2  Heat Transfer

2.1 Conduction

Conduction transfers heat through direct contact, as faster-vibrating molecules pass energy to slower ones. In solids, heat moves from molecule to molecule, and between objects in contact, it flows from the hotter region to the cooler one until thermal equilibrium is reached.

An example of conduction is an iron bar with one end placed in a flame. The other end soon becomes hot as heat is conducted along the bar from molecule to molecule through the metal.

The rate of heat transfer by conduction through a solid is given by:

\[ Q = \frac{k \, A \, t \, \Delta T}{s} \tag{2.1}\]

where:

  • \(Q\) = heat transferred (J)

  • \(k\) = thermal conductivity of the material (\(\text{W/m·K}\))

  • \(A\) = cross-sectional area through which heat flows (\(\text{m}^2\))

  • \(t\) = time of heat transfer (s)

  • \(\Delta T\) = temperature difference across the material (K or °C)

  • \(s\) = thickness or length of the material through which heat is conducted (m)

This equation shows that heat conduction increases with greater thermal conductivity, larger surface area, longer time, and larger temperature difference but decreases as the material’s thickness increases.

2.2 Temperature Difference Across a Heat Exchanger Endplate

A large heat exchanger has a shell diameter of 1.8 meters. One endplate, a flat cover, is made of 45 mm thick carbon steel and is uninsulated. The maximum allowable heat loss through this cover, without insulation, is 150 MJ/h. Given that the thermal conductivity of steel is 55 W/mK, calculate the permissible temperature difference across the endplate.

Given

  • Shell diameter: (D = 1.8 \(\mathrm{m}\))
  • Endplate thickness: (s = 45 \(\mathrm{mm}\) = 0.045 \(\mathrm{m}\))
  • Material: carbon steel, thermal conductivity: (k = 55 \(\mathrm{W/m\cdot K}\))
  • Maximum allowable heat loss: (\(Q_\mathrm{max} = 150\) \(\mathrm{MJ/h}\))
  • Endplate is uninsulated

Calculate the temperature difference \(\Delta T\) across the endplate to limit heat loss.

  1. Area of the circular endplate

\[ A = \pi \left( \frac{D}{2} \right)^2 = \pi \left( \frac{1.8}{2} \right)^2 \]

\[ A = \pi \times 0.9^2 = 2.5447\ \mathrm{m^2} \]

  1. Convert heat loss to Watts

\[ Q_\mathrm{max} = 150\ \mathrm{MJ/h} = 150 \times 10^6\ \mathrm{J/h} \]

\[ 1\ \mathrm{h} = 3600\ \mathrm{s} \quad \Rightarrow \quad Q = \frac{150 \times 10^6}{3600} = 41666.7\ \mathrm{W} \]

  1. Use law of conduction for a flat plate

\[ Q = \frac{k A \Delta T}{s} \quad \Rightarrow \quad \Delta T = \frac{Q s}{k A} \]

Substitute values:

\[ \Delta T = \frac{41666.7 \times 0.045}{55 \times 2.5447} \]

\[ \Delta T = \frac{1875.0}{139.96} = 13.39\ \mathrm{K} \]

Result

\[ \boxed{\Delta T = 13.4^\circ \mathrm{C}} \]

Notes:

  • The maximum allowable temperature difference across the endplate is 13.4°C to keep heat loss below 150 MJ/h.

  • If the temperature difference exceeds this, insulation would be required.

2.2.1 Code

# Given data
D = 1.8              # diameter of endplate, m
s = 0.045            # thickness, m
k = 55               # thermal conductivity, W/m.K
Q_MJ_per_h = 150     # maximum heat loss, MJ/h

# Convert heat loss to W
Q_W = Q_MJ_per_h * 1e6 / 3600  # W

# Area of the circular plate
import math
A = math.pi * (D/2)**2

# Temperature difference
delta_T = Q_W * s / (k * A)

# Display results
print(f"Endplate area: {A:.4f} m²")
print(f"Heat loss: {Q_W:.2f} W")
print(f"Temperature difference across the endplate: {delta_T:.2f} °C")

2.3 Heat Transfer Calculation for Cold Storage Room

The product storage room for a fresh frozen fish plant is 30 m long, 23 m wide, and 3 m high. The walls and ceiling are insulated with 25 cm of granulated cork (\(k = 0.049 \text{ W/mK}\)), and the floor with 15 cm of corkboard (\(k = 0.043 \text{ W/mK}\)). Calculate the heat transfer per hour due to conduction if the temperature inside the chamber is kept at -30°C and the ambient air temperature outside is 21°C. Take the ground temperature as 10°C.

2.3.1 Code

"""
Heat Transfer Calculation for Cold Storage Room

Calculate heat transfer per hour through walls, ceiling, and floor
due to conduction in a frozen fish storage room.
"""

# Room dimensions
length = 30  # m
width = 23  # m
height = 3  # m

# Insulation properties
cork_thickness_walls = 0.25  # m (25 cm)
cork_k_walls = 0.049  # W/mK
corkboard_thickness_floor = 0.15  # m (15 cm)
corkboard_k_floor = 0.043  # W/mK

# Temperatures
T_inside = -30  # °C
T_ambient = 21  # °C
T_ground = 10  # °C

# Calculate surface areas
area_floor = length * width
area_ceiling = length * width
area_wall_long = 2 * (length * height)  # Two long walls
area_wall_short = 2 * (width * height)  # Two short walls
area_walls_total = area_wall_long + area_wall_short

# Heat transfer equation: Q = k × A × ΔT / s
# where:
#   Q = heat transfer rate (W)
#   k = thermal conductivity (W/mK)
#   A = area (m²)
#   ΔT = temperature difference (K or °C)
#   s = thickness (m)

# 1. Heat transfer through walls (exposed to ambient air)
delta_T_walls = T_ambient - T_inside
Q_walls = (
    cork_k_walls * area_walls_total * delta_T_walls
) / cork_thickness_walls

# 2. Heat transfer through ceiling (exposed to ambient air)
delta_T_ceiling = T_ambient - T_inside
Q_ceiling = (
    cork_k_walls * area_ceiling * delta_T_ceiling
) / cork_thickness_walls

# 3. Heat transfer through floor (exposed to ground)
delta_T_floor = T_ground - T_inside
Q_floor = (
    corkboard_k_floor * area_floor * delta_T_floor
) / corkboard_thickness_floor

# Total heat transfer
Q_total_watts = Q_walls + Q_ceiling + Q_floor
Q_total_kW = Q_total_watts / 1000
Q_total_per_hour_kWh = Q_total_kW  # kW × 1 hour = kWh
Q_total_per_hour_MJ = Q_total_kW * 3.6  # kW × 3600 s = kJ/s × 3600 = MJ

# Display Results
print("\n" + "=" * 30)
print("HEAT TRANSFER PER HOUR")
print("=" * 30)
print(f"Q per hour: {Q_total_watts * 3600 / 1000:.2f} kJ")

pct_walls = (Q_walls / Q_total_watts) * 100
pct_ceiling = (Q_ceiling / Q_total_watts) * 100
pct_floor = (Q_floor / Q_total_watts) * 100

print(f"Walls:   {pct_walls:>6.2f}%")
print(f"Ceiling: {pct_ceiling:>6.2f}%")
print(f"Floor:   {pct_floor:>6.2f}%")
print("=" * 30)

2.4 Latent Heat of Vaporization

Water is boiling in an aluminum pan on a stovetop. The pan’s bottom is \(s = 0.7\ \text{cm}\) thick and has a diameter of \(D = 12\ \text{cm}\). Water evaporates at a rate of \(\dot m = 1.00\ \text{g/s}\).

Given:

  • Thermal conductivity of aluminum: \(k = 220\ \text{W/m·K}\)

  • Latent heat of vaporization: \(LHV = 2.256\times10^{6}\ \text{J/kg}\)

Calculate the temperature difference across the pan’s bottom.

The heat required to evaporate the water is

\[ \dot Q = \dot m LHV \]

Heat conduction through the pan bottom:

\[ \dot Q = \frac{kA}{s}\,\Delta T \]

Solving for the temperature difference:

\[ \Delta T = \frac{\dot Q\, s}{kA} \]

2.4.1 Code

import math

# Given values
m_dot = 1.0e-3      # kg/s
LHV = 2.256e6       # J/kg
k = 220             # W/m·K
s = 0.007           # m
D = 0.12            # m

# Heat transfer rate
Q_dot = m_dot * LHV

# Area of pan bottom
A = math.pi * (D / 2)**2

# Temperature difference
delta_T = Q_dot * s / (k * A)

# Display results
print(f"Heat transfer rate: {Q_dot:.4f} J/s")
print(f"Area of pan bottom: {A:.4f} m²")
print(f"Temperature difference: {delta_T:.4f} °C")

2.5 Convection Heat Transfer

Convection transfers heat through the movement of fluids (liquids or gases). When a fluid is heated, it expands, becomes less dense, and rises, while cooler, denser fluid moves in to replace it, creating a convection current that distributes heat.

Natural convection occurs without mechanical aid, whereas forced convection involves devices such as pumps or fans. When fluid movement is driven by a pump or fan, heat is transferred by forced convection. Examples include:

  • A pump circulating hot water through a building’s heating system,
  • A fan forcing air through an automobile radiator, or
  • A forced draft fan pushing hot gases through a boiler.

The total heat transferred between a solid surface and a moving fluid over a specified time is given by:

\[ Q = h_A \, A \, t \, (T_s - T_f) \tag{2.2}\]

where:

  • \(Q\) = total heat transferred (J)

  • \(h_A\) = surface (convective) heat transfer coefficient (\(\text{W/m}^2\text{·K}\))

  • \(A\) = surface area of heat transfer (\(\text{m}^2\))

  • \(t\) = time during which heat transfer occurs (s), with \(t = 1 \, \text{s}\) for unit time

  • \(T_s\) = surface temperature of the solid (K or °C)

  • \(T_f\) = temperature of the surrounding fluid (K or °C)

This equation describes how heat energy is transferred between a surface and a fluid over a unit time of 1 second, depending on the temperature difference, the surface area, the time, and the fluid’s heat-carrying effectiveness, represented by the convective heat transfer coefficient (\(h_A\))

2.6 Hot Metal Convection Heat Transfer

A hot metal plate measuring 1.2 m × 0.8 m is exposed to air at 25°C. The surface temperature of the plate is maintained at 85°C. If the convective heat transfer coefficient (surface heat transfer coefficient) between the plate and air is \(h_A = 18 \, \text{W/m}^2\text{·°C}\), calculate the rate of heat loss by convection from the entire plate surface.

Given:

  • Plate dimensions: (\(1.2\ \text{m} \times 0.8\ \text{m}\))

  • Plate surface area: \(A = 1.2 \times 0.8 = 0.96 \, \text{m}^2\)

  • Surface temperature: \(T_s = 85^\circ \text{C}\)

  • Air temperature: \(T_f = 25^\circ \text{C}\)

  • Convection coefficient: \(h_A = 18 \, \text{W/m}^2\text{·°C}\)

Solution

The rate of convective heat loss is given by:

\[ Q = h_A \, A \, (T_s - T_f) \]

Substituting the values:

\[ Q = 18 \times 0.96 \times (85 - 25) \]

\[ Q = 18 \times 0.96 \times 60 \]

\[ Q = 1036.8 \, \text{W} \]

Answer

\[ \boxed{Q = 1036.8 \, \text{W}} \]

2.6.1 Code

# Given data
L = 1.2        # length of the plate (m)
W = 0.8        # width of the plate (m)
T_surface = 85 # surface temperature (°C)
T_air = 25     # air temperature (°C)
hA = 18        # convective heat transfer coefficient (W/m²·°C)

# Calculate area of the plate
A = L * W

# Calculate heat loss rate
Q = hA * A * (T_surface - T_air)

# Display results
print(f"Plate area: {A:.2f} m²")
print(f"Temperature difference: {T_surface - T_air:.1f} °C")
print(f"Convective heat loss: {Q:.2f} W")

2.7 Radiation

Radiation transfers heat through electromagnetic waves that travel in straight lines and can pass through a vacuum. When these waves strike a surface, they may be absorbed (increasing temperature), reflected, or transmitted. Dark, rough surfaces absorb more radiation, while shiny, smooth ones reflect it.

The heat transfer by radiation is given by:

\[ Q = \sigma \, \varepsilon \, A \, t \, (T_1^4 - T_2^4) \tag{2.3}\]

where:

  • \(Q\) is the heat energy transferred by radiation (in kilojoules, kJ).

  • \(\sigma\) is the Stefan–Boltzmann constant, equal to \(5.6703 \times 10^{-11}\ \text{kW·m}^{-2}\text{·K}^{-4}\).

  • \(\varepsilon\) is the emissivity of the surface (dimensionless, between 0 and 1), indicating how efficiently the surface emits or absorbs radiation compared to a perfect blackbody.

  • \(A\) is the surface area of the radiating body (in square meters, m²).

  • \(t\) is the time during which heat transfer occurs (in seconds, s).

  • \(T_1\) is the absolute temperature of the radiating surface (in kelvin, K).

  • \(T_2\) is the absolute temperature of the surroundings (in kelvin, K).

This equation expresses the net radiant heat transfer between two bodies at different temperatures, taking into account the emissivity of the surface, its area, and the duration of heat exchange.

Examples include heat from the sun reaching Earth and radiant heat in a boiler furnace.

In a steam boiler, radiation occurs in the furnace. Any heating surfaces that are directly exposed to the furnace will receive heat directly by radiation from the flame. These include the waterwalls and some generating tubes of a watertube boiler, radiant superheater tubes (located at the outlet of the furnace), and the furnace walls of a firetube boiler.

2.8 Radiant Heat from a Flat Circular Plate

A flat circular plate is 500 mm in diameter. Calculate the theoretical quantity of heat radiated per hour when its temperature is 227 °C and the temperature of its surrounds is 27 °C.

Given:

  • Stefan–Boltzmann constant: \(\sigma = 5.6703\times10^{-11}\ \mathrm{kW\,m^{-2}\,K^{-4}}\)

  • Emissivity: \(\varepsilon = 1\) (ideal black body)

  1. Convert temperatures to Kelvin

\[ T_1 = 227 + 273 = 500\ \mathrm{K}, \quad T_2 = 27 + 273 = 300\ \mathrm{K} \]

  1. Calculate the area of the circular plate

\[ A = \pi \left( \frac{D}{2} \right)^2 = \pi \left( \frac{0.5}{2} \right)^2 \approx 0.196349\ \mathrm{m^2} \]

  1. Apply the formula

\[ Q = \sigma \times \varepsilon \times A \times t \times \left( T_1^4 - T_2^4 \right) \]

Substituting the values:

\[ Q = 5.6703 \times 10^{-11} \times 1 \times 0.196349 \times 3600 \times \left(500^4 - 300^4\right) \] \[ Q = 2180.3997\ \mathrm{kWh} \]

Result

\[ \text{Plate area: } 0.196349\ \mathrm{m^2} \] \[ \text{Emissivity: } \varepsilon = 1 \] \[ \boxed{\text{Heat radiated (per hour): } 2180.3997\ \mathrm{kWh}} \]

Note: This represents the theoretical maximum radiation for a perfect black body. Real materials would radiate less depending on their emissivity \(\varepsilon < 1\).

2.8.1 Code

import math

# Given data
sigma = 5.6703e-11   # kW/m^2/K^4
epsilon = 1.0       # emissivity (black body)
T1 = 227 + 273  # K (plate temperature)
T2 = 27 + 273   # K (surroundings)
d = 500 / 1000     # m (diameter)
A = math.pi * (d / 2)**2  # area in m²

# Heat radiated per second (kW)
Q_dot = sigma * epsilon * A * (T1**4 - T2**4)

# Heat radiated per hour (kWh)
Q_hour = Q_dot * 3600

# Display results
print(f"Plate area: {A:.6f} m²")
print(f"Emissivity (theoretical): {epsilon}")
print(f"Heat radiated per second: {Q_dot:.4f} kW")
print(f"Heat radiated per hour: {Q_hour:.2f} kWh")

2.9 Heat Exchangers

In an inert gas system, the boiler exhaust is cooled from 420°C to 130°C in a parallel flow heat exchanger. The gas flow rate is 0.55 kg/s, the cooling water flow rate is 0.65 kg/s, and the cooling water inlet temperature is 15°C. The overall heat transfer coefficient from the gas to the water is 150 W/m²K. Determine the cooling surface area required. Take \(c_p\) for exhaust gas to be 1130 J/kgK and \(c_p\) for water to be 4190 J/kgK.

2.9.1 Heat Transfer Rate

The heat transfer rate (\(Q\)) can be calculated using the formula:

\[ Q = \dot{m}_{\text{gas}} \cdot c_{p,\text{gas}} \cdot (T_{\text{gas,in}} - T_{\text{gas,out}}) \]

Substituting the given values:

\[ Q = 0.55 \cdot 1130 \cdot (420 - 130) \]

\[ Q = 0.55 \cdot 1130 \cdot 290 \]

\[ Q = \boxed{180,235 \, \text{W}} \]

2.9.2 Log Mean Temperature Difference (LMTD)

The temperature differences at the two ends of the heat exchanger are:

\[ \Delta T_1 = T_{\text{gas,in}} - T_{\text{water,in}} = 420 - 15 = 405 \, \text{°C} \]

\[ \Delta T_2 = T_{\text{gas,out}} - T_{\text{water,out}} \]

To find \(T_{\text{water,out}}\), use the heat balance equation:

\[ Q = \dot{m}_{\text{water}} \cdot c_{p,\text{water}} \cdot (T_{\text{water,out}} - T_{\text{water,in}}) \]

Rearranging for \(T_{\text{water,out}}\):

\[ T_{\text{water,out}} = \frac{Q}{\dot{m}_{\text{water}} \cdot c_{p,\text{water}}} + T_{\text{water,in}} \]

Substitute the known values:

\[ T_{\text{water,out}} = \frac{180,235}{0.65 \cdot 4190} + 15 \]

\[ T_{\text{water,out}} = 66.18 + 15 = 81.18 \, \text{°C} \]

Now calculate \(\Delta T_2\):

\[ \Delta T_2 = T_{\text{gas,out}} - T_{\text{water,out}} = 130 - 81.18 = 48.82 \, \text{°C} \]

The LMTD is given by:

\[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \]

Substitute the values:

\[ \text{LMTD} = \frac{405 - 48.82}{\ln\left(\frac{405}{48.82}\right)} = \frac{356.18}{\ln(8.30)} \]

\[ \text{LMTD} = \frac{356.18}{2.1157} \approx \boxed{168.35 \, \text{°C}} \]

2.9.3 Cooling Surface Area

The cooling surface area (\(A\)) is calculated using the formula:

\[ Q = U \cdot A \cdot \text{LMTD} \]

Rearranging for \(A\):

\[ A = \frac{Q}{U \cdot \text{LMTD}} \]

Substitute the known values:

\[ A = \frac{180,235}{150 \cdot 168.35} \]

\[ A = \frac{180,235}{25,252.5} \approx \boxed{7.14 \, \text{m}^2} \]

2.9.4 Final Answer

The required cooling surface area is approximately 7.14 m².

2.9.5 Code

#!/usr/bin/env python3
"""
Heat Exchanger

Parallel flow heat exchanger with exhaust gas and cooling water
"""

import math

# Given data
m_dot_gas = 0.55  # kg/s
m_dot_water = 0.65  # kg/s
T_gas_in = 420  # °C
T_gas_out = 130  # °C
T_water_in = 15  # °C
cp_gas = 1130  # J/kgK
cp_water = 4190  # J/kgK
U = 150  # W/m²K

# Heat transfer rate
Q = m_dot_gas * cp_gas * (T_gas_in - T_gas_out)

# Water outlet temperature
T_water_out = (Q / (m_dot_water * cp_water)) + T_water_in

# Temperature differences
Theta_1 = T_gas_in - T_water_in
Theta_2 = T_gas_out - T_water_out

# LMTD Theta_m
LMTD = (Theta_1 - Theta_2) / math.log(Theta_1 / Theta_2)

# Surface area
A = Q / (U * LMTD)

# Results
print(f"Heat Transfer Rate: {Q:,.2f} W")
print(f"Water Outlet Temp: {T_water_out:.2f}°C")
print(f"LMTD: {LMTD:.2f}°C")
print(f"Surface Area: {A:.2f} m²")