm) is 750 mmHg. Find the absolute pressure in the tank. Properties The density of mercury is given to be (ρ = 13,590 kg/m3). Solution : Analysis The atmospheric (or barometric) pressure can be expressed as kg m 𝑃𝑎𝑡𝑚 = 𝜌 𝑔 ℎ = 13,590 ∗ 9.81 ∗ 0.75𝑚 = 99,988.425 Pa = 99.988 kPa 𝑚 𝑠 Then the absolute pressure in the tank becomes 𝑃 =𝑃 −𝑃 = 99.988 kPa − 15 kPa ≅ 85 kPa Example 1.4. A pressure gage connected to a tank reads 500 kPa. Find the absolute pressure in the tank if the atmospheric pressure (Patm) is 94 kPa. Solution : 𝑃 =𝑃 +𝑃 = 94𝑘𝑃𝑎 + 500𝑘𝑃𝑎 = 594 𝑘𝑃𝑎 Example 1.5. A vertical piston–cylinder device contains a gas, the piston has a mass of 50 kg and a diameter of 22 cm. Find the absolute pressure (Pabs ) of the gas if the atmospheric pressure (Patm) is 0.97 bar. Solution : 𝑃 =𝑃 +𝑃 𝐹 𝑚∗𝑔 =𝑃 + =𝑃 + 𝐴 𝐴 𝑚 𝐹 𝑚∗𝑔 50 𝑘𝑔 ∗ 9.81 𝑃 = = = 𝜋 𝑠 𝐴 𝐴 (0.22) 𝑚 4 𝑁 1 𝑏𝑎𝑟 = 12,898 = 12,898 Pa = 12,898 ∗ = 0.129 𝑏𝑎𝑟 𝑚 10 ∴ 𝑃 = 0.97𝑏𝑎𝑟 + 0.129 𝑏𝑎𝑟 = 1.099 bar Where = 𝑃𝑎 , 1 𝑏𝑎𝑟 = 10 𝑃𝑎 ∴ 1 𝑃𝑎 = 20 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Example 1.6 A vertical piston–cylinder device contains a gas. Pressure of the gas is be increased by placing some weights on the piston. The piston with the weights has a mass of 2 kg and a diameter of 24 cm. The atmospheric pressure is be measured by a mercury barometric, the mercury high is 750 mm. Determine the pressure gage and the absolute pressure (Pabs ) of the gas in mmHg ،PSI ،bar & kPa units. The density of mercury is given to be (ρ = 13,600 kg/m3). Solution : 0.003253 mHg = 3.253 mmHg mHg = 753.2 mmHg 21 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Example 1.7 A U-tube mercury manometer with one arm open to atmosphere is used to measure pressure in a steam pipe. The level of mercury in open arm is 97.5 mm greater than that in the arm connected to the pipe. Some of steam in the pipe condenses in the manometer arm connected to the pipe. The height of this column is 34 mm. The atmospheric pressure is 760 mm of Hg. Find the absolute pressure of steam in mm Hg, Pa and bar unites. Solution : Equation the pressure in mmHg on both arms above the line X-X , we get 𝑃 + 𝑃 =𝑃 +𝑃 P in mmHg 𝜌 ℎ 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑃 ∶ 𝜌 𝑔 ℎ =𝜌 𝑔 ℎ , ℎ = 𝜌 𝜌 ℎ 1000 ∗ 34 ∴𝑃 𝑖𝑛 ℎ = = = 2.5 𝑚𝑚𝐻𝑔 𝜌 13600 ∴ 𝑃 + 2.5 = 760 + 97.5 ∴ 𝑃 = 855 𝑚𝑚𝐻𝑔 855 𝑚𝑚𝐻𝑔 𝑘𝑔 𝑚 𝑁 = ∗ 13600 ∗ 9.81 = 114,070.68 = 114,070.68 𝑃𝑎 𝑚𝑚𝐻𝑔 𝑚 𝑠 𝑚 1000 𝑚𝐻𝑔 𝑏𝑎𝑟 ∴ 𝑃 = 114,070.68 𝑃𝑎 ∗ 5 = 1.1407 𝑏𝑎𝑟 10 𝑃𝑎 22 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed Example 1.8 Solution : 23 Thermodynamics I lecture 1 Dr. Yasmeen H.Abed H.W 1. The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 kg/m3, what is the height of the building ? 2. A vacuum gage connected to a chamber reads 35 kPa at a location where the atmospheric pressure is 92 kPa. Determine the absolute pressure in the chamber. 3. The absolute pressure in water at a depth of 5 m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 0.85 at the same location. 4. The piston of a vertical piston–cylinder device containing a gas has a mass of 60 kg and a cross-sectional area of 𝟒×𝟏𝟎-2 𝒎𝟐, as shown in Fig. The local atmospheric pressure is 0.97 bar, and the gravitational acceleration is 9.81 m/s 2. (a) Determine the pressure inside the cylinder. (b) If some heat is transferred to the gas and its volume is doubled, do you expect the pressure inside the cylinder to change? 5. convert (20 m3) to ( ft3), Liter & gallon unites. convert (0.0025 lbm/in3) to ( kg/m3) & (slug/ft3) unites. convert (120W) to (Btu/h) & (hp) Note : Btu = British thermal unit of heat 6. The temperature is 45O C , find the Temperature in F , K and R. 24