Heat Transfer Excel Calculations

HEAT TRANSFER SPREADSHEET CALCULATIONS CONDUCTION Heat travels three ways: * Conduction - by spreading through solids. * Convection is heat transfer by the movement of heated gasses and liquids. * Radiation - is heat in the form of radiation and travels through space at the speed of light. Heat always travels from an area of higher temperature to an area of lower temperature. Heat transfer (Q) is the flow rate of heat and is measured in Watts or Btu's per hour.

UNITS Symbol

Units

Q Btu/hr Q Btu/hr w Btu K Btu/hr-ft-deg F h Btu/hr-ft^2-F ρ lbm/ft^2 Cp Btu/lbm-deg F μ lbm / sec-ft μ lbm / sec-ft L ft L in L in F lbf w ft-lbf ν ft^2 / sec Heat transfer is measured in feet and meter units.

Multiply by 0.2931 3.93E-04 778.2 1.730 5.5956 16.0185 4187 3600 1.488 0.3048 25.4 0.0254 4.4482 1.3558 0.0929

Units Obtained W hp ft-lbf W / m-C W / m^2-C kg / m^3 J/kg-C lbm / hr-ft kg / sec-m m mm m N N-m m^2 / s

Use the above units table from left to right: Input Data

Units = Multiply by = Units Obtained =

42 0.0254 Calculations 1.067

Use the above units table from right to left: Input Data

Units Obtained = Divide by = Units = Temperature is the intensity of heat:

22 5.5956 Calculations 3.932 Input Data

Input Data

T= T= T=

52 deg C + 273.2 325.2

deg C deg K

T= 50 T = deg F + 460 T= 510

Input Data

T= 62 T = 5*(deg F - 32)/9 T= 16.67

Area, Answer: Heat transfer, Low temperature, Answer:

A= A= Q= t2 = t2 =

deg F deg C

T= 60 deg C T = (9*deg C/5) + 32 T= 140 deg F

Input Data 12 Watts 100 deg C 111 W / m-K 4.00 in 4.00 in 0.375 in Calculation (L1*L2)*0.0254^2 0.0103 m^2 K*A*(t1 - t2) / (X*0.0254) t1 - (Q*X) / (K*A) 99.90 deg C

CONDUCTIVITIES & DENSITIES MATERIAL Copper, pure Brass, 70 Cu, 30 Zn Silver, 99.9% pure Duralumin, 3-5%Cu, trace Mg Carbon Steel, 1.0%C Bronze, 75 Cu, 25 Sn Stainless Steel,18 Cr, 8 Ni Concrete, stone, 1-2-4 mix Glass, window

deg R

Input Data

Conduction Example

Heat source on t1 side, Q = High temperature, t1 = Thermal conductivity, K = Dimension in inches, L1 = Dimension in inches, L2 = Thickness in inches, X =

deg F

Properties at 68 deg F ρ K lbm/ft^3 Btu/hr-ft-F 559 223 532 64 657 235 174 95 487 25 541 15 488 9.4 0.79 0.45

K W/m*C 386 111 407 164 43 26 16.3 1.37 0.78

Brick, common building Wood, fir Wood, white pine Glass Wool, 1.5 lb/ft^3

0.40 0.063 0.065 0.022

0.69 0.109 0.112 0.038

Heat Transmission Through Air Films and Solids Conduction through wall and air films on each side of the wall. Find the heat transfer rate Q and the inner and outer wall surface temperatures. OVERALL HEAT TRANSFER COEFFICIENT U

The heat flow rate, Q, is the same through each layer in the diagram above.

Thermal Resistances of Layers Heat transfer per unit area, Q / A = (t1 - t2) / (Xa / Ka) Heat intensity =

t1 -t2

Thermal resistance for layer a, Ra =

Xa / Ka

Thermal resistance for layer o, Ro =

1/ ho

Overall heat transfer rate, Q / A = (t1 -t4) / (Ra + Rb + Rc) Overall temperature difference, ∆T=

t1 - t4

Overall heat transfer coefficient, U =

1 / (A * Σ Rn)

Heat transfer per unit area, Q / A =

U*(∆T)

Equation 1

U.S. Units Wall length, L = Wall height, H = Wall thickness, X = Inside convective coefficient, ho = to = t5 = Thickness, Xa = Xb = Xc = Thermal conductivity, Ka = Thermal conductivity, Kb = Thermal conductivity, Kc = Outside convective coefficient, h5 =

Input Data 300.00 48.00 7.00 2 70.0 36.0 1.00 4.00 0.75 0.065 0.400 0.280 5.80 Calculation L*H / 144 100.00

in in in Btu/hr-ft^2-F deg F deg F in in in Btu/hr-ft-F Btu/hr-ft-F Btu/hr-ft-F Btu/hr-ft^2-F

Wall area, A = A= ft^2 Thermal Resistances Ro = 1 / ho Ro = 0.500 Ra = Xa / Ka Ra = 1.2821 Rb = Xb / Kb Rb = 0.8333 Rc = Xc / Kc Rc = 0.2232 R5 = 1 / h5 R5 = 0.172 Overall heat transfer coefficient 1 / U = 1/ho + Xa / Ka + Xb / Kb + Xc / Kc + 1/h5 1 / U = Ro+R1+Ra+Rb+Rc+R5 1/U= 3.0110 Answer: U = 0.332 ∆T = to - t5 Answer: ∆T = 34 deg F Heat transfer per unit area, Q / A = U*(∆T) Answer: Q / A = 11.29 Heat transfer, Q = U*A*(∆T) Answer: Q = 1129 Btu/hr Answer: Q = 331 Watts Surface temperature is found from Equation-1 above: Heat transfer per unit area, Q / A = (to - t1) / (1 / ho) t1 = to - (Q/A)*(1 / ho) Answer: t1 = 64.4 deg F Internal temperature follows: Heat transfer per unit area, Q / A = (t1 - t2) / (Xa / Ka) t2 = t1 - (Q/A)*(Xa / Ka) Answer: t2 = 49.9 deg F

Thermal Resistances of Layers - continued

Wall length, L = Wall height, H = Wall thickness, X = Inside temperature, t1 = Outside temperature, t4 = Wall material conductivity, K = Inside convective coefficient, ha = Outside convective coefficient, hc = Wall area, A = A= Ra = Ra = Rb = Rb = Rc = Rc = Overall thermal resistance, R = ΣR = Overall temperature difference, ∆T = ∆T =

Input Data 12.00 12.00 0.5 70 20 0.263 2.00 10.00 Calculations L*H / 144 1.00 1 / ha 0.50 X/K 1.90 1 / hc 0.10 Ra + Rb + Rc 2.50 t1 - t4 50.0

in in ft deg F deg F Btu/hr-ft-F Btu/hr-ft^2-F Btu/hr-ft^2-F

ft^2

deg F

Heat transfer per unit area, Q / A = ∆T / ΣR Q/A= 20.00 Btu / hr-ft^2 Heat transfer, Q = A*∆T / ΣR Q= 20.00 Btu/hr Q= 5.86 Watts Internal temperatures are found from Equation-1 above: Heat transfer per unit area, Q / A = t2 = Answer: t2 =

(t1 - t2) / (Ra) t1 -(Q/A)*(Ra) 60.00

deg F

Heat transfer per unit area, Q / A = (t3 - t4) / (Rc) t3 = t4 + (Q/A)*(Rc) Answer: t3 = 22.00

deg F

2D HEAT TRANSFER EXCEL'S SOLVER > see MATH TOOLS tab below. The steady state energy balance on the interior nodal point N is: 0 = Q1-N + Q2-N + Q3-N + Q4-N L = Thickness into page. Finite difference equations for each conductive flux: Q1-N = K*L*(∆Y)*(T1 - TN) / (∆X) Q2-N = K*L*(∆X)*(T2 - TN) / (∆Y)

Q3-N = K*L*(∆Y)*(T3 - TN) / (∆X) Q4-N = K*L*(∆X)*(T4 - TN) / (∆Y) If ∆X equals ∆Y: 0 = T1 +T2 + T3 + T4 -4*TN

Node 1, 0 = 400 + 500 + T2 + T4 - 4*T1 Node 2, 0 = T1 + 500 + 200 + T3 - 4*T2 Node 3, 0 = T4 + T2 + 200 + 300 - 4*T3 Node 4, 0 = 400 + T1 + T3 + 300 - 4*T4

Node 1, -900 = T2 + T4 - 4*T1 Node 2, -700 = T1 + T3 - 4*T2 Node 3, -500 = T4 + T2 - 4*T3 Node 4, -700 = T1 + T3 - 4*T4

Node 1 Node 2 Node 3 Node 4

Equations -900.0 -700.0 -500.0 -700.0

Constants -900 -700 -500 -700

Solution T1 = T2 = T3 = T4 =

Column E 400 350 300 350

deg C deg C deg C deg C

CLICK THE "MATH TOOLS" TAB BELOW FOR INFORMATION ABOUT "SOLVER"

Node 1 Node 2 Node 3 Node 4

Equations 0.0 0.0 0.0 0.0

Constants -900 -700 -500 -700

Solution T1 = T2 = T3 = T4 =

Column E deg C deg C deg C deg C

Node 1, Node 2, Node 3, Node 4, Node 5, Node 6, Node 7, Node 8, Node 9,

Node 1, Node 2, Node 3, Node 4, Node Node 6, Node 7, Node 8, Node 9,

0= 0= 0= 0= 0= 0= 0= 0= 0=

-500 = -100 = -200 = -400 = 5, 0 = -100 = -600 = -200 = -300 =

Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 Node 7 Node 8 Node 9

BOUNDARY CONDITIONS

400 + 100 + T2 + T4 - 4*T1 T1 + 100 + T3 + T5 - 4*T2 T2 + 100 + 100 + T6 - 4*T3 400 + T1 + T5 + T7 - 4*T4 T4 + T2 + T6 + T8 - 4*T5 T5 + T3 + 100 + T9 - 4*T6 400 + T4 + T8 + 200 - 4*T7 T7 + T5 + T9 + 200 - 4*T8 T8 + T6 + 100 + 200 - 4*T9

T2 + T4 - 4*T1 T1 + T3 + T5 - 4*T2 T2 + T6 - 4*T3 T1 + T5 + T7 - 4*T4 T4 + T2 + T6 + T8 - 4*T5 T5 + T3 + T9 - 4*T6 T4 + T8 - 4*T7 T7 + T5 + T9 - 4*T8 T8 + T6 - 4*T9 Equations -500.0 -100.0 -200.0 -400.0 0.0 -100.0 -600.0 -200.0 -300.0

Constants -500 -100 -200 -400 0 -100 -600 -200 -300

Solution T1 = T2 = T3 = T4 = T5 = T6 = T7 = T8 = T9 =

Column E 235.7 166.1 128.6 276.8 200.0 148.2 271.4 208.9 164.3

deg C deg C deg C deg C deg C deg C deg C deg C deg C

0 = (T1 + T2)/ 2 + (h*∆X / k)*T∞ - (h*∆X / k + 1)*Tn Input Data Ambient temperature, T∞ = 90.0 Convective heat transfer coefficient, hc = 9.1 Conductivity, k = 2.0 Grid spacing, ∆X = ∆Y = 1.5

0 = T1 + T4 + (T2 + T3)/ 2 + (h*∆X / k)*T∞ - (h*∆X / k + 3)*Tn

0 = (T1 + T2)/ 2 + T3 - 2*Tn

0 = T1 / a*(a + 1) + T2 / (b + 1) + T3 / (a + 1) + T4 / b*(b + 1) - (1/a + 1/b)*Tn

Linear Thermal Expansion

Length, L = Material Coefficient, α = Temperature Change, Δt =

Input Data 120 0.000012 100

Units in in/in deg F

Coefficients of Linear Expansion in the range 0 to 100C Aluminum α = 0.0000238 Bronze α = 0.0000175 Copper α = 0.0000165 Mild Steel α = 0.000012 Porcelain α = 0.000004

Length change, ΔL = Answer: ΔL = This is the end of this spread sheet.

Calculations L * α * Δt 0.144

in or mm

or mm mm/mm deg C

Metric Units

Input Data 300.00 in 48.00 in 6.00 in 9.38 Btu/hr-ft^2-F 22.2 deg F -1.0 deg F 1.00 in 0.66 in 0.50 in 0.10 Btu/hr-ft-F 0.10 Btu/hr-ft-F 0.52 Btu/hr-ft-F 34.10 Btu/hr-ft^2-F Calculation Wall area, A = L*H*0.0929 / 144 A= 9.29 m^2 Thermal Resistances: Ro = 1 / ho Ro = 0.107 Ra = Xa / Ka Ra = 0.2540 Rb = Xb / Kb Rb = 0.1676 Rc = Xc / Kc Rc = 0.0244 R5 = 1 / h5 R5 = 0.029 Overall heat transfer coefficient 1 / U = 1/ho + Xa / Ka + Xb / Kb + Xc / Kc + 1/h5 1 / U = Ro+R1+Ra+Rb+Rc+R5 1/U= 0.5820 Answer: U = 1.718 ∆T = to - t5 Answer: ∆T = 23.2 deg C Heat transfer per unit area, Q / A = U*(∆T) Answer: Q / A = 39.86 Heat transfer, Q = U*A*(∆T) Answer: Q = 370 Watts Wall length, L = Wall height, H = Wall thickness, X = Inside convective coefficient, ho = to = t5 = Xa = Xb = Xc = Ka = Kb = Kc = Outside convective coefficient, h5 =

Surface temperature is found from Equation-1 above: Heat transfer per unit area, Q / A = (to - t1) / (1 / ho) t1 = to - (Q/A)*(1 / ho) Answer: t1 = 18.0 deg F Internal temperature follows: Heat transfer per unit area, Q / A = (t1 - t2) / (Xa / Ka) t2 = t1 - (Q/A)*(Xa / Ka) Answer: t2 = 7.8 deg F

Node 1, 0 = 300 + 400 + T2 + T4 - 4*T1 Node 2, 0 = T1 + 400 + 100 + T3 - 4*T2 Node 3, 0 = T4 + T2 + 100 + 200 - 4*T3 Node 4, 0 = 300 + T1 + T3 + 200 - 4*T4

Node 1, -700 = T2 + T4 - 4*T1 Node 2, -500 = T1 + T3 - 4*T2 Node 3, -300 = T4 + T2 - 4*T3 Node 4, -500 = T1 + T3 - 4*T4

Node 1 Node 2 Node 3 Node 4

Equations 0.0 0.0 0.0 0.0

Constants -700 -500 -300 -500

Solution Column E T1 = T2 = T3 = T4 =

deg C deg C deg C deg C

HEAT TRANSFER SPREADSHEET CALCULATIONS

4 PDH

Copy write, © Heat Transfer Spreadsheet Calculations by John R Andrew, 12 June 2011

CONVECTION Convection is heat transfer by the movement of heated gasses and liquids.

Measuring Air Film Coefficient Heat source, Q = Surface area, A = Inside air temp. thermocouple, t1 = Inside surface temp. thermocouple, t2 =

Input Data 100 0.4 65.3 20.0

Watts sq m C C

Calculations Heat convection, air layer, Q = h * A * (t1 - t2) Watts h = Q / (A*(t1 - t2)) S.I. Answer: h = 5.52 W/m^2*C h = (W/m^2*K)/5.596 U.S. Answer: h = 0.986 Btu/hr-ft^2*F Boundary layer thickness = Air flow velocity =

X V

mm m/s

Convective Heat Transfer Coefficient Convective heat transfer coefficient, h = k*C*(Gr*Pr)^n / L Description Vertical Plate or Vertical Cylinder

Length L

Horizontal Plate hot surface facing up

(S1 + S2) / 2

Gr C < 10^4 1.36 10^4 Protection > Protect Sheet > OK

GOAL SEEK - Trial and Error by Excel Spread Sheet T4 = 38.7 deg. C in the calculation below when input heat flow Q = 5.0 Watts. The objective is to find the input heat Q that will result in the temperature T4 = 25 deg. C. Excel spread sheets will make a trial and error iteration automatically with the tool called, "Goal Seek". 1. Select the calculated answer at red cell, T4 = 38.7 below. 2. Select: Tools > Goal Seek > Pick "To value:" > 25 > By changing: > Pick green cell, Q = 16.0 > Okay.

EXAMPLE - LOCKED Q= A= T1 = h1 = h2 = k= L= R1 = R1 = R2 = R2 = R3 = R3 = Q/A= T4 = Answer: T4 =

Input Data 5.0 Watts 0.25 sq m 45 deg. C 9.00 W/sq m C 5.00 W/sq m C 164 W/mC 0.25 m Calculations 1 / h1 sq m C/W 0.111 sq m C/W L/k m*C / W 0.0015 m*C / W 1 / h2 sq m C/W 0.200 sq m C/W (T1 -T4) / (R1 + R2 + R3) T1 - ((Q / A)*(R1 + R2 + R3)) 38.7 deg. C

PROBLEM - UNLOCKED Practice Goal Seek below:

Input Data Q= 16.0 A= 0.25 T1 = 45

Watts sq m deg. C

h1 = h2 = k= L= R1 = R1 = R2 = R2 = R3 = R3 = Q/A= T4 = Answer: T4 =

9.00 W/sq m C 5.00 W/sq m C 164 W/mC 0.25 m Calculations 1 / h1 sq m C/W 0.111 sq m C/W L/k m*C / W 0.0015 m*C / W 1 / h2 sq m C/W 0.200 sq m C/W (T1 -T4) / (R1 + R2 + R3) T1 - ((Q / A)*(R1 + R2 + R3)) 25.0 deg. C

EXCEL'S SOLVER > see 2D CONDUCTION To install Solver, click the Microsoft Office Button, click Excel Options, and click Add-Ins. In the Manage box at the bottom of the window, select Excel Add-ins, and click Go. Check the Solver Add-in box in the Add-Ins dialog box, and click OK. After Solver is installed, you can run Solver by clicking Solver in the Analysis group on the Data tab.

Step-1 Clik the "Excel Button" top left > Click "Excel Options".

This is the end of this spread sheet.