Topic1. Thermal Principles1.1 Introduction to Refrigeration1.2 Work and Heat1.3 Temperature1.4 Pressure1.5 Thermodynamic Properties1.6 Liquid-vapor Properties1.7 Boiling Point1.8 Heat and Enthalpy1.9 Sensible and Latent Heat1.10 Entropy and Isentropic process1.11 General Gas Laws1.12 Thermodynamic Processes1.13 Conservation of mass1.14 Steady-flow energy equation

1

1.15 Heating and cooling1.16 Bernoulli’s equation1.17 Heat transfer1.18 Conduction1.19 Convection1.20 Radiation1.21 Thermal resistance1.22 Cylindrical cross section1.23 Heat exchangers1.24 Thermal Comfort

1.1 Introduction to Refrigeration

2

Refrigeration: the process of achieving and maintaining a temperature of product or space below that of the surroundings

Applications:Food processingFood storageTransportation refrigerationAir conditioningIndustrial processSpecial applications

: the science of moving heat from low temp. to high temp.Ex. Refrigerator, Freezer, Cooler, Ice maker, air conditioner etc.

: milk, ice cream, beverage

: meats, fish, fruits, vegetables, frozen food: bus, truck, train, ship, aircraft

: hospital, hotel, factory, residential AC

: chemical process, laboratories, printing, textile

: ice maker, ice sketch, construction

1.2 Work and Heat

3

Fig1.1 Reversible heat engine E driving a reversible refrigerator (heat pump) P

PW

Cold reservoir, T0

Hot reservoir, T1

E

Q1

Q2

Q1

Q2

The first laws of thermodynamicsWhen work and heat energy are exchanged there is no net gain or loss of energy.

A net flow of heat from the low temp. to the high temp. without any external work input, which is impossible. Wmin = Q2 (T1 –T0)/T0

Coefficient of performance (COP) of P = heat extracted/work input = Q2/W = T0/(T1 –T0)

Carnot Efficiency of E = W/Q1 = (T1 –T0)/T1

T in absolute scale K

Example 1.1: Heat is to be removed at a temperature of -5°C and rejected at a temperature of 35°C.What is the Carnot or Ideal COP?Problem: T0 = -5°C, T1 = 35°C, COPCarnot = ?

Solution: Carnot COP = T0/(T1 –T0) = 268/(308-268) = 6.7

4

1.3 TemperatureTemperature: its thermal state and its ability to exchange energyReference points on Celsius scale: 0C and 100CAbsolute temperature in Kelvin, T = T C + 273 K, T K = T CConversion units: C = (F - 32)/1.8Absolute temperature in Rankine, T = T F + 460 R

104°F86°F68°F50°F32°F14°F-4°F

-22°F-40°F

40°C30°C20°C10°C

0°C-10°C-20°C-30°C-40°C

Fig1.2 Temperature scale

1.4 Pressure

5

Pressure P: the normal force exertedby a fluid per unit areaAbsolute pressure, gauge pressure, standard pressure 101.325 kPaMeasurement device: pressuregauges, manometersPressure units conversionMetric unitsInch-Pound (I-P) units

Fig1.3 Pressure gauges, manometers

Fig1.4 Pressure units conversion

1.5 Thermodynamic Properties

6

Density : the mass occupying a unit volume

Specific volume : the volume occupying a unit mass of air at standard atmospheric pressure and 25C = 1.2 kg/m3

Properties: temperature T, pressure P, density ,specific volume , specific heat cv cp, enthalpy h, entropy s, liquid vapor propertyThermo. properties are not all independently variable.

A substance: two intensive thermodynamic properties define the state

A mixture: three intensive thermodynamic properties define the state, i.e. dry air and water vapor

Example 1.2: What is the mass of air contained in a room of dimensions 4 by 6 by 3 m if the specific volume of the air is 0.83 m3/kg?Problem: mair = ? for a WLH and air

Solution: mair = V/air = W*L*H/air = 4*6*3/0.83 = 72/0.83 = 86.7 kg

1.5 Thermodynamic Properties

7

Specific heat cv and cp: the quantity of energy required to raise the temperature of a unit mass by 1 Kcp is more applies to most of the heating and cooling processesApproximated cp for important substances are constant for small Tcp = 1.00 kJ/kg.K for dry aircp = 4.19 kJ/kg.K for liquid watercp = 1.88 kJ/kg.K for water vapor

Example 1.3: What is the rate of heat input to a water heater if 0.4 kg/s of water enters at 82C and leaves at 93C?Problem: Q = ? to a heater of ሶ𝑚 at Ti and Te

Solution: CV at the water volume1st Law SSSF: Q = W + ሶ𝑚cpTAssume: Boiler W = 0, cp = 4.19 kJ/kg.K for liquid waterQin = ሶ𝑚cpt = 0.4*4.19(93 – 82) = 0.4*46.1 = 18.44 kW

Q = qAs

Fig1.5 Ex1.3 Water heater

1.6 Liquid-vapor Properties

8

Liquid-vapor properties: Most cooling and heating system usesubstances that pass between liquid and vapor states in their cycle.T, P, h, s – Tables or Charts of properties, i.e. P-T, T-s, P-h charts

Figure1.6 Skeleton pressure-enthalpy P-h diagram for water

pressure-enthalpy p-h diagramseparated by saturation line(1) Subcool-liquid region(2) Liquid-vapor region (mixture)(3) Superheated-vapor regionConstant temperature lineWater boils at a higher temp. when the pressure is higher.

Constant T = 100C

T = 50C, P = 12.3 kPa

T = 150C, P = 476 kPa

9

Table of properties10 < T < 40 C; Ps = 0.0051T2 -0.0515T + 1.29 kPa

Saturated liquid for 10 < T < 150 C; hf = 4.19T kJ/kg

Saturated vapor 10 < T < 150 C; hg = 1.75T + 2501 kJ/kg

Superheated vapor at T (C); h = 1.88T-0.13Tsat+ 2501 kJ/kg

Dynamic viscosity of liquid for 10 < T < 150 C; 105 = -0.00018T3-0.044T2-3.4T+ 165 kg/m.s

1.7 Boiling Point

10

Water boils at a higher temp. when the pressure is higher.

Figure1.7 Change of state with pressure and temperature (P-T diagram)

T = 373C, P = 22 MPa

T = 0.01C, P = 0.61 kPa

10 < T < 40 C; Ps = 0.0051T2 -0.0515T + 1.29 kPa

1.8 Heat and Enthalpy

11

Heat - many forms of energy, generated from chemical sources. The heat of a body is its thermal or internal energy. A change in heat - a change of temperature or - a change between the solid, liquid and gaseous states.Enthalpy is the sum of its internal energy and flow work: H = u + Pv

Enthalpy : the amount of heat is the change in enthalpy at constant-pressure process and no work done W = 0Datum plane for water and steam: h = 0 kJ/kg, s = 0 kJ/kgK at 0C but datum plane for refrigerant: h = 200 kJ/kg , s = 1.0 kJ/kgK at 0C then at 100C, hf = 419.06 kJ/kg for liquid water = cpT = 4.19*100,

hg = 2676 kJ/kg for water vaporExample 1.3: Q = ሶ𝑚cpT = ሶ𝑚h, h = cpT for liquid water

1.8 Heat and Enthalpy

12

Example 1.4: A flow rate of 0.06 kg/s of water enters a boiler at 90C, at with temperature the enthalpy is 376.9 kJ/kg. The water leaves as steam at 100C. What is the rate of heat added by the boiler?Problem: a boiler, ሶ𝑚 at Ti (given hi), Te , Q = ?

Solution: CV at the boiler1st Law SSSF: Q = W+ ሶ𝑚hAssume: Boiler W = 0, at 100C, hg = 2676 kJ/kg for sat vaporQin = ሶ𝑚h = 0.06(2676 – 377) = 0.06*2299 = 137.6 kW

Q

m,Ti

Te

Fig1.8 Fire tube boiler

Using function: hf = 4.19T = 4.19*90 = 377 kJ/kghg = 1.75T + 2501 kJ/kg = 2676 kJ/kgQin = ሶ𝑚h = 0.06(2676 – 377) = 137.6 kW

1.9 Sensible and Latent Heat

13

Sensible heat - a change of temperatureQS = 419 kJ/kg for raise water from 273K to 373K (0C to 100C)Latent heat - a change of state (solid to liquid, liquid to gas)QL = 334 kJ/kg for melting ice water at 273K (0C)QL = 2257 kJ/kg for boiling liquid water at 373K (100C)

Fig1.9 Change of temperature (K) and state of water with enthalpy (T-H diagram)

1.9 Sensible and Latent Heat

14

Example 1.5: If 9 kg/s of liquid water at 50C, flow into a boiler, is heated, boiled, and superheated to a temperature of 150C and the entire process takes place at standard atmospheric pressure, what is the rate of heat transfer to the water? Problem: a boiler, m liquid at Ti and superheated vapor Te, at const. 101 kPa, Q = ?

Consist of 3 processes: (1) bringing Ti of subcool water at P to Tsat

(2) converting liquid to vapor at Tsat

(3) Superheating vapor from Tsat to Te

Solution: Boiler Q = ሶ𝑚h T = 50C, hi = 4.19*50 = 210 kJ/kgTe = 150C & Tsat = 100C at 101 kPahe = 1.88Te-0.13Tsat+ 2501 = 2770Q = ሶ𝑚h = 9(2770 – 210) = 23,045 kW

Qm,Ti

Te

Ti Tsat

Tsat Te

Tsat

hi he

Fig1.10 Water tube boiler

1.10 Entropy and Isentropic process

15

Entropy s: If a gas or vapor is compressed or expanded frictionlesslywith Q = 0 during the process, the entropy remains constant,h represents work required by compression or expansion

Isentropic compression: adiabatic and without friction, occurs at constant entropy s1 = s2. Wisen = ሶ𝑚 (h1 – h2) Wactual = ሶ𝑚 (h1 – h2’) > Wisen

Compression efficiency, c = Wisen /Wact

Example 1.6: Compute the power required to compress 1.5 kg/s of saturated water vapor from a pressure of 34 kPa to one of 150 kPa.Problem: W = ? to compress sat. water vapor m from P1 to P2

Solution: CV at compressor; Q = W + ሶ𝑚(h2 – h1), Assume compressor Q = 0h1 = hg@34 kPa = 2630 kJ/kg, s1 = 7.7 kJ/kg.K (p-h diagram)At P2 = 150 kPa, s2 = s1 h2 = 2930kJ/kgW = - ሶ𝑚(h2 – h1) = -1.5*(2930 - 2630) = -450 kW

P1 = 34 kPa

P2 = 150 kPa

h1 h2

16

P1 = 34 kPa

P2 = 150 kPa

h1 = hg@34 kPa

= 2630 kJ/kg

h2 = 2930kJ/kg

Isentropic compression: adiabatic and reversible, s1 = s2

Wisen = m(h1 – h2)

c = Wisen /Wact

2’

1

h2’

1.11 General Gas Laws

17

Perfect-gas law: PV = mRTP = absolute pressure, PaV = volume, m3; m = mass, kg; = density = m/V, kg/m3

R = gas constant = 287 J/kg.K for air and 462 J/kg.K for waterT = absolute temperature, KPerfect-gas equation is applicable to dry air and to highly superheatedwater vapor and not applicable to water and refrigerant vapors close totheir saturation conditions.

Example 1.7: What is the density of dry air at 101 kPa and 25C?Problem: dry air = ? at P and TSolution: Pv = RT, P/ = RT, = P/RTR = 287 J/kg.K for air, T = 25 + 273 = 298 K = P/RT = 101*1000/(287*298)= 1.18 kg/m3

Boyle’s Law: PV = constantCharles’s Law: V/T = constantDalton’s Law: Ptotal =Pi, P = RT

Example 1.8: If dewpoint at 15C for moist air at 101 kPa and 25C, what is the density of dry air?Problem: dry air = ? at Pt , T, Tdp = Tsat

Solution: a = Pa/RaT, Pa = Pt - Ps

Ps = 0.0051T2dp -0.0515Tdp + 1.29 = 1.67 kPa,

Pa = 101 – 1.67 = 99.34 kPaa = Pa/RaT = 1.20 kg/m3

w = Ps/RsT = 1.67/(462*298) = 0.0125 kg/m3

v = 1/w = 79.9 m3/kg > vsat@15C = 77.9

1.12 Thermodynamic Processes

18

Heating and cooling = thermodynamic processesEnergy is the central concept. Energy analysis is fundamental procedure in thermodynamics.System – object consideredControl volume – specified region in spaceSystem boundary (control surface) – encloses the control volumeSize and shape of the system are specified for simplify accounting for the change in energy storage within the system or energy transfers across a system boundary.Environment – not included in system

Problem: consider a pump or an entire buildingStep 1: Problem definitionStep 2: Schematic System (change in energy storage) ,

System boundary (energy transfers across)Step 3: Assumptions steady state, steady flow, no heat loss e.t.c.Step 4: Physical law mass conservation, energy balance (1st law thermo.)

1.13 Conservation of mass

19

Mass is neither crested nor destroyed in the process analyzed.Mass may be stored within a system or transferred between a system and its environment, but it must be accounted for in any analysis procedure.mt + m1 = mt+t + m2

Dividing by t: m1/t = (mt+t - mt)/t + m2/t

ሶ𝑚1 = m/t + ሶ𝑚2

If the rate of change of mass within the system is zero m/t = 0 ሶ𝑚1 = ሶ𝑚2 = ሶ𝑚 steady flow ሶ𝑚 = mass flow rate, kg/s

Newton’s second law: Force = ma = m(dV/dt)m = mass, kga = acceleration, m/s2

V = velocity, m/st = time, s

Fig1.11 Flow system for conservation of mass

1.14 Steady-flow energy equation

20

ሶ𝑚(h1 + V21/2 +gz1) + Q - ሶ𝑚(h2 + V2

2/2 +gz2) – W = dE/dtz = elevation, mg = gravitational acceleration = 9.81 m/s2

Q = rate of heat transfer in form of heat, WW = rate of heat transfer in form of work, WE = energy in system, J

dE/dt = 0 for steady flow process ሶ𝑚(h1 + v2

1/2 +gz1) + Q = ሶ𝑚(h2 + v22/2 +gz2) + W

Often V2/2 and gz is negligible compare to h, W or Qሶ𝑚h1 + Q = ሶ𝑚h2 + W

Q – W = ሶ𝑚(h2 - h1)

Fig1.12 Energy balance on a control volume

1.15 Heating and cooling

21

steady flow process: Q – W = ሶ𝑚(h2 - h1)If W = 0 by a pump, compressor or engine: Q = ሶ𝑚(h2 – h1)

Example 1.9: Water flowing at a steady rate of 1.2 kg/s is to be chilledfrom 10 to 4C to supply a cooling coil in an air-conditioning system.Determine the necessary rate of heat transfer.Problem:water low ሶ𝑚 chilled steadily from T1 to T2 , Q = ? Solution: CV water in shell Q = ሶ𝑚(h2 – h1) h2 = 4.19*4 = 16.8 kJ/kgh1 = 4.19*10 = 41.9 kJ/kg Q = ሶ𝑚(h2 – h1) Q = 1.2*(16.8 - 41.9) = -30.2 kWLiquid water Q = ሶ𝑚cp(T2 – T1) = -30.2 kW

Adiabatic processes - no heat is transferred; Q = 0-The walls of the system are thermally insulated.-The throughput rates of energy are large in relation to the energytransmitted to or from the environment in the form of heat.

T2ሶ𝑚, T1

Q

Fig1.13 Liquid chiller

1.16 Bernoulli’s equation

22

Bernoulli’s equation : derived from mechanical behavior of fluids,But it is also derivable as a special case of the energy equation through second-law considerations: Tds = du +Pdv –(1) (Gibb’s equation)

For adiabatic process Q = 0, and no mechanical work W = 0:

h + V2/2 +gz = const

Differentiation yields: dh + VdV + gdz = 0 –(2)

Enthalpy, h = u + Pv is differentiated to yield dh = du + Pdv + vdP –(3)

Combining eq (1) and (3) results in: Tds = dh – vdP

For an isentropic process (adiabatic+ no friction): ds = 0 = dh – vdPSubstituting dh = vdP = dP/ into eq.(2) dP/ + VdV + gdz = 0

For constant density, integrates to the Bernoulli equation:

P/ + V2/2 +gz = const

Bernoulli equation shall be used for liquid and gas flow in which the

density varies only slightly and may be treated as incompressible.

1.16 Bernoulli’s equation

23

Example 1.10: Water is pumped from a chiller in the basement, where z1 = 0 m to a cooling coil located on the twentieth floor of a building, where z1 = 80 m. What is the minimum pressure rise the pump must be capable of providing if the temperature of the water is 4C.Problem: water pumped from z1 to z2 at T, P (Pa) = ? Solution: ignore friction loss in pipe using Bernoulli equation; P1/ + V2

1/2 +gz1 = P2/ + V22/2 +gz2

Assume V1 = V2, @4C = 1000 kg/m3

P1 - P2 = g(z2 - z1) = 1000*9.81*(80 – 0) = 785000 Pa = 785 kPa

Fig1.14 water pipe

Modified Bernoulli’s equation

24

Example 1.11: Compute the pressure drop when 3.0 L/s of 4C water flows through a steel pipe with a nominal diameter of 50 mm (ID = 52.5 mm) that is 20 m long in horizontal and 80 m long in vertical with minor loss Leq = 0.2Ltotal.Problem: water flow ሶ𝑉 at T, pipe D and L pumped from z2 - z1, P (Pa) = ? Solution: pressure drop include major and minor loss in pipe L+Leq=1.2Ltot=120 mP1/ + V2

1/2 +gz1 = P2/ + V22/2 +gz2 + Ploss, Ploss = f(L/D)V2/2,

Assume smooth tube f = 0.184Re-0.2, Re = VD/, Water 4C, @4C = 1000 kg/m3,105 = -0.00018T3-0.044T2-3.4T+ 165 = 150 kg/ms →= 150 10-5 kg/msV = ሶ𝑉/((ID)2/4) = 1.386 m/s, Re = 1000*1.386(0.0525)/0.0015 = 48513, f = 0.184Re-0.2 = 0.02126, Ploss = 0.02126(120/0.0525)1000*1.3862/2 = 46.7 kPaP1 - P2 = g(z2 - z1) + Ploss = 1000*9.81*(80 – 0) + 46.7 kPa = 831 kPa

P1/ + V21/2 +gz1 = P2/ + V2

2/2 +gz2 + Ploss

Pressure loss in pipe; Ploss = f(L/D)V2/2, smooth tube for turbulent flow Re > 10,000; f = 0.184Re-0.2

Re = VD/Dynamic viscosity of liquid water for 10 < T < 150 C; 105 = -0.00018T3-0.044T2-3.4T+ 165 kg/m.s

1.17 Heat transfer

25

Heat transfer analysis s developed from thermodynamics laws of conservation of mass and energy, the second law of thermodynamics, and three rate equations describing conduction, convection and radiation.

1. Conduction: Direct from one body touching the other, or through a continuous mass2. Convection: By mean of a heat-conveying fluid moving between one and the other.3. Radiation: Mainly by infrared waves (but also in a visible band, e.g. solar radiation), which are independent of contact or an intermediate fluid.

26

1.18 ConductionFourier’s law of heat conduction (1822):Qcond = -kAT/L = -(k/L)ATA = cross-sectional area, m2, L = x = length, mT = temperature difference = T2 - T1, K or Ck = thermal conductivity, W/m.K – characteristic of the materialk/L conductance

Thermal conductivity and rate of conductive heat transfer is related to

the molecular structure of materials.

rate of heat transfer in differential form: Qcond = -kA(dT/dx)Table1.1 Thermal conductivity of some materials

xx

T1

T2

Td𝑇d𝑥

ሶ𝑄𝑐𝑜𝑛𝑑

A

1.19 Convection

27

Newton (1701): Qconv = hcA(Ts – Tf)Ts = surface temperature, CTf = fluid temperature, Chc = convection coefficient, W/m2.K- widely used in engineering- Essence of convective heat transfer analysis is the evaluation of hc

hc = function of (flow velocity, fluid properties, geometry of the surface)in flow over plan surface, in pipes and ducts, and across tubes

Extensive theory has been developed to support the experimentally

observed correlations and to extend them to predict behavior in untested flow configurations.

It is the correlations rather than the theories used in practical

engineering analysis.

Dimensionless analysis: basis of most of the pertinent correlations

Hot iron block Ts = 400 C

Tf = 15 C

Vair = 0

Ts

T=Tair

T(y)

T

ሶQ𝐜𝐨𝐧𝐝

Thermal boundary layer

d𝑇

d𝑦

ሶ𝑞 = −k d𝑇d𝑦Vair = 3 m/s

ሶQconv

ሶ𝑞 = hcT

1.19 Convection

28

Dimensionless parameters: understanding the physical phenomena involved and ability to construct reasonable models for basic flow Reynold number Re = VD/ or VL/Prandtl number Pr = cp/kNusselt number Nu = hcD/k or hcL/kGrafshop number Gr = RaPr = g2x3T/2

Flow Correlation: Nu = C(Ren)(Prm)

C, n and m = determined experimentally

Fig 1.15 Correlation for forced convection in smooth tubes, turbulent flow

Nu = 0.023Re0.8Pr0.4

→ Forced convection, inertia/viscous→ Fluid properties, momentum/thermal diffusion→ Convection/conduction→ Natural convection

29

Nusselt number of flow in circular tubes

30

Boiling and condensation heat transfer

1.20 Radiation

31

Radiant-energy transfer results when protons emitted from one surface travel to other surfaces, then either absorbed, reflected, or transmitted through the surfaces.

Emissive power: E = T4 Stefan–Boltzmann law = 5.669 10−8 W/m2 · K4 = Stefan–Boltzmann constantT = Absolute temperature, K = emissivity, = 1 for a black body

In many real materials, emissivity = absorptivity, = → as gray bodies

Radiation leaving a surface is distributed uniformly in all direction →

the geometry relationship between two surfaces affects the radiant-energy exchange between them. → shape factor FA

If F expresses effects of emissivity, absorptivity, reflectivity,

transmissivity, and rate of radiant heat transfer of the surfaces,

The radiant heat exchange: Qrad = AFFA(T41- T4

2)

1.21 Thermal resistance

32

Qcond = -(k/L)AT

Qconv = hcAT

Qrad = AFFA(T41- T4

2)

Qrad hrAT --linearized equation, where hr = FFA(T41- T4

2)/(T1- T2)

Ohm’s law: I = E/R, Current = Potential difference/Resistance

Analogy to Ohm’s law: Q = T/RT , RT = thermal resistance

Qcond = -T /(L/kA), RT,cond = L/(kA)

Qconv = T/(1/hcA), RT,conv = 1/(hcA)

Qrad T/(1/hrA), RT,rad = 1/(hrA)

→ Qcond linear with temperature

→ Qconv linear with temperature

→ Qrad non-linear with temperature

T1 T2

T1 T2

RT = Rs1+RA+1/(1/RB+1/RC)+RD+RS2

Q = (T2-T1)/RT

1D steady state heat transfer through buildings

qin = (qconv,o + qrad,o) = (qconv,i + qrad,i)

qin = ho(To - Tw,o) = hi(Tw,i - Ti)

ho, hi = inner, outer surfaceconductances, account both convection and radiation heat transfers

1.21 Thermal resistance

34

Q = (Ti – To)/Rtot = UA(Ti – To)U = overall heat coefficient, W/m2.KUA = 1/Rtot, U = 1/(RtotA)RA = L/k = Rw – conduction through wallRA = 1/(hc + hr) = Rs --surface

1.22 Cylindrical cross section

35

T2T1

r2

T2

h1

h2

T3T2

Rcomb

kins

Find Rcyl by Qcond = -kArdT/dr

r1r2(Qcond/(2rL))dr = T1

T2(-k)dT

Qcond = -(T2–T1)[2Lk/(ln(r2/r1))]

Rcyl = ln(r2/r1)/(2Lk)

Rcyl = ln(ro/ri)/(2Lk) x/(Amk)

x/(Amk) = (ro - ri)/(2Lk(ro + ri)/2)

= [2(ro - ri)/(ro + ri)]/(2Lk)

For thin tube x = ro - r1;

ln(ro/ri) 2(ro - ri)/(ro + ri) (< 0.1%)

Rcyl x/(Amk)

for thin copper tubes and steel pipes

1.23 Heat exchangers

36

A device in which energy is transferred from one fluid stream to another across a solid surface.Be used extensively in refrigeration and air conditioningQ = UA(LMTD), UA = 1/Rtot

Since the temperature of one or both fluids may vary as they flow through the heat exchanger, mean temperature difference - determinedLMTD = Tlm = (TA - TB)/[ln(TA/TB)]T1 = temperature difference between two fluids at position A, CT2 = temperature difference between two fluids at position B, C

T1 T2

1.23 Heat exchangers

37

Example 1.13: Determine the heat transfer rate of the shell and tube heat exchanger: h1 = 50W/m2.K, h2 = 80W/m2.K, T1,i = 60C, T1,o = 40C, T2,i = 20C, T2,o = 30C, ro = 11 mm, ri = 10 mm, length = 1 m, and for the metal k = 386 W/m.K, Problem: [CF HX,1 in shell, 2 in tube] Q = ? h1, h2, T1,i, T1,e, T2,i, T2,o, ro, ri, L, k Solution: Q = UATlm = Tlm/Rtot

Tlm = (TA - TB)/[ln(TA/TB)],Rtot = 1/(h1A1)+ln(ro/ri)/(2Lk)+1/(h2A2)

A1= 2roL = 2(0.011)(1) = 0.069 m2

A2= 2riL = 2(0.011)(1) = 0.063 m2

Rtot =1/(50*0.069)+ln(11/10)/(2*1*386)+1/(80*0.063)=0.289+0.00004+0.199=0.487W/K

TA = T1,o – T2,I = 40 – 20 = 20CTB = T1,i – T2,o = 60 – 30 = 30CTlm = (30-20)/[ln(30/20)]=24.7CQ = 24.7/0.487 = 50.7 W

1.23 Heat exchangers

38

Fig 1.16 Temperature change.

(a) Refrigerant cooling fluid. (b) Fluid cooling refrigerant. (c) Two fluids.

Tlm = (Tmax – Tmin)/[ln(Tmax/Tmin)]

Example 1.14: A fluid evaporates at 3C and cools water from 11.5C to 6.4 C. What is the logarithmic mean temperature difference and what is the heat transfer if it has a surface area of 420 m2 and the thermal transmittance is 110 W/m2K? Problem: Evaporator As and U at TR and cools water from Ti to To ,Q = ? Solution: Q = UATlm , Tlm = (TA - TB)/[ln(TA/TB)]Tmax = Ti – TR = 11.5 – 3 = 8.5C, Tmin = To – TR = 6.4 – 3 = 3.4CTlm = (8.5-3.4)/[ln(8.5/3.4)]=5.566CQ = 110*420*5.566 = 257000 W = 257 kW

39

1.24 Thermal Comfort

40-500 W/m2

30 – 70 %RH

Fresh air 8 – 15 L/s3 – 7 m/min

23 - 31 C

Temperature difference < 3C between head and foot levels

Skin temp., tskin 33.7C

tcore < 31C→ Lethal/deadlytcore 35C → Slight discomforttcore 36.8Ctcore 39C → Slight discomforttcore > 43C → Lethal/deadly

1.24 Thermal Comfort

A living human body as a heat engineConsumes food → metabolism → heat and work

Generates heat at a rate 100 – 2000 W100 W → sedentary person, sitting2000 W → strenuous exercise

Qgenerate = Qskin + Qrespiration + Qstore

Qgenerate M = c*met* As,naked

met = 58.2 W/m2

As,naked = 0.202mass0.425height0.725