In the case of turbulent flow through a horizontal isothermal cylinder of diameter ‘D’, the free convection heat transfer coefficient for the cylinder will

Option 1 : be independent of diameter

__ Explanation__:

For Turbulent flow in natural convection, the Nusselt number (Nu) is given by:

Nu = C [Gr, Pr]1/3

The Grashoff number (Gr) is given by

\(\left( {Gr} \right) = \frac{{g \beta \left( {{\rm{\Delta }}T} \right)L_e^3}}{{{\nu ^2}}}\)

i.e. Gr ∝ L3

where L = characteristic length.

For various setups, the characteristic length changes and some are given below:

For vertical plate = L (length of the plate)

For Vertical cylinder = L (length of the cylinder)

For Horizontal cylinder = D (diameter of the cylinder)

For square plate - 0.25 a (a is side of the square plate)

Now;

Nu = C [Gr, Pr]1/3

\(\frac{{hD}}{k} = C{\left[ {\frac{{g \beta \left( {{\rm{\Delta }}T} \right){D^3}}}{{{\nu ^2}}}} \right]^{\frac{1}{3}}}{\left( {Pr} \right)^{1/3}}\)

\(\therefore \frac{{hD}}{k} \propto {D}\)

i.e. h is independent of characteristic length or diameter of the cylinder.

Option 1 : Diameter of the cylinder

__Concept:__

Grashoff Number

- It indicates the relative strength of the buoyant to viscous forces.

\({G_r} = \frac{{gB\left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{\nu ^2}}}\)

- It is used in free convection.
- It has a role in free convection similar to that played by Reynold's number in forced convection.
- For calculation of h, we need to calculate grashof number.
- For various setups, the characteristic length changes and some are given below

For vertical plate - L (length of the plate)

For Vertical cylinder - L (length of the cylinder)

**For Horizontal cylinder - D (diameter of the cylinder)**

For square plate - 0.25 a (a is side of the square plate)

A flat plate of length 1 m and width 50 cm is placed in an air stream at 30°C blowing parallel to it. The convective heat transfer coefficient is 30 W/(m^{2}K). The heat transfer if the plate is maintained at a temperature of 400°C is

Option 1 : 5.55 kW

**Concept:**

The heat transfer is given as, Q = hAΔT

where, h = convective heat transfer coefficient, A = Area of plate, ΔT = temperature difference

**Calculation:**

**Given:**

Area of plate = 1m × 0.5 m = 0.5 m^{2}, h = 30 W/m^{2}K, ΔT = 400 - 30 = 370°C

Q = hAdT = 30 × 0.5 × (400 – 30) = 5550 W = 5.55 kW

Water (specific heat, C_{p} = 4.18 kJ/kgK) enters a pipe at a rate of 0.01 kg/s and a temperature of 20°C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux \(q_w''\ in\ \frac{W}{{{m^2}}}\) .

Option 2 : 62

**Calculation:**

__Given:__

\(q_w'' = 2500x\), ṁ = 0.01 kg/s, T1 = 20°C, d = 50 mm = 0.05 m, l = 3 m, c_{p} = 4.18 kJ/kgK = 4.18 × 10^{3} K/kgK

\(x = 0,\;q_{w}'' = 0\)

\(x = 3,\;q_{w}'' = 2500 × 3 = 7500 ~W/m^2\)

\({q_{avg}} = \frac{{0 + 7500}}{2} = 3750~W/m^2\)

q_{avg} × Area = ṁ × c_{p} × ΔT

3750 × π × 0.05 × 3 = 0.01 × 4.18 × 10^{3} × (T - 20) {∵ Area = π × d × l}

⇒ T = 42.2 + 20 = **62.2°C**

Option 2 : Grashof number

__Explanation:__

In convection studies, it is common practice to nondimensional the governing equations and combines the variables which group together into dimensionless numbers in order to reduce the number of total variables.

Nusselt number is a parameter to non-dimensional heat transfer coefficient.

Non-dimensional groupings

- Nusselt No. Nu = hLc/k = (convection heat transfer strength)/ (conduction heat transfer strength)
- Prandtl No. Pr = n/a = (momentum diffusivity)/ (thermal diffusivity)
- Reynolds No. Re = Ux/ν = (inertia force)/ (viscous force)

**The dimensionless parameter which represents the natural convection effects and is called the Grashof number.**

Grashof number, Gr, as the ratio between the buoyancy force and the viscous force:

\(Gr = \frac{{g\beta \left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{\nu ^2}}}\)

Nusselt number is a function of the Grashof number and the Prandtl number alone. Nu = f (Gr, Pr)

Option 3 : Fluid over a heated horizontal plate

__Explanation:__

- The
**mechanism of heat transfer**or fluid motion due to density difference is known as free convection. - The warmer fluid is replaced by colder fluid creating convection current. The force which induces these convection is called
**a buoyancy force**which is due to the presence of a density gradient within the fluid and a body force. - For free convection
**Nusselt number**is a function of**Prandlt number**and**Grashof Number.** - For fluid over a heated horizontal plate,
**Grashof number**is maximum so**Nusselt number will also be maximum i.e. convection dominates**. So minimum temperature difference will be required fluid over a horizontal plate.

Option 2 : The diameter of the cylinder

__Concept:__

The dimensionless parameter which represents the natural convection effects and is called the Grashof number.

Grashof number, Gr, is the ratio between the **buoyancy force and the viscous force:**

Grashof Number (Gr) = (gβΔTL^{3}) / v^{2}

Where g is acceleration due to gravity

β is coefficient of thermal expansion

ΔT is temperature difference, V is kinematic viscosity

L is characteristic length

For vertical plate and cylinder L is height

For horizontal cylinder and sphere L is diameter

The back surface of a plate is heated at the rate of 50 W/m^{2} and the front surface is cooled by airflow at 30°C. It is given that the heat transfer coefficient between the air and the plate surface is 50 W/m^{2}–K. Considering steady-state heat transfer, what is the temperature (in °C) of the front surface of the plate.

Option 2 : 31

__Concept:-__

According to Newton's law of cooling

**Q = hA∆T **

__Calculation:__

__Given:__

Q/A = 50 W/m^{2}, h = 50 W/m^{2}k, T_{b }= 30°C

Q/A = hΔT

Q/A = h × ( T_{F} – T_{b })

50 = 50 × (T_{F }- 30)

**T _{F} = 31°C**

Option 1 : 2.156 kW

**Concept:**

According to Newton's Law cooling,

Q = h A dT

where, Q = heat transfer, h = Convection heat Transfer coefficient, A = surface area, dT = temperature gradient

**Calculation:**

**GIven:**

T_{∞ }= 20°C, Area of plate = 50 × 75 cm^{2}, T_{s} = 250°C, h = 25 W/m^{2}°C

Q = 25 × 50 × 75 × 10^{-4} × (250 – 20)

Water (specific heat, C_{p} = 4.18 kJ/kgK) enters a pipe at a rate of 0.01 kg/s and a temperature of 20°C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux \(q_w''\ in\ \frac{W}{{{m^2}}}\) .

Option 4 : 81

**Concept:**

Heat transfer rate between the entire pipe is

\(q_w'' × A = mC_p{\rm{\Delta }}T\)

**Given:**

q"_{w} = 5000 = constant, h = 1000 W/m^{2}K, ṁ = 0.01 kg/s, T_{1} = 20°C, d = 50 mm = 0.05 m, l = 3 m, c_{p} = 4.18 kJ/kgK = 4.18 × 103 K/kgK

**Calculation:**

**\(q_w'' × A = mC_p{\rm{\Delta }}T\)**

5000 × π × 0.05 × 3 = 0.01 × 4.18 × 10^{3} × (T_{0} - 20)

T_{0} – 20 = 56.3 = T_{0} = 76.3°C

Heat flux between any two sections is same

Q = hA(T_{p} – T_{0})

But \(\frac{Q}{A} = q_w'' = 5000 = \left( {{T_p} - {T_0}} \right)\)

5000 = 1000(T_{p} – 76.3)

⇒ T_{p} = 76.3 + 5 =** 81.3°C**

1. Some height

2. Both vertical

3. Same fluid

4. Same fluid flow pattern

Select the correct answer using the code given below

Option 1 : 1 only

__Concept:__

In convection studies, it is common practice to non dimensionalize the governing equations and combine the variables which group together into dimensionless numbers in order to reduce the number of total variables.

Nusselt number is a parameter to non dimensionalize the heat transfer coefficient.

Non-dimensional groupings

- Nusselt No. Nu = hLc/k = (convection heat transfer strength)/ (conduction heat transfer strength)
- Prandtl No. Pr = ν/α = (momentum diffusivity)/ (thermal diffusivity)
- Reynolds No. Re = ρVx/μ = (inertia force)/ (viscous force)

The dimensionless parameter which represents the natural convection effects and is called the Grashof number.

Grashof number, Gr, as the ratio between the buoyancy force and the viscous force:

\(Gr = \frac{{g\beta \left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{ν ^2}}}\)

Nusselt number is a function of the Grashof number and the Prandtl number alone. Nu = f (Gr, Pr)

In the case of vertical pipe and vertical cylinder, if the **height is same**, then the characteristic dimension (L_{c}) is same, then Grashoff number is same, then nusselt number is same, **then heat transfer coefficient will be same**.

The convective heat transfer coefficient h is defined by the relationship (m = mass, c_{p} = specific heat, k = conductivity, q = heat flow rate, Nu = Nusselt Number, L = Thickness, ΔT = Temperature difference)

Option 3 : \(h = \frac{q}{{{\rm{\Delta }}T}}\)

**Concept:**

According to Newton’s Law of cooling

q = hAΔT

where, Q = heat transfer by convection, h = convective heat transfer coefficient, ΔT = temperature difference

\(\Rightarrow h = \frac{Q}{{{\rm{Δ T}}}}\)

For A = 1 unit

A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/m-K, density 9000 kg/m^{3}and specific heat 385 J/kg-K. If the heat transfer coefficient is 250 W/m^{2}-K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s.

Option 3 : 17.3

**Concept:**

**Rate of heat transfer by convection = Rate of fall of the heat content of ball**

∴ \( - hA\left( {T - {T_∞ }} \right) = ρ \;C\;V\frac{{dT}}{{dt}}\)

**Calculation:**

**Given:**

diameter, d = 5mm, ⇒ radius, r = 2.5 mm = 2.5 × 10^{-3} m

surrounding temperature, T_{∞ }= 300 K, initial temperature, T = 500 K,

k = 400 W/m-K, ρ = 9000 kg/m^{3}, C = 385 J/kg-K, h = 250 W/m^{2}-K

As Rate of heat transfer by convection = Rate of fall of the heat content of ball

\( - hA\left( {T - {T_∞ }} \right) = ρ \;C\;V\frac{{dT}}{{dt}}\)

\(⇒ \frac{{dT}}{{dt}} = - \frac{{250 × 4\pi {r^2}\left( {500 - 300} \right)}}{{9000 × 385 × \frac{4}{3}\pi {r^3}}}\)

\(⇒ \frac{{dT}}{{dt}} = - \frac{{250 × 3× 200}}{{9000 × 385 × r}}\)

\(⇒ \frac{{dT}}{{dt}} = - \frac{{250 × 3× 200}}{{9000 × 385 × (2.5\times 10^{-3})}}\)

\(⇒\frac{{dT}}{{dt}} = 17.3\ K/s\)

Hence, **the rate of fall of the temperature of the ball at the beginning of cooling will be 17.3 K/s.**