GTU 3151909 Heat Transfer Summer 2026 Solved Paper & Solutions

1. Conduction heat transfer: the transfer of heat through a solid, or a stationary fluid, from the hotter part to the colder part, by the collision of molecules and the movement of free electrons, with no bulk motion. Fourier's law: $q = -kA\dfrac{dT}{dx}$.
2. Thermal diffusivity ($\alpha$): the ratio of the thermal conductivity to the heat capacity per unit volume, $\alpha = \dfrac{k}{\rho c_p}$, in m²/s. It shows how fast heat spreads in a material.
3. Thermal resistance: the opposition that a body or a surface offers to the flow of heat. It is the temperature difference per unit heat flow, $R = \dfrac{\Delta T}{Q}$. For a plane wall, $R = \dfrac{L}{kA}$, and for convection, $R = \dfrac{1}{hA}$. Unit: K/W.

Mean film temperature ($T_f$): the average of the surface (wall) temperature and the free stream (fluid) temperature, used in external flow and free convection, to find the properties of the fluid.

The properties of the fluid, such as density, viscosity and conductivity, change with temperature, so they are taken at $T_f$, which represents the layer of fluid near the wall.

Bulk mean temperature ($T_b$): the average temperature of a fluid that flows in a tube or a duct, over a cross section. It is weighted by the flow of energy, so it is the temperature that a thermometer would show if the fluid were collected and mixed.

It changes along the length of the tube, as the fluid is heated or cooled. For a rough estimate, it is taken as the average of the inlet and the outlet values, $T_b = \dfrac{T_{b,in} + T_{b,out}}{2}$, and it is used for internal flow.

Derivation of the general heat conduction equation

Take a small rectangular element of the material, with sides $dx$, $dy$ and $dz$. The material has a density $\rho$, a specific heat $c$ and a thermal conductivity $k$. Heat is generated inside at the rate $\dot{q}$ per unit volume.

Energy balance:

The heat leaving the right face (by a Taylor series) is

The net heat conducted into the element in the $x$ direction is

For a constant $k$, divide by $k$ and use $\alpha = \dfrac{k}{\rho c}$.

This is the general heat conduction equation in Cartesian coordinates.

Steady state, no heat generation

• Steady state: the temperature does not change with time, so $\dfrac{\partial T}{\partial t} = 0$.
• No heat generation: $\dot{q} = 0$.

The equation becomes

This is the Laplace equation, which is the required result.

• Hydrodynamic (velocity) boundary layer: the thin layer of fluid next to a surface, in which the velocity of the fluid changes from zero at the surface to about 99 percent of the free stream velocity $U$. Its thickness is $\delta$. The effect of viscosity is limited to this layer.
• Thermal boundary layer: the thin layer of fluid next to a surface, in which the temperature changes from the surface temperature $T_w$ to about 99 percent of the free stream temperature $T_\infty$. Its thickness is $\delta_t$. The temperature gradient, and so the convection heat transfer, is in this layer.

The velocity profile is $\dfrac{u}{U} = \dfrac{y}{\delta}$, a linear profile. Let $\eta = \dfrac{y}{\delta}$, so $\dfrac{u}{U} = \eta$ and $dy = \delta\,d\eta$.

(iv) The ratio $\dfrac{\delta^*}{\theta}$ (the shape factor)

Given: inner diameter of the steel pipe $d_1 = 20$ mm, so $r_1 = 10$ mm. The wall is 2 mm thick, so $r_2 = 12$ mm. The insulation is 20 mm thick, so $r_3 = 32$ mm. $k_{ins} = 0.05$ W/mK, $h_i = 10$ W/m²K, $h_o = 5$ W/m²K.

Assumption: the conductivity of steel is not given. Steel conducts heat very well, and the wall is thin, so its resistance is neglected.

Take a length of 1 m of the pipe. The resistances are in series.

• Biot number ($Bi$): the ratio of the internal conduction resistance of a body to its surface convection resistance.

Here $L_c = V/A_s$ is the characteristic length. If $Bi < 0.1$, the temperature inside the body is almost uniform, and the lumped system analysis can be used.

• Fourier number ($Fo$): the dimensionless time, the ratio of the rate of conduction to the rate of storage of heat.

A larger $Fo$ means that the body has reached nearly the temperature of its surroundings. In the lumped analysis, $\dfrac{T - T_\infty}{T_i - T_\infty} = e^{-Bi\,Fo}$.

Applications of fins

1. The cylinders of air-cooled engines, such as motorcycles and scooters.
2. Radiators of vehicles, and the condenser and evaporator coils of refrigerators and air conditioners.
3. Heat sinks for electronic chips and power transistors.
4. Transformers and electric motors, which have fins on their body.
5. Economisers and heat exchangers, to improve the heat transfer on the gas side.

How a fin enhances heat transfer

The heat transfer from a surface by convection is $Q = hA(T_s - T_\infty)$. A fin is an extended surface that increases the area $A$ for the same base. Heat flows by conduction along the fin from the base, and it is lost by convection to the surrounding fluid from the large area of the fin.

• A fin is useful when $h$ is low, as for air or gas, so the area is the factor that limits the heat transfer.
• The fin should have a high conductivity $k$, such as aluminium or copper, so that its temperature stays near that of the base. It should be thin, and have a large number of fins, but they should not be so close that they block the flow.

Buckingham $\pi$ theorem: if a problem has $n$ variables, which involve $m$ fundamental dimensions, then it can be written with $(n - m)$ dimensionless groups ($\pi$ terms).

Variables of free convection

The heat transfer coefficient $h$ depends on the following variables. In free convection, the fluid moves because of buoyancy, so the product $g\beta\Delta T$ (gravity, expansion coefficient, temperature difference) takes the place of the velocity.

There are $n = 7$ variables and $m = 4$ dimensions ($M$, $L$, $T$, $\theta$). So the number of $\pi$ terms is $7 - 4 = 3$.

Repeating variables: $L$, $\mu$, $k$ and $g\beta\Delta T$ (four, one for each dimension).

Finding the $\pi$ terms

$\pi_1$ with $h$: $\pi_1 = h\,L^{a}\mu^{b}k^{c}(g\beta\Delta T)^{d}$. Making each dimension zero:

• $\theta$: $-1 - c = 0$, so $c = -1$.
• $M$: $1 + b + c = 0$, so $b = 0$.
• $T$: $-3 - b - 3c - 2d = 0$, so $d = 0$.
• $L$: $a - b + c + d = 0$, so $a = 1$.

$\pi_2$ with $\rho$: $\pi_2 = \rho\,L^{a}\mu^{b}k^{c}(g\beta\Delta T)^{d}$.

• $\theta$: $c = 0$.
• $M$: $1 + b = 0$, so $b = -1$.
• $T$: $-b - 2d = 0$, so $d = \tfrac{1}{2}$.
• $L$: $-3 + a - b + d = 0$, so $a = \tfrac{3}{2}$.

$\pi_3$ with $c_p$: $\pi_3 = c_p\,L^{a}\mu^{b}k^{c}(g\beta\Delta T)^{d}$.

• $\theta$: $-1 - c = 0$, so $c = -1$.
• $M$: $b + c = 0$, so $b = 1$.
• $T$: $-2 - b - 3c - 2d = 0$, so $d = 0$.
• $L$: $2 + a - b + c = 0$, so $a = 0$.

Result

A common form is $Nu = C\,(Gr\,Pr)^n$, where $C$ and $n$ come from experiments.

• Black body: a body that absorbs all the radiation that falls on it, at all wavelengths and from all directions. Its absorptivity is $\alpha = 1$. It is also the best emitter, with an emissivity $\varepsilon = 1$.
• White body: a body that reflects all the radiation that falls on it. Its reflectivity is $\rho = 1$, and it absorbs nothing.
• Gray body: a body whose emissivity is the same at all wavelengths, and less than $1$. It absorbs a fixed fraction of the radiation, and $\alpha = \varepsilon < 1$.

A radiation shield is a thin sheet of a material with a low emissivity (high reflectivity), such as polished aluminium, which is placed between two surfaces that exchange heat by radiation. It reduces the net radiation heat transfer between them.

Working: the shield receives radiation from the hot surface, and it reflects most of it. It heats up a little, and it emits a small amount to the cold surface. So the radiation resistance between the surfaces rises. For $N$ shields of the same emissivity as the surfaces, the heat transfer falls to $\dfrac{1}{N+1}$ of its value without shields.

Utility

1. Cryogenic storage: the vessels for liquid nitrogen or oxygen are protected from the heat of the surroundings.
2. Thermocouples: a shield stops errors in the measurement of the temperature of a gas, by radiation from hot walls.
3. Furnaces and boilers, to protect the people and the instruments.
4. Space vehicles, and thermos flasks, which have a silvered surface.

The thin hollow cylinder has two circular ends, surfaces 1 and 2, and the curved surface, surface 3. $D = 8$ cm, $L = 10$ cm, and the shape factor between the circular surfaces is $F_{1-2} = 0.172$. We need $F_{3-3}$.

By symmetry, $F_{3-2} = 0.1656$.

Dropwise condensation has the higher heat transfer coefficient. In filmwise condensation, the liquid film covers the surface and acts as a resistance to the heat flow, and its thickness grows down the surface. In dropwise condensation, much of the surface is bare, and the drops fall off quickly, so there is no film resistance.

Heat exchangers are classified in these ways.

1. By the nature of the heat exchange process

• Direct contact type: the hot and cold fluids mix. Example: a cooling tower.
• Indirect contact type: a wall separates the fluids. These are the recuperators (the usual type), and the regenerators (the hot and the cold fluid use the same space, one after the other).

2. By the flow arrangement

• Parallel flow: both fluids flow in the same direction.
• Counter flow: the fluids flow in opposite directions.
• Cross flow: the fluids flow at right angles.

3. By the construction

• Double pipe, shell and tube, plate type and finned tube (compact) exchangers.

4. By the phase of the fluids

• Single-phase (liquid or gas) exchangers, and two-phase exchangers such as condensers and evaporators.

Given: hot water: $\dot{m}_h = 4$ kg/min $= 0.06667$ kg/s, $c_h = 4200$ J/kgK, $T_{h,i} = 100^\circ$C. Cold fluid: $\dot{m}_c = 8$ kg/min $= 0.13333$ kg/s, $c_c = 2400$ J/kgK, $T_{c,i} = 20^\circ$C.

The maximum possible effectiveness is the value when the exchanger area is very large ($NTU \to \infty$).

(a) Parallel flow

The heat transfer is $Q = \varepsilon C_{min}(T_{h,i} - T_{c,i}) = 0.533 \times 280 \times 80 = 11947$ W. Both fluids leave at the same temperature, $100 - \dfrac{11947}{280} = 57.3^\circ$C.

(b) Counter flow

The heat transfer is $Q = 1 \times 280 \times 80 = 22400$ W. The hot fluid leaves at $20^\circ$C (the cold inlet temperature), and the cold fluid leaves at $20 + \dfrac{22400}{320} = 90^\circ$C.