Sensible Heat: Definition, Diagram, Formula, Explanation, Example

 

What is Sensible heat?

Sensible heat is the heat exchanged by the object that causes a change in temperature of the object without changing its phase.

Hence sensible heating affects the system by changing its temperature. As the name says, we can sense the changes happening during the sensible heating.

Example:- If you start heating to the water kept in a container, then in this case you can sense the temperature changes just by dipping your fingers in the water. As time-lapse, you can feel the rise in temperature, so this process is known as sensible heating.

The unit of the sensible heat in the SI system is the joule (J) and in the FPS system is Btu.

Sensible heat diagram:

In the above graph, the region a-b, region c-d, and region after e, show the sensible heating process.

Sensible heat explanation:

The above figure shows the heating of the ice at -5°C to convert it into water vapors at 100°C.

The complete process consists of a combination of sensible heating and latent heating.

Here ice from -5°C is heated to raise its temperature up to 0°C. Hence the heat supplied is known as Sensible heat.

Again water from 0°C is heated to raise its temperature up to 100°C, hence in this case also the heat supplied is known as Sensible heat.

Sensible heat equation:

The amount of sensible heat supplied or rejected by the substance is given by,

QSensitive = mC(TFinal – TInitial)`

Where,
m = Mass of substance
C = Specific heat of a substance
TFinal = Final temperature
TInitial = Initial temperature

Sensible heat example:

10 Kg of water inside the container at the atmospheric pressure and 25°C is heated to the 80°C. Find the amount of sensible heat required for the heating of the water if Cp(water) = 4.186 KJ/Kg.

Given:
T1 = 25°C
T2 = 80°C
Cp = 4.186 KJ/Kg
m = 10 Kg

Solution:-

The amount of sensible heat required for the heating of water is given by,

Qsensible = m.Cp.[T2 – T1]

Qsensible = 10 x 4.186 x [80 – 25]

Qsensible = 2302.3 KJ

Amount of sensible heat = 2302.3 KJ

THANKS TO

Pratik

FAQ

Why is sensible heating important?
Sensible heating helps to increase or decrease the temperature of the system without changing the phase of the system.


What is sensible heat in air conditioning?
In the case of air conditioning, the amount of heat required to raise or lower the temperature of the air in the cabin is known as sensible heat.

Latent Heat: Definition, Formula, Types, Diagram, Example

If we observe the temperature changes of the melting ice or boiling water with the thermometer, then we know that even with continuous heating the temperature of the system never changes.

The heat gained by the system during these processes is known as latent heat. Let’s know about the term latent heat in detail in the below article.

What is Latent heat?

Latent heat is the heat exchange by the material at a constant temperature while changing its phase.

It is also known as heat absorbed or rejected by the matter at a constant temperature and is denoted by the symbol ‘L’.

It can be observed during the transition between the solid and liquid state (melting or freezing), liquid and vapor state (vaporization or condensation), and direct transition between the solid and vapor state (sublimation or deposition).

During the phase transition from solid → liquid → gas, the material gains the latent heat to break the intermolecular bond of attraction. While during the phase transitions from vapor → liquid → solid, the molecules of the material lose the latent heat so that the intermolecular forces pull the molecules together that causes a change in phase from vapor to liquid and to solid.

It is sometimes referred to as hidden heat as it never shows the temperature change during its exchange. As it indicates energy, it is measured in terms of a unit of energy i.e. Joules (J).

Specific latent heat is the amount of heat required to change the phase of the unit mass of material. Thus in the SI system, the unit of specific latent heat is J/Kg.

Latent heat equation:

The amount of latent heat required for a substance is given by,

QLatent = L x m

Where
L = Specific latent heat
m = Mass of a substance

Types of latent heat:

Based on the phase change, the latent heat can be classified as follows:

A] Latent heat of fusion/ Latent heat of melting, lf:-

It is the amount of heat gained or rejected by the object to change its phase from solid to liquid or from liquid to solid.

E.g. During the melting, the ice gains the latent heat of melting/fusion and while freezing the water rejects the latent heat of fusion/melting.

The latent heat of fusion is much higher than the sensible heat, that’s the reason why we use ice instead of water to keep beverages cold for a longer time.

Consider the solid material is subjected to heating. Till the melting point, the heat supplied to the material is utilized to raise the kinetic energy of the molecules i.e. to raise the temperature of the material (sensible heating). When the temperature reaches the melting point, the material gains the heat not to increase the kinetic energy but to weaken the intermolecular bonds (Force of attraction between the molecules).

As the intermolecular forces become weaker, the ordered molecules in the solid state become less disordered and get converted into the liquid state.

During the process of freezing, the molecules of the liquid lose the latent heat. It strengthens the intermolecular forces to pull the molecules together to get converted into a solid state.

The heat supplied at the melting point does not increase the kinetic energy of the molecules, hence the temperature of the solid material during the melting remains constant.

B] Latent heat of vaporization/ Latent heat of condensation, lvap:-

It is the amount of heat gained or rejected by the object while changing its phase from liquid to vapor or from vapor to liquid. It is also known as a latent heat of evaporation.

E.g. When water vaporizes, the molecules of water receive the latent heat of condensation/vaporization, whereas when steam condenses into water, the molecules of steam reject the latent heat of condensation/vaporization.

When the liquid reaches boiling point, it gains the heat to break the intermolecular bonds between the liquid molecules. At the boiling point, the heat gained by the liquid is utilized only to overcome the intermolecular forces not to raise the kinetic energy of the molecules.

Thus the molecules become more disordered and it has a significantly less intermolecular force of attraction, which causes the transition of liquid into the vapor state.

During condensation, the vapor molecules lose the latent heat, as a result, the intermolecular forces pull the molecules closer to attain a liquid state.

The value of the latent heat of vaporization is significantly higher than the sensible heat, that’s why steam is used for the heating instead of hot water.

C] Latent heat of sublimation, lsub:-

The latent heat of sublimation is the amount of heat gained by a substance to convert its phase from solid to vapor or it is heat rejected by a substance while converting its phase from vapor to solid.

Sublimation indicates the transition of the phase of the material from solid to vapor without entering into the liquid phase. The reverse of this is known as desublimation/deposition, during which the vapors are converted into a solid state without entering into the liquid state.

E.g. During sublimation, the naphthalene gains the latent heat of sublimation while during deposition, the naphthalene vapors lose the latent heat to the surrounding.

The latent heat of sublimation is equivalent to the sum of its latent heat of melting and the latent heat of vaporization.

Example of latent heat:

The above figures show the conversion of ice crystals at -5°C to the vapors at 100°C.

The heat supplied for the complete process can be divided into the following parts:-

Ice [-5°C] → Ice [0°C] :- Sensible heat
Ice [0°C] → Water [0°C] :- Latent heat of melting
Water [0°C] → Water [100°C] :- Sensible heat
Water [100°C] → Steam [100°C] :- Latent heat of vaporization

Here at 0 °C, ice gains the latent heat of melting to convert into water at 0 °C.

And at 100 °C, water gains the latent heat of vapourization to convert it into steam.

Latent heat solved problems:

1] Consider 1 Kg of ice at 0° Celsius is heated at atmospheric pressure. Find the amount of heat required to vaporize the ice. (For water, lm = 333.5 KJ/Kg, Iv = 2.25 x 103 KJ/Kg, CP = 4.2 KJ/Kg.K)

Given:
Latent heat of fusion/melting, lm = 333.5 KJ/Kg
Latent heat of vaporization, Iv = 2.25 x 103 KJ/Kg
CP of water = 4.2 KJ/Kg.K
Mass, m = 1 Kg

Solution:-

The heat required to melt 1 Kg of ice is given by,

LMelting = m x lm = 1 x 333.5 = 333.5 KJ

The amount of heat required to heat water from melting point to boiling point at atmospheric pressure is given by,

QSensible = m.Cp. Δt = m.Cp . [TBoiling – TMelting]

As for water, TBoiling = 373 K, TMelting = 273 K,

∴ QSensible = 1 x 4.2 x [373 – 273]

QSensible = 420 KJ

The heat required to vaporize the water at the boiling point is given by,

LVaporization = m x lv = 1 x [2.25 x 103] = 2.25 x 103 KJ

 The total heat supplied to the ice to vaporize is given by,

Qtotal = LMelting + QSensible + LVaporization

Qtotal = 333.5 + 420 + 2.25 x 103

Qtotal = 3003.5 KJ

Thanks To

PRATIK

FAQ’s

1. Why latent heat is significant?

   The amount of latent heat is much higher than the sensible heat. Thus, it is used for the heating or cooling processes.

2. What factors affect latent heat?

   The factors like mass, pressure affect the amount of latent heat required.

3. What are the different types of latent heat?

   Based on the phase changes the latent heat is of the following types:
   Latent heat of melting
   Latent heat of vaporization
   Latent heat of sublimation

4. Does evaporation cause the release of latent heat?

   No, during evaporation, the liquid absorbs the latent heat of evaporation.

5. When is the latent heat considered positive or negative?

   The latent heat is considered negative if the material rejects it and it is positive if the material gains it.

6. What mean by latent heat of steam?

   It is the amount of heat rejected by the saturated steam to get condense into the saturated liquid.

7. What mean by the latent heat of ice?

   It is the amount of heat gained by the ice at melting temperature to change its phase from solid to liquid.

8. What is the effect of latent heat on temperature?

   The temperature remains constant during the transfer of latent heat.

9. Is latent heat also a thermal energy?

   Yes, latent heat is thermal energy.

10. Are latent heat and enthalpy the same thing?

    No, The latent heat indicates the change in enthalpy of the material during the phase change.

Critical Radius & Thickness of Insulation

The critical radius of insulation & the critical thickness of insulation are the necessary factors in the case of insulation over the cylindrical or spherical surface.

For the insulation applied over a spherical or cylindrical surface, up to a certain radius the insulation the rate of heat transfer from the object increases.

The radius of insulation where the heat transfer from the object is maximum is known as the critical radius of insulation & The thickness of insulation at a critical radius is known as the critical thickness of insulation.

What is Critical radius of insulation?

The critical radius of insulation is the outer radius of insulation at which the rate of heat transfer through the body is maximum. Or it is the radius of insulation at which the insulation has a minimum thermal resistance.

As shown in the below figure, the rate of heat transfer increases with an increase in the thickness of insulation up to the critical radius of insulation and then it starts decreasing.

The insulation is used to reduce the heat loss from the object. But in the case of insulation to the cylinder or sphere, there is a certain insulation thickness that is not helpful in decreasing the rate of heat transfer from the surface of the object.

It means that up to the critical radius, instead of decreasing the heat transfer, the insulation increases the heat transfer rate.

If we increase the thickness more than the critical thickness value, then the heat transfer rate starts to decrease.

The critical radius of insulation depends on the thermal conductivity (K) of insulation and convective heat transfer coefficient (h) at the surround of the insulation.

For the cylinder, the critical radius of insulation (rcr) is given by,

$r_{cr}=\frac{K}{h}$

For the sphere, the critical radius of insulation is given by,

$r_{cr}=\frac{2K}{h}$

Critical thickness of insulation:

The thickness of insulation at which the heat transfer reaches the maximum value is the critical thickness of insulation.

It is the thickness of the insulation up to the critical radius. It is given by, the difference between the critical radius and the inner radius of insulation.

Critical thickness = Critical radius of insulation – Inner radius of insulation

$t=r_{cr}-r_i$

Critical radius of insulation graph:

This graph indicates the variation in heat transfer rate with the increase in the thickness of insulation.

In the below graph at a certain radius of insulation (rcr), the rate of heat transfer stops increasing (Qmax), this radius indicates the critical radius of insulation. While the thickness of the insulation is called the critical thickness of insulation.

The variation of outer radius of insulation (ro) with respect to the critical radius rcr, affects the heat transfer rate as follows:

A] For ro < rcr [outer radius is below critical radius]:-

The insulation on a surface up to the radius rcr causes a rise in the rate of heat transfer through the insulating body.

The rise in insulation up to the critical radius causes thermal resistance to decrease. Thus it results in a rise in heat transfer rate.

B] For ro = rcr [outer radius is equal to critical radius]:-

At the critical radius of insulation, the thermal resistance is minimum hence the heat transfer rate is maximum (Qmax).

Up to the critical radius, as the outside area exposed to the surroundings increases, the increase in conduction resistance is dominated by an increase in convective heat transfer.

C] For ro > rcr [outer radius more than critical radius]:-

If the radius of insulation increases further of the rcr, the conduction resistance starts to increase, therefore, the rate of heat transfer also goes on decreasing.

In this case, the increase in the rate of convective heat transfer is dominated by an increase in conductive resistance.

Significance of Critical radius & thickness of insulation:

The critical thickness of insulation and critical radius of insulation has the following significance based on their purpose:-

1) For thermal insulation:

Thermal insulation is used to avoid or lower the rate of heat transfer from objects. Hence, in this case, the thickness of insulation should be greater than the critical thickness of insulation.

2) For electric insulators:

The insulations used for the electric conductors are good insulators for electricity and good conductor for heat.

Therefore in this case the radius of insulation should be equal to or less than the critical radius of insulation. So that it will resist only the flow of electricity but allow the transfer of heat generated while electricity conduction.

Critical radius and thickness of insulation for cylinder:

The formula for the critical radius of insulation for the cylindrical surface is given by,

$r_{cr}=\frac{K}{h}$

Here,

K= Thermal conductivity, w/mK
h= Convective heat transfer coefficient, w/m2.K

The critical thickness of insulation for the cylinder is given by,

$\text{Critical thickness,}{t}_{\text{cylinder}}=r_{cr}-r_i$

$t_{\text{cylinder}}=\frac{K}{h}-r_i$

Where, ri = inner radius of insulation.

Derivation for Critical radius of insulation for cylinder:

The below figure shows the insulated cylindrical object with insulation having an inner radius ri and outer radius ro.

The total thermal resistance by the insulation is given by,

Rth = Rcond. + Rconv.

As for cylinders, $R_{cond.}=\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi KL}$ and $R_{conv.}=\frac{1}{h.(2\pi r_o.L)}$,

$\therefore R_{th}=\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi KL}+\frac{1}{h.(2\pi r_o.L)}$

At the critical radius of insulation (ro = rcr), the thermal resistance of the insulation is minimum.

Therefore at ro = rcr,

$\frac{d{R}_{th}}{d{r}_o}=0$

$\frac{d}{d{r}_o}[\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi KL}+\frac{1}{h.(2\pi r_o.L)}]=0$

$\frac{1}{2\pi .L}.\frac{d}{d{r}_o}[\frac{\ln \left(\frac{r_o}{r_i}\right)}{K}+\frac{1}{h.r_o}]=0$

$\frac{d}{d{r}_o}[\frac{\ln \left(\frac{r_o}{r_i}\right)}{K}+\frac{1}{h.r_o}]=0$

$\frac{d}{d{r}_o}\left[\frac{\ln \left(\frac{r_o}{r_i}\right)}{K}\right]+\frac{d}{d{r}_o}\left[\frac{1}{h.r_o}\right]=0$

$[\frac{1}{K}.\frac{1}{\frac{r_o}{r_i}}.\frac{1}{r_i}]+[\frac{1}{h}.\frac{-1}{r_o^2}]=0$

$\frac{1}{K.r_o}–\frac{1}{r_o^2.h}=0$

$\frac{1}{K.r_o}=\frac{1}{r_o^2.h}$

$r_o=\frac{K}{h}$

$\therefore r_{cr}=r_o=\frac{K}{h}$

Critical radius and thickness of insulation for sphere:

The formula for the critical radius of insulation for the spherical surface is given by,

$r_{cr}=\frac{2K}{h}$

The critical thickness of insulation for the sphere is given by,

$\text{Critical thickness,}{t}_{\text{sphere}}=r_{cr}-r_i$

$t_{\text{sphere}}=\frac{2K}{h}-r_i$

Derivation for Critical radius of insulation for sphere:

The below figure shows the insulated sphere with the insulation having an inner radius of ri and an outer radius of ro.

The conduction resistance of the spherical insulation is given by,

$R_{cond.}=\frac{r_o-r_i}{4\pi .K{r}_i{r}_o}$

The convective resistance outside of the sphere is given by,

$R_{conv.}=\frac{1}{h.A}=\frac{1}{h.\left(4\pi r_o^2\right)}$

The total thermal resistance of insulation is given by,

Rth = Rcond. + Rconv.

$R_{th}=\frac{r_o-r_i}{4\pi .K{r}_i{r}_o}+\frac{1}{h.\left(4\pi r_o^2\right)}$

At the critical radius of insulation, Rth is minimum.

$\therefore \frac{d{R}_{th}}{d{r}_o}=0$

$\frac{d}{d{r}_o}[\frac{r_o-r_i}{4\pi .K{r}_i{r}_o}+\frac{1}{h.\left(4\pi r_o^2\right)}]=0$

$\frac{1}{4\pi }.\frac{d}{d{r}_o}[\frac{r_o}{K{r}_i{r}_o}-\frac{r_i}{K{r}_i{r}_o}+\frac{1}{h.r_o^2}]=0$

$\frac{d}{d{r}_o}[\frac{1}{K{r}_i}-\frac{1}{K{r}_o}+\frac{1}{h.r_o^2}]=0$

$\frac{1}{K}.\frac{-1}{r_o^2}+\frac{1}{h}.\frac{-2}{r_0^3}=0$

$\frac{1}{K{r}_o^2}=\frac{2}{h{r}_o^3}$

$r_o=\frac{2K}{h}$

$\therefore r_{cr}=r_o=\frac{2K}{h}$

Why rate of heat transfer increases till the critical radius of insulation?

As we increase the layer of insulation on a cylindrical or spherical surface, the area exposed to the surrounding also increases.

The convective heat transfer from outer surface is given by,

$Q_{conv.}=\frac{\Delta t}{h.A}$

As per the above equation, the increased surface area (A), increases the convective heat transfer.

Up to the critical radius, the conductive resistance of the added insulation is dominated by rising convective heat transfer. Therefore up to the critical radius of insulation, the rate of heat transfer increases.

Let’s explain with an example.

Consider a cylindrical insulation with inner radius ri and outer radius ro.

The total thermal resistance is the sum of the conduction resistance of insulation and convective resistance at the outer surface,

Rth = Rcond. + Rconv.

For the cylinder, the conduction resistance and convection resistance are as follows,

$R_{cond.}=\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi KL}\cdots \left[1\right]$

$R_{conv.}=\frac{1}{h.A}=\frac{1}{h.\left(2\pi r_{o}L\right)}\cdots \left[2\right]$

As per equation [1], as the outer radius (ro) increases, the conduction resistance of the insulation increases.

While as per equation [2], as the outer radius (ro) increases, the convection resistance decreases.

The addition of insulation over cylindrical or spherical surfaces increases the outer surface area subjected to the atmosphere. This causes, the convective heat transfer to increase.

The overall effect of these resistances (Rcond. & Rconv.) causes the value of the overall thermal resistance (Rth) to decrease up to a certain radius and then it starts to increase.

Therefore the heat transfer rate increases up to a certain radius of insulation and then it starts to decrease.

Solved Numerical:

Consider a steam flowing pipe is insulated with the material having a thermal conductivity of 0.055 W/mK. If the convective heat transfer coefficient is 2.5 w/m2K, find the critical radius of the insulation of the pipe. If the pipe diameter is 25 mm, find the critical thickness of the insulation.

Given:
K = 0.055 W/m.K
h = 2.5 w/m2.K
d1 = 25 mm, r1 = 12.5 mm

Solution:

For cylindrical pipe, the critical radius rcr is given by,

$r_{cr}=\frac{0.055}{2.5}$

rcr = 0.022 m = 22 mm

The critical thickness of insulation (t) for the steam pipe is given by,

t = rcr – r1 = 22 – 12.5 = 9.5 mm

THANKS TO

Pratik