Heat Transfer, Refrigeration and Air Conditioning — ESE Mechanical

Weightage: Heat transfer and refrigeration and air conditioning are each a dependable block in the Mechanical papers. They are formula-based, so the winning habit is to draw the thermal resistance network before writing any equation.

1. Conduction

Fourier's law gives the heat flow through a medium: $q=-kdT/dx$, with conductivity $k$. Metals conduct best, followed by non-metals, liquids and gases, and insulators have low $k$.

For a plane wall of thickness $L$ and area $A$:

$Q=\frac{kA(T_1-T_2)}{L}=\frac{\Delta T}{R},R=\frac{L}{kA}$

Resistances in series add, and for convection at a surface $R=1/(hA)$.

Worked example. A wall has 0.2 m of brick ($k=0.8$) and 0.1 m of insulation ($k=0.04$) with a 30 K temperature difference and negligible surface resistance. Per square metre, $R=0.25+2.5=2.75{\text{m}}^2\text{K/W}$, so $q=30/2.75=10.9{\text{W/m}}^2$. The insulation supplies 90 percent of the resistance.

For a cylinder, $Q=\frac{2\pi kL(T_1-T_2)}{\ln (r_2/r_1)}$, and for a sphere, $Q=\frac{4\pi k(T_1-T_2)}{1/r_1-1/r_2}$.

Critical radius of insulation is $r_c=k/h$ for a cylinder and $2k/h$ for a sphere. Insulation of an outer radius below $r_c$ increases heat loss, which matters for thin wires and small pipes.

2. Fins

A fin extends a surface to enhance heat transfer. With $m=\sqrt{hP/(k{A}_c)}$, an infinitely long fin dissipates $Q=\sqrt{hPk{A}_c}(T_b-T_{\infty })$. The fin efficiency is the actual heat over the heat that would flow if the whole fin were at the base temperature. A fin helps when $h$ is low (gases) and $k$ high, and it is of little use when $hP/k{A}_c$ is already large.

3. Transient conduction

If the Biot number $Bi=h{L}_c/k$ (with $L_c=V/A$) is below 0.1, the body's temperature is nearly uniform, and the lumped capacitance model applies:

$\frac{T-T_{\infty }}{T_i-T_{\infty }}=e^{-t/\tau },\tau =\frac{\rho Vc}{hA}$

The time constant $\tau$ is the time to reach 63.2 percent of the total change. A thermocouple needs a small $\tau$, so thin wires respond quickly.

4. Convection

Convection combines conduction and fluid motion. It is forced when a pump or fan drives the flow and natural when buoyancy does. The key numbers are:

• Nusselt $Nu=hL/k$.
• Reynolds $Re=VL/\nu$.
• Prandtl $Pr=\nu /\alpha =\mu c_p/k$.
• Grashof $Gr=g\beta \Delta T{L}^3/{\nu }^2$, for natural convection.

Correlations to remember:

For laminar flow, $Nu$ is independent of $Re$ once the flow is fully developed. The thermal boundary layer is thicker than the velocity one when $Pr<1$ (gases are near 0.7, liquid metals far below), and thinner when $Pr>1$ (oils).

5. Radiation

A black body absorbs all radiation and emits $E_b=\sigma T^4$, with $\sigma =5.67\times 10^{-8}{\text{W/m}}^2{\text{K}}^4$. A real surface emits $\epsilon \sigma T^4$. Kirchhoff's law says that at thermal equilibrium emissivity equals absorptivity. Wien's law gives the wavelength of maximum emission, ${\lambda }_{max}T=2898\mu \text{m⋅K}$.

Exchange between two large parallel grey plates is:

$q=\frac{\sigma (T_1^4-T_2^4)}{\frac{1}{{\epsilon }_1}+\frac{1}{{\epsilon }_2}-1}$

A thin radiation shield between them lowers the exchange, and with $n$ identical shields of the same emissivity it falls to $1/(n+1)$ of the original. The view factor $F_{12}$ is the fraction of radiation leaving surface 1 that strikes surface 2, with the reciprocity relation $A_1{F}_{12}=A_2{F}_{21}$.

6. Heat exchangers

For a heat exchanger, $Q=UA\Delta T_{lm}$, with the log mean temperature difference:

$\Delta T_{lm}=\frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}$

Counterflow gives a higher LMTD than parallel flow for the same terminal temperatures, and is therefore more efficient.

Worked example. A counterflow exchanger has hot fluid cooling from 100 to 60 C and cold fluid heating from 20 to 50 C. The terminal differences are $100-50=50$ and $60-20=40$, so $\Delta T_{lm}=10/\ln (1.25)=44.8$ K.

When outlet temperatures are unknown, use the effectiveness-NTU method, with $NTU=UA/C_{min}$ and effectiveness $\epsilon =Q/Q_{max}$. For a condenser or boiler ($C_r=0$), $\epsilon =1-e^{-NTU}$ whatever the flow arrangement.

7. Vapour-compression refrigeration

The standard cycle has four steps: compression (isentropic, in the vapour region), condensation at high pressure, throttling through an expansion valve, and evaporation at low pressure. The COP is:

$COP=\frac{h_1-h_4}{h_2-h_1}$

with state 1 at the compressor inlet. Lowering the condenser temperature or raising the evaporator temperature raises COP. Subcooling the liquid and superheating the vapour in a controlled way change the refrigeration effect. A ton of refrigeration is 3.5167 kW (12,660 kJ/h).

Refrigerants are chosen for pressure, latent heat and environmental effect. CFCs and HCFCs deplete the ozone layer and are phased out under the Montreal Protocol, and the Kigali Amendment phases down HFCs because of their high global warming potential. Natural refrigerants such as ammonia, CO2 and hydrocarbons are used where suitable.

Absorption refrigeration replaces the compressor with an absorber, a pump and a generator, driven by heat. It suits waste heat, but its COP is lower.

8. Psychrometrics and air conditioning

Moist air is described by dry-bulb temperature, wet-bulb temperature, dew-point temperature, relative humidity and specific humidity $\omega =0.622p_v/(p-p_v)$ kg of vapour per kg of dry air. At saturation, the three temperatures coincide.

Basic processes on the psychrometric chart:

The sensible heat factor is $SHF=Q_s/(Q_s+Q_L)$. The bypass factor of a cooling coil is the fraction of air that passes unaffected. For comfort, about 24 C dry-bulb and 50 percent relative humidity is a common design condition.

Common traps

• Using Celsius in radiation formulas. Use kelvin.
• Applying lumped analysis without checking $Bi<0.1$.
• Choosing parallel flow for a better LMTD. Counterflow is better.
• Insulating below the critical radius and expecting less loss.
• Making Nusselt number depend on Reynolds number for fully developed laminar pipe flow, where it is a constant.

Memory aids

• "Resistances in series add": wall problems.
• "Bi under 0.1": lumped analysis.
• "Counterflow wins": exchanger arrangement.

Summary

Conduction and convection problems reduce to thermal resistance networks, with fins and transient cases handled by $m$ and the Biot number. Radiation uses kelvin and the grey-body formula, and heat exchangers use LMTD or effectiveness-NTU.

Vapour-compression cycles are rated by COP, refrigerants by environmental effect, and air conditioning by psychrometric processes and the sensible heat factor.

Exam protocol

• Draw the resistance network first.
• Check $Bi$ before using lumped capacitance.
• Use kelvin for radiation and absolute temperatures.
• Pick LMTD when terminal temperatures are known and NTU when they are not.