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 $x = x<em>1$, $T = T</em>1$

  • At $x = x<em>2$, $T = T</em>2$

• Integrating Fourier's Law:

  • $\int<em>{x</em>1}^{x<em>2} dx = -\frac{kA}{q</em>x} \int<em>{T</em>1}^{T_2} dT$

  • $q<em>x (x</em>2 - x<em>1) = -kA (T</em>2 - T_1)$

• Temperature Distribution:

  • $T(x) = T<em>1 - \frac{q</em>x}{kA} (x - x_1)$

• Thermal Resistance Concept:

  • $R<em>t = \frac{T</em>1 - T<em>2}{q</em>x} = \frac{x<em>2 - x</em>1}{kA}$

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:

  • $R<em>t = \frac{x</em>2 - x_1}{kA} = \frac{0.01 \, m}{17 \, W/(m \cdot K) \cdot 1 \, m^2} = 5.88 \times 10^{-4} \, °C/W$

  • $\frac{q<em>x}{A} = \frac{T</em>1 - T<em>2}{R</em>t} = \frac{110 - 90}{5.88 \times 10^{-4}} = 34013 \, W/m^2$

  • $R_{t,0.5} = \frac{0.005 \, m}{17 \, W/(m \cdot K) \cdot 1 \, m^2} = 2.94 \times 10^{-4} \, °C/W$

  • $T = T<em>1 - \frac{q</em>x}{A} R_{t,0.5} = 110 - 34013 \times 2.94 \times 10^{-4} = 100 \, °C$

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 $r = r<em>i$, $T = T</em>i$

  • At $r = r<em>o$, $T = T</em>o$

• Temperature Distribution:

  • $\int<em>{r</em>i}^{r<em>o} \frac{dr}{r} = -\frac{2 \pi k L}{q</em>r} \int<em>{T</em>i}^{T_o} dT$

  • $\ln \left( \frac{r<em>o}{r</em>i} \right) = \frac{2 \pi k L}{q<em>r} (T</em>i - T_o)$

• Thermal Resistance:

  • $R<em>t = \frac{T</em>i - T<em>o}{q</em>r} = \frac{\ln(r<em>o/r</em>i)}{2 \pi k L}$

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:

  • $R<em>t = \frac{\ln(r</em>o/r_i)}{2 \pi k L} = \frac{\ln(0.05/0.03)}{2 \pi \times 43 \times 40} = 4.727 \times 10^{-5} \, °C/W$

  • $q<em>r = \frac{T</em>i - T<em>o}{R</em>t} = \frac{115 - 90}{4.727 \times 10^{-5}} = 528903 \, W$

Multilayered Systems

• Heat transfer through multiple layers of different materials.

• Consider 3 layers: B, C and D having thickness $\delta x<em>B$, $\delta x</em>C$ and $\delta x<em>D$ and thermal conductivities $K</em>B$, $K<em>C$ and $K</em>D$ respectively.

• The total temperature difference is the sum of the temperature drops across each layer: $\Delta T = \Delta T<em>B + \Delta T</em>C + \Delta T_D$

• Heat transfer through each layer:

  • $q = -K<em>B A \frac{\Delta T</em>B}{\Delta x<em>B} = -K</em>C A \frac{\Delta T<em>C}{\Delta x</em>C} = -K<em>D A \frac{\Delta T</em>D}{\Delta x_D}$, where A is the area.

• Total thermal resistance is the sum of individual resistances:

  • $R<em>{total} = R</em>B + R<em>C + R</em>D$

• Heat flux is calculated as follows:

  • $q = \frac{T<em>1 - T</em>4}{\frac{\delta x<em>B}{K</em>B A} + \frac{\delta x<em>C}{K</em>C A} + \frac{\delta x<em>D}{K</em>D A}}$

• Two-Layer System:

  • $R<em>t = R</em>{tA} + R_{tB}$

  • Cylindrical: $R<em>t = \frac{\ln(r</em>2/r<em>1)}{2 \pi L K</em>A} + \frac{\ln(r<em>3/r</em>2)}{2 \pi L K_B}$

Example

• Calculate the interface temperature between steel and insulation.

• Given: Steel pipe with $k<em>1 = 17 \, W/(m \cdot K)$, insulation with $k</em>2 = 0.035 \, W/(m \cdot K)$, $L = 1 \, m$, $T<em>\infty = 25 \, °C$, $T</em>1 = 130 \, °C$, $r<em>1 = 0.04 \, m$, $r</em>2 = 0.06 \, m$, $r_3 = 0.1 \, m$

• Calculations:

  • $R<em>{t1} = \frac{\ln(r</em>2/r<em>1)}{2 \pi L k</em>1} = \frac{\ln(0.06/0.04)}{2 \pi \times 1 \times 17} = 0.0038 \, °C/W$

  • $R<em>{t2} = \frac{\ln(r</em>3/r<em>2)}{2 \pi L k</em>2} = \frac{\ln(0.1/0.06)}{2 \pi \times 1 \times 0.035} = 2.3229 \, °C/W$

  • $q = \frac{T<em>1 - T</em>\infty}{R<em>{t1} + R</em>{t2}} = \frac{130 - 25}{0.0038 + 2.3229} = 45.13 \, W$

  • $T<em>i = T</em>1 - q R_{t1} = 130 - 45.13 \times 0.0038 = 129.83 \, °C$

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: $Gr = \frac{g \beta (T<em>s - T</em>\infty) L^3}{\nu^2}$

• Film Temperature: Average of the surface and fluid temperatures.

  • $T<em>f = \frac{T</em>s + T_\infty}{2}$

• 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:

  1. Heat loss is by free convection.

  2. Film temperature: $T_f = \frac{130 + 30}{2} = 80 °C$

  3. Properties of air at 80°C:

     • $\rho = 0.968 \, kg/m^3$

     • $\beta = 2.83 \times 10^{-3} \, K^{-1}$

     • $c_p = 1.019 \, kJ/(kg \cdot °C)$

     • $k = 0.0293 \, W/(m \cdot °C)$

     • $\mu = 20.79 \times 10^{-6} \, N \cdot s/m^2$

     • $Pr = 0.71$

     • $g = 9.81 \, m/s^2$

  4. Rayleigh number:

     • $Gr = \frac{d^3 \rho^2 g \beta \Delta T}{\mu^2} = \frac{(0.1)^3 (0.968)^2 (9.81) (2.83 \times 10^{-3}) (130-30)}{(20.79 \times 10^{-6})^2} = 6.019 \times 10^6$

     • $Ra = Gr \times Pr = (6.019 \times 10^6)(0.71) = 4.27 \times 10^6$

  5. From Table 4.2, for a horizontal cylinder:

  6. Nusselt number:

     • $Nu = 0.6 + \frac{0.387 Ra^{1/6}}{\left[1 + (0.559/Pr)^{9/16}\right]^{8/27}} = 0.6 + \frac{0.387(4.27 \times 10^6)^{1/6}}{\left[1 + (0.559/0.71)^{9/16}\right]^{8/27}} = 22$

  7. 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.

• $q = \frac{T<em>\infty - T</em>o}{R_{total}}$, where:

  • $R<em>{total} = R</em>{convection,inside} + R<em>{conduction} + R</em>{convection,outside}$

  • $R_{convection} = \frac{1}{hA}$

  • $R<em>{conduction} = \frac{\ln(r</em>o/r_i)}{2 \pi k L}$

• $q = U<em>i A</em>i (T<em>i - T</em>\infty)$, where $U_i$ is the overall heat transfer coefficient based on the inside area.

  • $\frac{1}{U<em>i A</em>i} = \frac{1}{h<em>i A</em>i} + \frac{\ln(r<em>o/r</em>i)}{2 \pi k L} + \frac{1}{h<em>o A</em>o}$

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:

  1. Steady state.

  2. Overall heat transfer coefficient is constant.

  3. No axial conduction.

  4. Well insulated.

• $dq = U<em>i (\Delta T)</em>{overall} dA_i$

• $dq = m<em>c c</em>{pc} dT<em>c = m</em>h c<em>{ph} dT</em>h = U<em>i (T</em>h - T<em>c) dA</em>i$

• Log Mean Temperature Difference (LMTD):

  • $\Delta T<em>{lm} = \frac{\Delta T</em>1 - \Delta T<em>2}{\ln(\Delta T</em>1/\Delta T_2)}$

• $q = U A (\Delta T)_{lm}$

Example

• Liquid food (inlet 20°C, exit 60°C, $c<em>p$ = 4 kJ/kg°C, 0.5 kg/s) is heated by hot water (inlet 90°C, $c</em>p$ = 4.18 kJ/kg°C, 1.0 kg/s).

• Solution:

  1. Heat balance: $q = m<em>c c</em>{pc} \Delta T = m<em>h c</em>{ph} \Delta T$

     • $(0.5)(4)(60-20) = (1)(4.18)(90-T_2)$

     • $T_2 = 70.9 °C$

  2. The exit temperature of water is 70.9°C.

  3. Log-mean temperature difference:

     • $\Delta T_{lm} = \frac{(90-60) - (70.9-20)}{\ln((90-60)/(70.9-20))} = \frac{30 - 50.9}{\ln(30/50.9)} = 39.5 °C$

  4. The log-mean temperature difference is 39.5°C

  5. Heat transfer rate: $q = (0.5)(4)(60-20) = 80 kJ/s$

  6. Heat exchanger length in the countercurrent case: $L = \frac{(80 \, kJ/s)(1000 \, J/kJ)}{\pi (0.05 \, m) (39.5°C) (2000 \, W/[m^2°C])} = 6.45 \, m$

  7. For parallel-flow operation, assume the exit temperature will be the same as for counterflow, $T_2 = 70.9°C$.

  8. Log-mean temperature difference for parallel flow:

     • $\Delta T_{lm} = \frac{(90-20) - (70.9-60)}{\ln((90-20)/(70.9-60))} = 31.8 °C$

  9. The log-mean temperature difference for parallel flow is 31.8°C, about 8°C less than that for the countercurrent flow arrangement.

  10. Heat exchanger length in the parallel-current case: $L = \frac{(80 \, kJ/s)(1000 \, J/kJ)}{\pi (0.05 \, m) (31.8°C) (2000 \, W/[m^2°C])} = 8 \, m$