Question

Air at ${20}^{\circ }\mathrm{C}$ flows over a 4-m-long and 3-m-wide surface of a plate whose temperature is ${80}^{\circ }\mathrm{C}$ with a velocity of $5\mathrm{m}/\mathrm{s}$. The rate of heat transfer from the surface is (a) $7383\mathrm{W}$ (b) $8985\mathrm{W}$ (c) $11,231\mathrm{W}$ (d) 14,672 W (e) $20,402\mathrm{W}$ (For air, use $k=0.02735\mathrm{W}/\mathrm{m}\cdot \mathrm{K},\mathrm{Pr}=0.7228,\nu =1.798\times$ ${10}^{-5}{\mathrm{m}}^2/\mathrm{s})$

Short answer

Answer: The rate of heat transfer from the surface is 8985 W.

Step-by-step solution

Calculate Reynolds Number

First, we need to calculate the Reynolds number for the airflow over the plate. The Reynolds number is given by the formula: $Re=\frac{V\cdot L}{\nu }$ where $V$ is the fluid velocity, $L$ is the length of the plate, and $\nu$ is the kinematic viscosity of the fluid. For this exercise, we have $V=5\mathrm{m}/\mathrm{s}$, $L=4\text{m}$, and $\nu =1.798\times {10}^{-5}{\mathrm{m}}^2/\mathrm{s}$. So, we can find the Reynolds number as: $Re=\frac{5\text{m/s}\cdot 4\text{m}}{1.798\times {10}^{-5}{\text{m}}^2/\text{s}}=1,113,460.51$

Calculate Nusselt Number

The Nusselt number can be determined with the following empirical correlation: $Nu=0.664R{e}^{1/2}P{r}^{1/3}$ Here, $Nu$ represents Nusselt number, and $Pr$ refers to the Prandtl number. With $Re=1,113,460.51$ and $Pr=0.7228$, we can calculate the Nusselt number as: $Nu=0.664\cdot (1,113,460.51{)}^{1/2}\cdot (0.7228{)}^{1/3}=375.27$

Calculate Convective Heat Transfer Coefficient

Now, we can find the convective heat transfer coefficient, $h$, by using the formula: $h=\frac{k\cdot Nu}{L}$ where $k$ is the thermal conductivity of air. For this problem, $k=0.02735\mathrm{W}/\mathrm{m}\cdot \mathrm{K}$. So, the convective heat transfer coefficient is: $h=\frac{0.02735\text{W/mK}\cdot 375.27}{4\text{m}}=2.58{\text{W/m}}^2\cdot \mathrm{K}$

Calculate Rate of Heat Transfer

Finally, we can calculate the rate of heat transfer using the convective heat transfer coefficient and given surface temperature and dimensions. The rate of heat transfer, $Q$, is given by: $Q=h\cdot A\cdot \mathrm{\Delta }T$ where $A$ is the surface area of the plate and $\mathrm{\Delta }T$ is the temperature difference between the surface and air. For this problem, $A=4\text{m}\cdot 3\text{m}=12m^2$, and $\mathrm{\Delta }T=(80-20{)}^{\circ }C={60}^{\circ }\mathrm{C}$. Now we can find the rate of heat transfer: $Q=2.58{\text{W/m}}^2\mathrm{K}\cdot 12{\text{m}}^2\cdot {60}^{\circ }\mathrm{C}=8985\text{W}$ Based on our calculations, the rate of heat transfer from the surface is 8985 W, which corresponds to option (b).

Key concepts

Reynolds Number

In the realm of fluid dynamics, the Reynolds number is pivotal in determining the flow characteristics over surfaces. It is a dimensionless quantity that helps predict whether the flow will be laminar or turbulent. The formula for calculating the Reynolds number is $Re=\frac{V\cdot L}{u}$. Here, $V$ stands for fluid velocity, $L$ represents the characteristic length (in this exercise, it's the length of the plate), and $u$ is the kinematic viscosity of the fluid.
A higher Reynolds number signifies a more turbulent flow, while a lower value indicates laminar flow. For example, in the provided exercise, the Reynolds number calculated was $1,113,460.51$, suggesting a turbulent flow over the plate. It's essential to compute this number accurately since it aids in determining other factors like drag coefficient and heat transfer rates.

Nusselt Number

The Nusselt number is another important dimensionless parameter in convective heat transfer. It expresses the ratio between the convective and conductive heat transfer across a boundary. The formula utilized in the exercise is an empirical one: $Nu=0.664R{e}^{1/2}P{r}^{1/3}$. This calculation involves the Reynolds number, further connecting it to the flow dynamics, alongside the Prandtl number ($Pr$), which describes fluid flow properties.
In this context, a higher Nusselt number means more effective convective heat transfer across the plate. It indicates the enhancement of thermal energy transfer due to the flow. For our case, $Nu=375.27$ was calculated, showcasing efficient heat transfer from the surface, thanks to turbulent flow and favorable thermal properties of the air.

Kinematic Viscosity

Kinematic viscosity $(u)$ is a measure of a fluid's resistance to flow and shear under gravitational forces. It is essentially dynamic viscosity divided by the fluid's density, providing insight into the fluid's pouring and spreading behavior. In the equation $Re=\frac{V\cdot L}{u}$, kinematic viscosity plays a crucial role in determining the Reynolds number.
Specifically, in air at ${20}^{\circ }C$, the kinematic viscosity is given as $1.798\times {10}^{-5}{\text{m}}^2/\text{s}$. This low value suggests that air flows relatively easily, resulting in lower internal resistance to flow. Kinematic viscosity is vital in determining flow regimes and is a constant factor in planning and predicting heat transfer scenarios.

Thermal Conductivity of Air

Thermal conductivity $(k)$ refers to the ability of a material, in this case air, to conduct heat. It is measured in $\text{W/mK}$ and is crucial for designing and analyzing heat transfer applications. In the given exercise, air's thermal conductivity is $0.02735\text{W/mK}$.
Thermal conductivity plays a key role when calculating the convective heat transfer coefficient $h$, using the formula $h=\frac{k\cdot Nu}{L}$. This coefficient helps quantify the heat transfer rate from the plate surface to air flowing over it. Air's relatively low thermal conductivity denotes that it is not the best conductor of heat; however, when combined with proper surface area and temperature gradient, it can facilitate significant convective heat transfer, as seen in the exercise's outcome.