Steady State Heat Transfer Notes
Steady State Heat Transfer
Conduction
Conduction: Heat transfer through a solid material due to a temperature gradient.
Thermal resistance circuit analogy.
Fourier's Law of Conduction: \frac{q_x}{A} = -k \frac{dT}{dx} , where:
q_x is the heat transfer rate in the x-direction.
A is the cross-sectional area perpendicular to the heat flow.
k is the thermal conductivity of the material.
\frac{dT}{dx} is the temperature gradient in the x-direction.
Assumptions:
Steady-state conditions (temperature does not change with time).
One-dimensional heat transfer.
Constant thermal conductivity.
Boundary Conditions:
At x = x1, T = T1
At x = x2, T = T2
Integrating Fourier's Law:
\int{x1}^{x2} dx = -\frac{kA}{qx} \int{T1}^{T_2} dT
qx (x2 - x1) = -kA (T2 - T_1)
Temperature Distribution:
T(x) = T1 - \frac{qx}{kA} (x - x_1)
Thermal Resistance Concept:
Rt = \frac{T1 - T2}{qx} = \frac{x2 - x1}{kA}
Example
Given a material with a temperature of 110°C on one side and 90°C on the other side, with a thickness of 1 cm, determine the temperature at 0.5 cm from the 110°C surface.
Calculations:
Rt = \frac{x2 - x_1}{kA} = \frac{0.01 \, m}{17 \, W/(m \cdot K) \cdot 1 \, m^2} = 5.88 \times 10^{-4} \, °C/W
\frac{qx}{A} = \frac{T1 - T2}{Rt} = \frac{110 - 90}{5.88 \times 10^{-4}} = 34013 \, W/m^2
R_{t,0.5} = \frac{0.005 \, m}{17 \, W/(m \cdot K) \cdot 1 \, m^2} = 2.94 \times 10^{-4} \, °C/W
T = T1 - \frac{qx}{A} R_{t,0.5} = 110 - 34013 \times 2.94 \times 10^{-4} = 100 \, °C
Cylindrical Coordinates
Heat transfer through a cylindrical geometry.
Fourier's Law in Cylindrical Coordinates: q_r = -kA \frac{dT}{dr} = -k (2 \pi r L) \frac{dT}{dr}, where:
r is the radial distance.
L is the length of the cylinder.
Boundary Conditions:
At r = ri, T = Ti
At r = ro, T = To
Temperature Distribution:
\int{ri}^{ro} \frac{dr}{r} = -\frac{2 \pi k L}{qr} \int{Ti}^{T_o} dT
\ln \left( \frac{ro}{ri} \right) = \frac{2 \pi k L}{qr} (Ti - T_o)
Thermal Resistance:
Rt = \frac{Ti - To}{qr} = \frac{\ln(ro/ri)}{2 \pi k L}
Example
Given a pipe with inner radius 0.03 m, outer radius 0.05 m, length 40 m, and thermal conductivity 43 W/m°C, with inner temperature 115°C and outer temperature 90°C, calculate the conductive heat loss.
Calculations:
Rt = \frac{\ln(ro/r_i)}{2 \pi k L} = \frac{\ln(0.05/0.03)}{2 \pi \times 43 \times 40} = 4.727 \times 10^{-5} \, °C/W
qr = \frac{Ti - To}{Rt} = \frac{115 - 90}{4.727 \times 10^{-5}} = 528903 \, W
Multilayered Systems
Heat transfer through multiple layers of different materials.
Consider 3 layers: B, C and D having thickness \delta xB, \delta xC and \delta xD and thermal conductivities KB, KC and KD respectively.
The total temperature difference is the sum of the temperature drops across each layer: \Delta T = \Delta TB + \Delta TC + \Delta T_D
Heat transfer through each layer:
q = -KB A \frac{\Delta TB}{\Delta xB} = -KC A \frac{\Delta TC}{\Delta xC} = -KD A \frac{\Delta TD}{\Delta x_D}, where A is the area.
Total thermal resistance is the sum of individual resistances:
R{total} = RB + RC + RD
Heat flux is calculated as follows:
q = \frac{T1 - T4}{\frac{\delta xB}{KB A} + \frac{\delta xC}{KC A} + \frac{\delta xD}{KD A}}
Two-Layer System:
Rt = R{tA} + R_{tB}
Cylindrical: Rt = \frac{\ln(r2/r1)}{2 \pi L KA} + \frac{\ln(r3/r2)}{2 \pi L K_B}
Example
Calculate the interface temperature between steel and insulation.
Given: Steel pipe with k1 = 17 \, W/(m \cdot K), insulation with k2 = 0.035 \, W/(m \cdot K), L = 1 \, m, T\infty = 25 \, °C, T1 = 130 \, °C, r1 = 0.04 \, m, r2 = 0.06 \, m, r_3 = 0.1 \, m
Calculations:
R{t1} = \frac{\ln(r2/r1)}{2 \pi L k1} = \frac{\ln(0.06/0.04)}{2 \pi \times 1 \times 17} = 0.0038 \, °C/W
R{t2} = \frac{\ln(r3/r2)}{2 \pi L k2} = \frac{\ln(0.1/0.06)}{2 \pi \times 1 \times 0.035} = 2.3229 \, °C/W
q = \frac{T1 - T\infty}{R{t1} + R{t2}} = \frac{130 - 25}{0.0038 + 2.3229} = 45.13 \, W
Ti = T1 - q R_{t1} = 130 - 45.13 \times 0.0038 = 129.83 \, °C
Convection
Convection: Heat transfer between a surface and a moving fluid.
Convective Heat Transfer Coefficient (h): Energy transmitted per unit time from/to a surface of unit cross-sectional area.
Thermal resistance to convection: \frac{1}{hA}.
Nusselt Number (Nu): Enhancement in rate of heat transfer caused by convection over the conduction mode.
Nu = 5: Rate of convective heat transfer is five times more than rate of heat transfer by conduction alone.
Dimensionless Numbers
Prandtl Number (Pr): Ratio of momentum diffusivity to thermal diffusivity.
\Pr = \frac{\nu}{\alpha} = \frac{\mu c_p}{k}
Indicates the relative thickness of the hydrodynamic and thermal boundary layers.
\Pr << 1: Molecular diffusivity of heat is larger than momentum diffusivity. Heat dissipates faster.
Gases: \Pr \approx 0.7
Water: \Pr \approx 10
Forced Convection
Forced Convection: Fluid motion is induced by external means (e.g., pump, fan).
Flow Regimes:
Laminar: Smooth, orderly fluid motion.
Transition: Intermediate regime.
Turbulent: Chaotic, disordered fluid motion.
Reynolds Number (Re): Dimensionless number indicating the flow regime.
Re = \frac{\rho V D}{\mu}
Laminar: Re < 2100
Transition: 2100 < Re < 10000
Turbulent: Re > 10000
Example 4.11
Water flowing at 0.02 kg/s is heated from 20°C to 60°C in a horizontal pipe (ID = 2.5 cm). The inside pipe surface temperature is 90°C. Estimate the convective heat transfer coefficient if the pipe is 1m long.
The solution is given by:
h = \frac{NNuk}{D} = \frac{(11.2)(0.633 \, W/[m°C])}{(0.025 \, m)} = 284 \, W/(m^2 °C)
The convective heat-transfer coefficient is estimated to be 284 W/(m² °C).
Free Convection
Free Convection: Fluid motion is induced by density differences due to temperature gradients.
Rayleigh Number (Ra): Dimensionless number characterizing free convection.
Grashof Number (Gr): Ratio of buoyancy forces to viscous forces.
It's defined as: Gr = \frac{g \beta (Ts - T\infty) L^3}{\nu^2}
Film Temperature: Average of the surface and fluid temperatures.
Tf = \frac{Ts + T_\infty}{2}
Table 4.2: Coefficients for free convection calculations based on geometry and Rayleigh number range.
Example 4.15
Estimate the convective heat transfer coefficient for convective heat loss from a horizontal 10 cm diameter steam pipe. The surface temperature is 130°C, and the air temperature is 30°C.
Solution:
Heat loss is by free convection.
Film temperature: T_f = \frac{130 + 30}{2} = 80 °C
Properties of air at 80°C:
\rho = 0.968 \, kg/m^3
\beta = 2.83 \times 10^{-3} \, K^{-1}
c_p = 1.019 \, kJ/(kg \cdot °C)
k = 0.0293 \, W/(m \cdot °C)
\mu = 20.79 \times 10^{-6} \, N \cdot s/m^2
Pr = 0.71
g = 9.81 \, m/s^2
Rayleigh number:
Gr = \frac{d^3 \rho^2 g \beta \Delta T}{\mu^2} = \frac{(0.1)^3 (0.968)^2 (9.81) (2.83 \times 10^{-3}) (130-30)}{(20.79 \times 10^{-6})^2} = 6.019 \times 10^6
Ra = Gr \times Pr = (6.019 \times 10^6)(0.71) = 4.27 \times 10^6
From Table 4.2, for a horizontal cylinder:
Nusselt number:
Nu = 0.6 + \frac{0.387 Ra^{1/6}}{\left[1 + (0.559/Pr)^{9/16}\right]^{8/27}} = 0.6 + \frac{0.387(4.27 \times 10^6)^{1/6}}{\left[1 + (0.559/0.71)^{9/16}\right]^{8/27}} = 22
Convective heat transfer coefficient:
h = \frac{Nu \cdot k}{d} = \frac{22 \cdot 0.0293}{0.1} = 6.5 \, W/(m^2 \cdot °C)
Overall Heat Transfer Coefficient
Combining convection and conduction resistances.
q = \frac{T\infty - To}{R_{total}}, where:
R{total} = R{convection,inside} + R{conduction} + R{convection,outside}
R_{convection} = \frac{1}{hA}
R{conduction} = \frac{\ln(ro/r_i)}{2 \pi k L}
q = Ui Ai (Ti - T\infty), where U_i is the overall heat transfer coefficient based on the inside area.
\frac{1}{Ui Ai} = \frac{1}{hi Ai} + \frac{\ln(ro/ri)}{2 \pi k L} + \frac{1}{ho Ao}
Example 4.16
A 2.5 cm ID pipe conveys liquid food at 80°C. The inside convective heat transfer coefficient is 10 W/m²C. The pipe (0.5 cm thick) is steel (k = 43 W/mC). The outside ambient temperature is 20°C, and the outside convective heat-transfer coefficient is 100 W/m²C. Calculate the overall heat transfer coefficient and the heat loss from 1 m length of the pipe.
Heat Exchangers
Devices designed for efficient heat transfer between two fluids.
Assumptions:
Steady state.
Overall heat transfer coefficient is constant.
No axial conduction.
Well insulated.
dq = Ui (\Delta T){overall} dA_i
dq = mc c{pc} dTc = mh c{ph} dTh = Ui (Th - Tc) dAi
Log Mean Temperature Difference (LMTD):
\Delta T{lm} = \frac{\Delta T1 - \Delta T2}{\ln(\Delta T1/\Delta T_2)}
q = U A (\Delta T)_{lm}
Example
Liquid food (inlet 20°C, exit 60°C, cp = 4 kJ/kg°C, 0.5 kg/s) is heated by hot water (inlet 90°C, cp = 4.18 kJ/kg°C, 1.0 kg/s).
Solution:
Heat balance: q = mc c{pc} \Delta T = mh c{ph} \Delta T
(0.5)(4)(60-20) = (1)(4.18)(90-T_2)
T_2 = 70.9 °C
The exit temperature of water is 70.9°C.
Log-mean temperature difference:
\Delta T_{lm} = \frac{(90-60) - (70.9-20)}{\ln((90-60)/(70.9-20))} = \frac{30 - 50.9}{\ln(30/50.9)} = 39.5 °C
The log-mean temperature difference is 39.5°C
Heat transfer rate: q = (0.5)(4)(60-20) = 80 kJ/s
Heat exchanger length in the countercurrent case: L = \frac{(80 \, kJ/s)(1000 \, J/kJ)}{\pi (0.05 \, m) (39.5°C) (2000 \, W/[m^2°C])} = 6.45 \, m
For parallel-flow operation, assume the exit temperature will be the same as for counterflow, T_2 = 70.9°C.
Log-mean temperature difference for parallel flow:
\Delta T_{lm} = \frac{(90-20) - (70.9-60)}{\ln((90-20)/(70.9-60))} = 31.8 °C
The log-mean temperature difference for parallel flow is 31.8°C, about 8°C less than that for the countercurrent flow arrangement.
Heat exchanger length in the parallel-current case: L = \frac{(80 \, kJ/s)(1000 \, J/kJ)}{\pi (0.05 \, m) (31.8°C) (2000 \, W/[m^2°C])} = 8 \, m