Heat Transfer Theory. Jennie Borgström

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2 Heat Transfer Theory Jennie Borgström

3 Modes of heat transfer Law of physics Heat = Energy If you take a hot spot and a cold spot the heat will always be transferred from the hot to the cold

4 Three ways to transfer heat Radiation Electromagnetic waves Reflected When it reaches a body it has 3 options: Absorbed Conduction Molecular or atomic vibrations No material transport Transmitted Convection Energy is transferred by the motion and intermixing of small mass elements Natural convection caused by density difference Forced convection is man-made (ex., pump)

5 Example, a day at the beach Convection Radiation Conduction

6 Heat Transferred in a HE The temperature profile at one point of the plate wall T 1 1, Bulk temperature on hot side Wall Steam T 3 T 4 Flow direction Cold side T 2, Bulk temperature on cold side Heat transfer (Q) driven by temperature difference

7 The Heat Transfer Equation Q = k x A x MTD

8 The Heat Transfer equation Q = k * A * LMTD Q = Heat Load, W (same as Q 1 = Q 2 before) k = k-value, overall heat transfer coefficient (OHTC), W/m² C Higher k-value = More efficient heat transfer A = heat transfer area (m²) LMTD= Logarithmic mean temperature difference

9 The Heat Transfer Equation Q = k x A x MTD

10 Flow principles Two types of flow Laminar Conduction Flow profile Velocity profile Orderly flow throughout the fluid Parabolic flow profile Fluid at the wall moves slower Due to the friction from the wall surface Ex., viscous fluids or water at low velocity

11 Flow principles Two types of flow Turbulent Conduction Convection Flow profile Velocity profile No orderly flow Random eddy motion mixes the fluid Always a laminar film closest to the wall Ex., water at higher velocity

12 Shear Stress Shear stress, τ Wall Velocity profile on hot side Velocity profile on cold side Shear stress, τ

13 Shear stress versus fouling rate 45 g resistan nce growth rate ( 10E-8 m2k/wh) Foulin Shear stess ( pa ) Rule of thumb: Try to keep the shear stress >50 Pa

14 Fouling Layer of fouling Major debris Biological growth Scaling Sedimentation Burn-on

15 Scaling Reversed solubility of dissolved salts Reduced solubility at higher temperature Solubility Sugar Ex CaCO CaCO 3 3 and Ca(PO 4 ) 2 Common problem Keep wall temperature low Design with high shear stress (τ) Temperature

16 Design Safety Factors k-value margin (for PHEs) Defined as a % margin between k Clean and k Service Margin = k clean k k service service (%) Margin = A act A A req req (%)

17 S&T use Rf the Fouling Factor Normal Rf for S&T: 1, m 2 ºC/W 5, ft 2, ºF,h/Btu Normal k Clean for PHE: 6000 W/ m 2 ºC 1000 Btu/ft 2, ºF,h k = + R f = + 10 k k 6000 Service = 3750 W/ m 2 ºC Service Clean Corresponding k-value margin kclean kservice M arg in = 100 = 100 = 60% k 3750 k service Much too high margin Too many plates Less turbulence Fouling! Maybe not competitive?

18 The Heat Balance Liquid-to-liquid Cold fluid in at T 2 In Mass flowrate m 2 Cold fluid out at T 2 Out Mass flowrate m 2 Definitions Q = Heat load, W (rate of heat transfer) Hot fluid in at T 1 In m = Mass flow rate, kg/s Mass flowrate m 1 Cp = Specific heat, J/kg C (the energy needed Hot fluid out at T 1 Out to heat 1 kg of the fluid Mass flowrate m 1 with 1 C) Heat released by the hot fluid: Q 1 =m 1 *Cp 1 *(T 1 In -T 1 Out ) Heat absorbed by the cold fluid: Q 2 =m 2 *Cp 2 *(T 2 Out -T 2 In ) Heat losses are negligible ibl Q 1 = Q 2

19 The Heat Balance - calculation What is the cold fluid outlet temperature? T 2In = 20 C m 2 = 120 kg/s Cp 2 = 4,2 kj/(kg C) T 2 Out = X T 1 In = 80 C m 1 = 100kg/s T 1 Out = 40 C Cp 1 = 4,0 kj/(kg C) Heat Load: Q 1 = m 1 *Cp 1 *(T 1 In -T 1 Out ) = 100 kg/s * 4.0 kj/(kg C) * (80-40) 40) C Q 1 = kj/s = kw Q 1 = Q 2 : kw = Q 2 =m 2 *Cp 2 *(T 2 Out -T 2 In ) kw = 120 kg/s * 4.2 kj/(kg C) * (X-20) C X / (120 * 4 2) C

20 The Heat Balance Steam-to-liquid Everyone knows what steam is? - The three three phases of water: Solid (s) Liquid (l) Gas (g) Ice, Snow Liquid water Steam = water vapour Crystals, Molecules l move Molecules l leave all molecules apart the liquid and tight together vaporise Melting energy Heat absorbed by the water (Cp) Heat of vaporisation / Heat of condensation

21 The Heat Balance Steam described in diagrams Temperature-Pressure Enthalpy-Temperature Enthalpy Saturated Temperature ( C) Superheated vapour 120 C Saturated steam ΔH vap Liquidid T boil Temperature ( C) 2 bar Pressure (bar) ΔH vap = Heat needed to vaporise 1 kg of a fluid (kj/kg) The same amount of energy is released during condensation

22 The Heat Balance Steam-to-liquid Cold fluid out at T 2 Out Cold fluid in at T 2 In Mass flowrate m 2 Mass flowrate m 2 Steam (g) at T 1 In Mass flowrate m 1 ΔH vap (kj/kg) Condensate (l) out Heat released by the steam: Q 1 =m 1 * ΔH vap Heat absorbed by the cold fluid: Q 2 =m 2 *Cp 2 *(T 2 Out -T 2 In ) Heat losses are negligible ibl Q 1 = Q 2

23 The Heat Balance - calculation How much steam is needed to heat the water? Water out at 90 C Water in at 60 C Mass flowrate 50 kg/s Cp 2 = 4,2 kj/(kg C) Steam (g) at 120 C 2 bar (1 bar gauge) Mass flowrate X ΔH vap = 2200 kj/kg Condensate (l) out Heat Load: Q 2 = m 2 *Cp 2 *(T 2 Out -T 2 In ) = 50 kg/s * 4.2 kj/(kg C) (90-60) 60) C Q 2 = kw Q 2 = Q 1 : 6300 kw = Q 1 = m 1 * ΔH vap 6300 kw = X kg/s * 2200 kj/kg X 286k /

24 The Heat Balance So why is steam used? Less than 3 kg steam can heat 50 kg water from 90 to 60 C It is a high-energy carrier! Steam is the most common way in industry to distribute energy for heating purposes

25 Physical properties Density (ρ) Specific heat (Cp) Thermal conductivity (λ) Small variations One value acceptable Viscosity (μ) μ Three values! Temperature

26 Example phys prop Heating of water 60 to 80 C, 500 kw, steam 2 bar(a) 1. Constant t viscosity it 2. True viscosity μ = 1 cp 16 1,6 1,2 0,8 0, TS6 22 plates TS6 16 plates

27 Example phys prop Heating of HFO 30 to 120 C, 400 kw, steam 6 bar(a) 1. Constant viscosity μ = 10 cp 2. True viscosity Visc Temp TS6 42 plates TS6 58 plates

28 Conventional PHE vs Steam PHE Heating of water 60 to 80 C, 500 kw, steam 2 bar (a) M10M TS6M Area: 2,6 m 2 Area: 1,9 m 2 Margin: 25% Margin: 6% P-drop: 45 kpa P-drop: 35 kpa

29 Sizing HE s correctly... If you ignore the margin Margin = 72 % 7 Bar(a) D = 2 Bar 5 Bar(a) 60 ºC HE outlet: t B Bar(a) 10 ºC 1 Bar(a) (Atmospheric) D = 3.5 Bar M6M kw Example of sizing a HE for condensing at 5 Bar(a) Result: M6M, surface = 1,7 m², margin = 72 %

30 Sizing HE s correctly... You could get serious problems! Actual conditions Typical margin related problems are: 7 Bar(a) D = 4.8 Bar 2.4 Bar(a) 60 ºC Over sized control valve Under sized steam trap Stalled heat exchanger HE outlet: 1 Bar(a) 10 ºC High steam velocities 1 Bar(a) (Atmospheric) Dp 0 M6M kw

31 PHE versus S&T in steam applications Size and weight Fatigue Thermal efficiency i Condensation temperature Subcooling of condensate Temperature control Operation in stall condition

32 Size and weight PHE requires less floor area PHE weighs less Easy installation Low PxV For small heat exchangers pressure vessel certificates and regular inspection are unnecessary Weight ratio Space ratio Hold-up volume ratio PHE S&T 6 7 6

33 Fatigue Thanks to the elasticity of the gaskets, there will be no thermal fatigue problems in the PHE plate pack

34 Thermal efficiency A close temperature approach is possible in a PHE PHE higher k-value T out T in

35 A lower condensation temperature Condensation at 150 C Load: 1 MW, ºC TS6-M with 20 plates Condensation at 120 C Load: 1 MW, ºC TS6-M with 28 plates Very small difference in no of plates Why design with a low condensation temperature? Minimise flash steam Reduce scaling No pressure build up in the condensate system

36 PHE provides minimum of flash steam Max condensation temp. 160 ºC Average condensation temp. 140 ºC Max condensation temp. 120 ºC Average condensation temp. 110 ºC Average steam flow = 0.3 kg/s 110 ºC 1MW 140 ºC / hf = 589 kj/kg 100 ºC 0.5 Bar(g) / hf = 468 kj/kg Sub-cooler (optional) 100 ºC Average operation time; 6,000 hr/year Energy cost; 25 Euro / MWh Q=m*(hfin (hfin-hfout) hfout)*time time = 0.3*( )*6,000 = 185,400 kwh Energy cost = 25* = 4635 EURO / Year No losses Minimum scaling No need to sub cool the condensate No pressure build up in the condensate system

37 Benefits compared to S&T Minimised condensation temperature reduce scaling Relative scaling rate Surface temp

38 Subcooling of condensate Possibility of introducing condensate level control and utilising the lower part as a subcooler Vertical orientation of the plates makes drainage easy in a PHE

39 Temperature control accuracy Set point Low hold-up volume, high thermal efficiency and low weight provides a short response time and an accurate temperature control PHE S&T

40 PHE can handle stall conditions Large contact area between steam and water Implosions, Noise Water hammer Poor temperature control Hot steam on cold condensate Small contact area between steam and water No implosions No noise Acceptable temperature control

41 Summary PHE versus S&T PHE compact and easily extendable Temperature control accuracy PHE short response time Low condensation temperature minimises scaling and flash steam PxV

42

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