Question

Using the material derivative \[ \frac{D T}{D t}=\frac{\partial T}{\partial t}+\mathbf{v} \cdot \nabla T \] Since the flow is steady, \( \frac{\partial T}{\partial t}=0 \), and one-dimensional in z-direction \[ \frac{D T}{D t}=v_{z} \frac{\partial T}{\partial z} \]

          Using the material derivative
\[
\frac{D T}{D t}=\frac{\partial T}{\partial t}+\mathbf{v} \cdot \nabla T
\]

Since the flow is steady, \( \frac{\partial T}{\partial t}=0 \), and one-dimensional in z-direction
\[
\frac{D T}{D t}=v_{z} \frac{\partial T}{\partial z}
\]

Using the material derivative

(D T)/(D t)=(∂ T)/(∂ t)+𝐯·∇ T

Since the flow is steady, (∂ T)/(∂ t)=0, and one-dimensional in z-direction

(D T)/(D t)=vz(∂ T)/(∂ z)

Added by Angela V.

Calculus

Ron Larson, Robert P. Hostetler, Bruce… 8th Edition

Chapter 13

Instant Answer

Solved by Expert Ma. Theresa Alin

Step 1

We are dealing with a flow that is steady, meaning that the temperature \(T\) does not change with time, i.e., \(\frac{\partial T}{\partial t} = 0\). The flow is also one-dimensional, occurring only in the z-direction. This simplifies our analysis significantly. Show more…

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Using the material derivative \[ \frac{D T}{D t}=\frac{\partial T}{\partial t}+\mathbf{v} \cdot \nabla T \] Since the flow is steady, \( \frac{\partial T}{\partial t}=0 \), and one-dimensional in z-direction \[ \frac{D T}{D t}=v_{z} \frac{\partial T}{\partial z} \]