Heat Transfer Course 2 nd level –Air Conditioning and Refrigeration Branch Power Mechanics Techniques Department Mosul Technical Institute Northern Technical University اعداد: م. احمدسل ابراهيم باتقدم الحرارة المنتقالذ علم استا ا( متقاعد) ح من قبل نق: م. م. ر محمود احمدنما ا مهندس تقني منتصر عبد محمدعملينب اللجا ا: محمد يحيىحكام ا
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Heat Transfer Course2nd level –Air Conditioning and Refrigeration Branch
Power Mechanics Techniques DepartmentMosul Technical Institute
Northern Technical University
:اعداد
باسل ابراهيم احمد.م
(متقاعد)استاذ علم انتقال الحرارة المتقدم
:نقح من قبل
انمار محمود احمد. م.م
مهندس تقني منتصر عبد محمد
:الجانب العملي
احكام محمد يحيى
HEAT TRANSFER 1 الحرارهانتقال
2nd AIR CONDITIONING AND REFRIGERATION
القدرةميكانيكقسم
يدفرع التكييفوالتبر
2018-2019 * Mosul tech. Institute Basil Ibraheem 1
Anmmar
Sticky Note
MigrationConfirmed set by Anmmar
INTRODUCTIONمقدمة
• Heat can be transferred from hot (high
• temperature )to cold (lower temperature) positions .
• Heat transfer methods :
• 1- Conduction; happens in solid bodies and in high viscosity liquids by the following onedimensional steady state equation :
• Q= -k *A*dt/dx …….. Fourier’s Law فوريرقانون
لحرارةا انيمكن تنتقل من ساخن مرتفعةحرارةدرجة
بارد اقلحرارةدرجةمناطق
الحرارةانتقالانواعاوطرق
لتوصيلا يحدث اللزوجةذاتالسوائلوالصلبةالاجسامف
العالية
باتجاهالانتقالمعادلةدرجاتبثبوتوواحد
زمنياالحرارة
لىا
1
* Mosul tech. Institute Basil Ibraheem 2
• Where: انحيث
• Q is heat transferred w
• k is thermal conductivity of the material w/m ⁰c
• A is area normal to the heat direction m²
• dt is the differential (very small) of temperature
⁰c
• dx is the differential of distance m
2
واتالمنتقلةالحرارة
للمادةالحراريالتوصيلمعامل
الحرارةاتجاهعلىالعموديةالمساحة
الحرارةلدرجةالتفاضليجداالصغيرلتغيرا
التيجداالصغيرةالتفاضليةالمسافة
الحرارهتقطعها
* Mosul tech. Institute Basil Ibraheem 3
2- Convection الحراريالحملتيارات
• Convection happens between solid body surface and fluid when there is a temperature difference and can be calculated by the following equation:
• Q=h*A*∆t
• Q is heat transferred w
• h is heat transfer coefficient w/m² c
• A is the area normal to heat direction m²
• ∆t is the temperature difference
• between surface and fluid ⁰c
• الحراريالحملتياراتوصلبسطحبي تحدث فرقوجودعندمائع
ف يمكنوالحرارةدرجات
للمعادلةوفقاحسابهالتالية
• المنتقلهالحرارة
• الحرارةانتقالمعامل
• لاتجاهالعموديةالمساحهالحرارةانتقال
• بي الحرارةدرجاتفرقالمائعوالسطح
3
* Mosul tech. Institute Basil Ibraheem 4
3 – Heat radiation الحراريالاشعاع• Heat radiation can be transferred through vacuum and
transparent medias (air, clear water and glass) .radiation can be calculated by the following equation :
• Q = σ *ϵ*A *(T₁⁴ -T₂⁴)
• Where
• Q is heat transferred by radiation w
• σ is Stefan-Boltzman constant =5.67*10⁻⁸ w/m² K⁴
• ϵ is the body emissivity ( =1 for black bodies )
• A is body surface area m²
• T₁ is the absolute hot body temperature =⁰c+273 K
• T₂ is the absolute cold body temperature =⁰c+273 K
واغالفر خلالينتقلالاشعاع•اءالهو مثلالشفافةالاجسام
ويحسبالزجاجوالماءوالمعادلةحسب
يالاشعاعالمنتقلةالحرارة•
مان-ستيفنثابت• بولت
مللجسالحراريةالانبعاثية•السوداءللاسطح 1=
للجسمالسطحيةالمساحة•
مللجسالمطلقةالحرارةدرجة•الساخن
مللجسالمطلقةالحرارةدرجة•للاشعاعالمستلمالبارد
4
* Mosul tech. Institute Basil Ibraheem 5
Tutorial (examples)❑ 300 watt Electrical heater of an area (A) 0.1m² and temperature (t1)
=200 c is in contact to a metal of same area and has a thickness (X) =
0.01m. The outer surface temperature (t2) 195 c. Calculate the thermal
conductivity (K).
❑ Answer:
❑ Q= - K*A* Δt /X
❑ Q = 300 watt , A= 0.1m² , t1 =200 c , t2 = 195 c , X = 0.01m
Heat conduction through composite plane wall الجدرانخلالالموصلهالحراره
المركبةالمستوية
• For Three layers composite plane wall there will be three thermal resistance .
• R1 =x1/k1 x1 k1 x2 k2 x3 k3
• R2= x2/k2
• R3 =x3/k3
• Total wall thermal resistance
• ∑ R= R1+R2+R3
• q= (t1-t4) / ∑ R• الحراريهالمقاوماتمجموع
• Where
• x1,x2 and x3 are the layers
• thicknesses t1 t2 t3 t4
• k1,k2 and k3 are the layers thermal conductivities
• t1,t2 t3 and t4 are temperature at each layer surface
• t1 is the highest temperature
• Q= q*A
المركبالمستويللجدار•توجدطبقاتثلاثمن
حراريةمقاوماتثلاث
=الكليةالحراريةالمقاومة•مجموع
المقاومات
الحرارية
طبقةكلسمك
لكلالحراريالتوصيلمعاملطبقة
طبقةلكلالسطححرارةدرجةةالطبقوجهعندحراربةدرجةاعلى
الاولى
الموصلهالكليةالحرارة
* Mosul tech. Institute Basil Ibraheem 12
Example: A composite plane wall has an inner layer thickness of(x₁) 0.025m with (k₁) thermal conductivity =3.1w/m⁰c,the intermediate layer has 0.2m
thickness and thermal conductivity 5w/m⁰c .the thickness of outer layer =0.03m and thermal conductivity =4.5w/m⁰c.for inner temperature was 25⁰c and outer temperature =2⁰c. Find out total heat conduction (Q) if wall width
=4m and wall height =3m
• x1=0.025 x2=0.2 x3=0.3m• To find layers thermal resistance
• R₁=x₁/k₁=0.025/3.1=0.008m² ⁰c/w
• R₂=x₂/k2=0.2/5=0.04m² ⁰c/w
• R₃=x₃/k₃=0.03/4.5=0.066m² ⁰c/w
• q = (t₁ -t₄)/∑ R
• q = (25 – 2)/(0.008+0.04+0.066)
• q =23/0.114 =201.75 w/m²
• Q=q*A=201.75*(4*3) = 2421 w
الجدارنرسم•
والسمكنثبت•التوصيلمعاملطبقةلكل
المقاوماتنحسب•ةطبقلكلالحرارية
درجةنستخدم•t₁المرتفعةالحرارة
معالمعادلةنطبق•فرقانملاحظةالحرارةدرجاتموجب
مساحةه 3*4•الجدار
الارتفاع*العرض)
k2k1 k3
t1 t2 t3 t4
وسطي
* 13Mosul tech. Institute Basil Ibraheem
* 14
Q1: A-Calculate the heat flux(q) and the total heat transferred from a brick plane wall has
area (A)=10m2 thermal conductivity(k) = 0.6 w/m ⁰c with inner temperature of(t1)=26⁰c
and outer temperature of(t2)= 6 ⁰c and wall thickness(x) =0.2m .
B-Radiation can be calculated by the following equation: -------
Q2: A- Heat can be transferred from ----------to------ positions.
B- Temperature is taken in------- for calculating conduction heat transfer.
Q3: A composite plane wall has inner layer thickness of(x₁) 0.035m with (k₁) thermal
=5w/m⁰c.for inner temperature (t1) = 35⁰c and outer temperature (t3) =2⁰c find out heat
conduction flux (q).
المقدمهوالحراريالتوصيلعناسئله
Mosul tech. Institute Basil Ibraheem
* 15
Q4: A brick plane wall of thermal conductivity (k) = 0.6 w/m ⁰c with inner temperature of
(t1) = 24⁰c and an outer temperature (t2) of 3⁰c and wall thickness (x) =0.3m. Calculate:
A- The conduction heat flux (q).
B- The total heat transferred (Q) for wall area (A) =12m2 .
Q5: Complete the following sentences:
A; The unit of thermal resistance for cylindrical walls is --------.
B; Convection can be calculated by the following equation ----------.
C; Conduction happens in --------.
D; Heat radiation is calculated by the equation -------------.
Q6: A furnace has an inside surface temperature (t1) = 400⁰C and outer steel shield
temperature (t3) of 50⁰c. The furnace is to be constructed by Firebrick of thermal
conductivity (k1) =1 w/m.⁰c), thickness (x1) =0.08m and steel shield thickness (x2) of 0.01m
(k2 =63w/m.⁰c). Find out heat conducted (q) and the intermediate temperature (t2).
المقدمهوالحراريالتوصيلعناسئله
Mosul tech. Institute Basil Ibraheem
HEAT TRANSFER 2 الحرارهانتقال
2nd AIR CONDITIONING AND REFREGERATION
القدرةميكانيكقسم
يدفرع التكييفوالتبر
* Mosul tech. Institute Basil Ibraheem 1
How to find the intermediate temperatures t₂and t₃ الطبقاتبينالحرارةدرجاتنجدكيف
* 2Mosul tech.inst
• Example: A composite plane wall has inner layer thickness of (x₁) 0.025m with thermal conductivity(k₁) =3.1w/m⁰c, the intermediate layer has thickness (x2) = 0.2m and thermal conductivity(k2) 5w/m⁰c .The thickness of outer layer(x3) =0.03m and thermal conductivity(k3) =4.5w/m⁰c.For inner temperature was (t1) 25⁰c and outer temperature (t4) =2⁰c. Find out 1- The conduction heat flux (q) 2- the intermediate temperatures (t2) and (t3).
Conduction through cylindrical walls(pipes) خلالالحراريالتوصيل
الاسطوانيةالجدران (الانابيب)• Thermal resistance of cylindrical
(pipes) walls is :
• R=ln(r₂/r₁) /2*π*k*L• Where:
• ln is the natural logarithm
• r₂ is the outer pipe radius m
• r₁ is the inner pipe radius m
• π =22/7=3.142
• k is the pipe thermal conductivity
• w/m⁰c
• L is the pipe length m
• Total heat conducted through pipe is: Q= (t₁ - t₂)/R W
• t₁ is the high temperature ⁰c• t₂ is the lower temperature ⁰c
• الاسطوانيةللجدرانالحراريةالمقاومة
• الطبيع للاساسلوغاريتم
• الخارجر القطرنصف
• الداخل القطرنصف
• دائرةايقطرومحيطبي الثابتهالنسبة
• الانبوبلمادةالتوصيلمعامل
• الانبوبطول
• الكليةالحرارة
• العاليةالحرارةدرجة r₁
r₂k
L
* 4Mosul tech.inst basil ibraheem
Example : Calculate the heat transferred through a steel pipe with thermal conductivity of 63 w/m⁰c , inner radius =0.1m and outer radius of 0.12m if the pipe length is 20m and inner temperature=60⁰c and outer one is 50⁰c.
• k=63w/m⁰c , r₁=0.1m , r₂=0.12m
• L=20m , t₁=60⁰c , t₂=50⁰c
• Thermal resistance=R=ln(r₂/r₁) /2*π*k*L
• R= ln(0.12/0.1)/2*π*63*20
• R=0.182/7920=2.29e-5 ⁰c/w
• Q=(t₁-t₂)/R
• Q= (60-50)/2.29*10⁻⁵ =436681 w• Home work : calculate the heat loss for similar in dimensions copper
pipe (k=398 w/m⁰c) and same inner and outer temperatures. Ans . :(2673831w)
• Thermal resistance of cylindrical (pipes) walls is-----------
• The unit of Thermal resistance of cylindrical (pipes) walls is --------
• Total heat conducted through pipe can be calculated by the equation ---------
• نبوبلا نحاسبنفسوالابعاد
الفرقنفس
درجاتفالحرارة
* 5Mosul tech.inst
* 6Mosul tech.inst
Conduction through Composite cylindrical walls المركبةالاسطوانيةالجدرانخلالالتوصيل
• Inner pipe thermal resistance R₁
• R₁=ln(r₂/r₁)/2*π*k₁*L
• Outer pipe thermal resistance R₂
• R₂=ln(r₃/r₂)/2*π*k₂*L
• Where:
• r₂ is inner pipe outer radius الخارجر القطرنصفالداخل للانبوب
• r₁ is inner pipe inner radius الداخل القطرنصفالداخل للانبوب
• r₃ is the outer radius of outer pipe نصفالخارجر للانبوبالخارجر القطر
• k₁ is thermal conductivity of inner pipe الداخل للانبوبالحراريالتوصيلمعامل
• k₂ is thermal conductivity of outer pipe الخارجر للانبوبالحراريالتوصيلمعامل
• Outer pipe
• Inner pipe
• Q=(t₁-t₃)/(R₁+R₂)• t₁ is the high temperature
• t₃ is the lower temperature
r₁
r₂
r₃
L
* 7Mosul tech.inst
R1 R2
qt1 t2 t3
Example: calculate the conduction heat transfer (Q) from a steel pipe (k₁=64w/m⁰c) has inner radius(r₁) of 0.02m and outer radius (r₂)=0.022m.The pipe was covered with insulation material glass wool (k₂= 0.04w/m⁰c)the outer insulating pipe radius =0.07m .The steel inside surface temperature=80⁰c and outside cover temperature =25⁰c .The pipe length (L)=60m
• Ans:
• R₁ =ln(r₂/r₁)/2*π*k₁*L
• R₁ =ln(O.O22/0.02)/2π*64*60
• =0.095/2*3.142*64*60
• = 3.95*10⁻⁶ ⁰c/w
• R₂= ln(r₃/r₂)/2*π*k₂*L
• = ln(0.07/0.022)/2*3.142*0.04*60
• =1.157/15.079
• =0.076 ⁰c/w
• Q= (t₁-t₃)/(R₁+R₂)
• Q=(80-25)/(3.95*10⁻⁶ + 0.076)
• Q=55/0.076003=723.64w
• انبوبي السؤالهذاف
• ساخنسائللنقلالفولاذمنالداخل
• العزللاغراضزجاجبصوفمغطىالحراري
• الانبوبالعازلالغطاءيشكل الثان
• الداخل قطرهنصف القطرنصف = للانبوبالخارجر
المعدن
• Home work: Find out the intermediate temperature (t₂) for the previous example.
• Insulating materials are used to decrease heatconduction and thus decrease surface temperatures. a gas, like air, can be a good insulator if it can be kept from moving when it is
heated or cooled. A vacuum is an excellent
insulator
• the engineering approach to insulation is the addition of a low-conducting material to the surface…
• Insulation materials types:• 1-alumina-silica thermal conductivities in the
range of 0.1–0.2 W/m⁰c
• 2-glass fibers insulation thermal conductivities can range from about 0.03–0.06 W/m⁰c
• 3-polyurethane, is light in weight, shows a very low thermal conductivity (about 0.02 W/m⁰c)
Heat Transfer by Convection2nd air-conditioning branch
اعذاد
تاسم اتزاهى احذ
اار يحىد احذ
يرصز عثذ محمد
2/27/2020 free conv. MOSUL Tech. Inst. 1
Heat transfer by convectionارقال انحزارج
تراراخ انحم
Convection
تارات الحمل
Free convection (natural)
تارات الحمل الحراري
نتجة )الطبع(اختلاف الكثافة
Forced convection
تارات حمل جبرة
بواسطة المراوح او المضخات
• The aim of studying convection is : دراسة الهدف من
• How to find heat transfer coefficient (h)كفة اجاد معامل انتمال الحرارة
• Heat transfer coefficient(h) depends onعتمد معامل الانتمال على
• surface shape شكل السطح افم عمودي
اسطوان
• ,temperature T,درجة الحرارة
• fluid density ρ كثافة المائع,
• Thermal conductivity k التوصل معامل
الحراري للمائع
• Fluid viscosity μ لزوجة المائع
2free conv. MOSUL Tech. Inst.2/27/2020
3
Heat convection
Natural or Free Heat
ConvectionForced Heat Convection (fan moved air or pump moved water)
free conv. MOSUL Tech. Inst.2/27/2020
Free (natural) convectiondimensionless groups انجايع انلاتعذح انلاسيح
نحساب يعايم ارقال انحزارج تانحم انحز• 1- Grashof number (Gr)• The ratio of buoyancy and viscous force.• Gr=ρ² *L³ *g *β *∆t/첕 ρ is fluid density kg/m³ • L is the height for vertical wall m• or average of width (a)and length (b) for horizontal
wall =(a+b)/2
• g =9.8m/s²• β is thermal expansion coefficient for
gases =1/Tf K⁻ᶦ
• Tf film temperature is the average of surface and fluid temperatures =(Ts+T∞ )/2 K
• ∆t is the temperature difference of surface and fluid = Ts-T∞ or T∞ -Ts
• μ is the fluid viscosity kg/m.s
• ( بدون وحدات)اهم المجامع اللابعدةهو عدد كراشوف و الذي مثل النسبة
نتجة لاختلاف درجة )بن لوة الطوفان .و لوة الناتجة عن اللزوجة ( الحرارة
• كثافة المائع عند درجة حرارة (المعدل)
• الارتفاع ف حالة الاسطج العمودة
• او معدل الطول و العرض للاسطح الافمة
• التعجل الارض
• معامل التمدد الحجم للمائع
• مملوب معدل درجة =وهو للغازات الحرارة المطلمة المتاخمة
• درجة = درجة الحرارة المتاخمة المطلمةدرجة حرارة +حرارة السطح المطلمة
2(/المائع المطلمة
• الفرق الموجب بن درجة حرارة السطح و المائع
• لزوجة المائع عند درجة حرارة المعدل
4free conv. MOSUL Tech. Inst.2/27/2020
Example: find out Grashof number (Gr) for air having temperature (t∞)=25⁰c with vertical wall of
temp.(ts)=100⁰c if wall height =3m• Tf =(Ts+T∞ )/2 • Ts=100+273=373k• T∞ =25+273=298k• Tf =(373+298)/2=335.5k • β=1/335.5=0.0029 K⁻ᶦ• Gr= ρ² *L³ *g *β *∆t/첕 ρ = 0.995kg/m³ FROM AIR TABLE• μ = 208.2*10⁻⁷kg/m.s FROM AIR TABLE at
350K the nearest value to 335.5K• ∆t=373-298= 75K= 75⁰C• Gr= 0.995² *3³ *9.8 *0.0029 *75 /(208.2*10⁻⁷)²• 56.97/ 0.000000000433= 131427052795=13*10¹⁰
• FOR Gr ˃ 10⁹ the flow is turbulent
• احسب درجة الحرارة المطلمة المتاخمة
• = معامل التمدد الحجم مملوب الدرجة المطلمة المتاخمة
• لاجاد عدد كراشوف• تؤخذ الكثافةمن جدول
خصائص الهواءعند درجة حرارة مماربة لدرجة الحرارة المطلمة المتاخمة وكذلن بالنسبة للزوجة
• اذا كان عدد كراشوف اكثر من
• 10⁹ • فهذا عن ان حركة المائع
اضطرابة
5free conv. MOSUL Tech. Inst.2/27/2020
T Temp.
(K)
ρdensity
(kg/m3)
Cp
specific heat
(kJ/Kg.K)
μviscosity
(10-7kg/m.s)
K thermal
conductivity
(10-3W/m.K)
Pr Prandtlnumber
100 3.5562 1.032 71.1 9.34 0.786
150 2.3364 1.012 103.4 13.8 0.758
200 1.7458 1.007 132.5 18.1 0.737
250 1.3947 1.006 159.6 22.3 0.72
300 1.1614 1.007 184.6 26.3 0.707
350 0.995 1.009 208.2 30.0 0.700
400 0.8711 1.014 230.1 33.8 0.690
450 0.7740 1.021 250.7 37.3 0.686
500 0.6964 1.030 270.1 40.7 0.684
Thermo physical properties of airالهواء
6free conv. MOSUL Tech. Inst.2/27/2020
T Temp.
(K)
ρdensity
(kg/m3)
Cp
specific heat
(kJ/Kg.K)
μviscosity
(kg .m/s)
k thermal
conductivity
(W/m.K)
Pr Prandtlnumber
βvolume
expansion coefficient
(10-4/K)
260 908 1.76 12.23 0.149 144500 7
280 896 1.83 2.17 0.146 27200 7
300 884 1.91 0.486 0.144 6450 7
320 872 1.99 0.141 0.141 1990 7
340 860 2.08 0.053 0.139 795 7
360 848 2.16 0.025 0.137 395 7
380 836 2.25 0.014 0.136 230 7
400 824 2.34 0.009 0.134 155 7
Thermo physical properties of engine oil زت المحركات
7free conv. MOSUL Tech. Inst.2/27/2020
2-Prandtl number (Pr)
• Prandtl number (Pr): is ratio between
momentum diffusivity and thermal diffusivity
• Pr =μ *cp /k
• cp is specific heat at constant pressure at fluid
film temperature Tf
J/kg.⁰c
• عذد تزاذذم ثم انسثح ت الارشار
اناشء ي انعشو انحزك و الارشار
انحزاري
• انحزارج انىعح تثثىخ انضغظ و
ذؤخذ هذ انخاصح ي انجذاول عذ
درجح انحزارج انراخح
8free conv. MOSUL Tech. Inst.2/27/2020
Example: Calculate Prandtl number(Pr) for the previous exampleاحسة عذد تزاذذم نهثال انساتك
• Ans.:
• Pr =μ *cp /k
• μ = 208.2*10⁻⁷kg/m.s FROM AIR TABLE at 350K the nearest
value to 335.5K
• Cp= 1.009kJ/kg⁰c =1009J/kg⁰c FROM AIR TABLE at 350K
the nearest value to 335.5K
• k= 30.0*10⁻³ w/m⁰c
• Pr = 208.2*10⁻⁷ *1009 / 30.0*10⁻³ = 0.700246
• Home work: find out Gr and Pr for the same previous conditions but with
engine oil احسة عذدي كزاشىف و تزاذذم نفس انضزوف انساتقح ونك انجذار انعىدي يلايس نشد
• 3-Nusselt number (Nu) : is the ratio between total heat transfer by convection and the estimated conductive heat transfer.
• Nu=h*L/k• Then : HEAT TRANSFER COEFFICIENT (h) is
• h=Nu*k/L
• k is fluid thermal conductivity at Tf (FROM TABLE)
• From experiments (empirical formula)
• Nu= c *(Gr*Pr)ⁿ
• c & n are experimental constants
• For vertical wall
• Nu= 0.15 *(Gr*Pr)⁰῾³³ for Gr*Pr 10⁷ to 10¹¹
10free conv. MOSUL Tech. Inst.2/27/2020
Example: Find out the heat transfer coefficient for the vertical wall described in the previous example then calculate the heat transferred by
free convection for wall area of 9 m²
• Ans.
• Gr=13*10¹⁰
• Pr = 0.700246
• Gr*Pr= 91031980000= 9.1*10¹⁰
• Nu= 0.15 *(Gr*Pr)⁰῾³³
• Nu= 0.15 *(9.1*10¹⁰)⁰῾³³
• Nu= 0.15*4135= 620.2
• h=Nu*k/L
• h=620.2* 30.0*10⁻³/3
• h=18.79/3 = 6.26 w/m².⁰c
• Q=h*A*∆t
• Q= 6.26* 9* (100 – 25) = 4228 w
• Home work:
• Find out heat transfer coefficient (h) for a horizontal plate (3m length,4m width) exposed from the top to air at temperature =25⁰c and the plate temperature =100⁰c.
• Note: ( Nu= 0.27*(Gr*Pr)⁰²⁵ )
• احسب معامل انتمال الحرارة لصفحة
• متر 4متر و عرضها 3طولها
• معرضة من الاعلى لهواء
11free conv. MOSUL Tech. Inst.2/27/2020
Empirical formulas for free convection calculations
• Reynolds number (Re) : Ratio between inertial and viscous forces. النسبة بن لوةالزخم و لوة اللزوجة
• Re= V*ρ*D/μ• Where• V is fluid velocity m/s السرعة • ρ is fluid density at fluid temperature (t∞) kg/m³• D is the pipe diameter (in case of pipes) m • is the flat plate length ( in case of flat plates)• μ is the fluid viscosity at (t∞) kg/m.s
• The following empirical equations can be used for finding forced convection heat transfer coefficient (h) for flow inside pipes
• Nu= 0.023*Re⁰∙⁸ *Pr⁰∙⁴ when pipe surface temperature (ts) ˃ fluid temp. t∞)
• Nu= 0.023*Re⁰∙⁸ *Pr⁰∙³ if ts ˂ t∞ ( cooling)
2/27/2020 13forced convection mosul technical institute
Calculate heat transfer coefficient (h) for air flowing inside pipe has diameter (D)=0.1m and temperature (ts) of 40⁰c .the air
temperature(t∞) =20⁰c and velocity(V) =3m/s
• Ans.
• velocity(V) =3m/s , diameter (D)=0.1m , air temperature(t∞) =20⁰c FROM AIR PROPERTIES TABLES AT T∞ =20+273 =293 K ≈300 K
• Question: Parallel flow heat exchanger has hot side fluid cph=
6000j/kg.⁰c inlet temperature th1=80⁰c,outlet temperature th2=58⁰c with a flow rate mh = 1kg/sec. the cold side fluid specific heat cpc=4200j/kg.⁰c ,flow rate mc =0.6 kg/s, inlet temperature tc1=2⁰c .
• A- calculate LMTD .
• B- calculate heat exchanger area if the overall heat transfer coefficient U =300w/m².⁰c
• C –if the pipe diameter =0.03m find out the heat exchanger length.
• Question : repeat the above question for counter flow heat exchanger .
2/27/2020forced convection Mosul technical
institute 22
Heat transfer in extended surface (fins)ارقال انحزارج ف الاسطح انرذج انشعاف
2/27/2020forced convection Mosul
technical institute23
Q=√(h*p*k*Aь)*(tь-t∞)
معامل انتمال الحرارة
(الزعنفة)محط السطح
معامل التوصل للزعنفة
مساحة لاعدة الزعنفة
فرق درجات الحرارة بن لاعدة الزعنفة و المحط
الحرارة المنتملة
Example: Calculate the heat transferred through rectangular fin has thickness(t)=0.001m and width (w) of 0.05m ,fin metal
thermal conductivity (k) =64w/m.⁰c heat transfer coefficient (h)=5w/m².⁰c , fin base temperature (tь)=180⁰c and
atmospheric temperature(t∞)=20⁰c.
2/27/2020forced convection Mosul
technical institute24
Ans.
Fin base area (Aь)= w*t =0.05*0.001 =0.00005 m²
Fin perimeter (p)= (w+t)*2 =(0.05+0.001)*2 =0.102 m
Heat transfer by the fin (Q) =√(h*p*k*Aь) *(tь-t∞)
=√5*0.102*64*0.00005 *(180-20)
= 0.04 *160
=6.46 watt
HOME WORK• Q: A- Calculate Grashof number (Gr) for air at temperature (t∞) = 25 oc surrounding a
vertical wall of height (L) =2m and surface temperature (ts) =29 oc.
• B- For calculation of free convection in horizontal walls L= ---------.
•
• Q: Find out heat transfer coefficient (h) if Grashof number (Gr)=1010 , Prandtl number Pr= 0.69, thermal conductivity (k)= 33.8*10 -3w/moc and height (L)= 2m. If
• (Nu= 0.13 *(Gr*Pr) 0.33)
•
• Q: Calculate the overall heat transfer coefficient (U) for a plane wall which has 1st layer thickness(x₁) =0.2m, thermal conductivity (k₁) =3.5 w/ m.⁰c . The 2nd layer thickness (x2) =0.025m, thermal conductivity (k₂) =2.3 w/m.⁰c if inner heat transfer coefficient (hi=5.2w/m².⁰c) and outer heat transfer coefficient (ho= 35w/m².⁰c).
•
• Q: Calculate Nusselt number for air at temperature =27oc flowing inside pipe of diameter (D) =0.3m at a velocity (v)=10m/sec if the empirical formula for this forced convection case is: Nu= 0.023*Re⁰∙⁸ *Pr⁰∙⁴