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# The 60-W fan of a central heating system is to circulate air through the ducts. The analysis of the flow shows that the fan needs to raise the pressure of air by 50 Pa to maintain flow. The fan is located in a horizontal flow section whose diameter is 30 cm at both the inlet and the outlet. Determine the highest possible average flow velocity in the duct.

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To determine the $\textbf{highest possible velocity}$ of the air in the duct $V$ we need to first look at the $\textbf{energy change }$of the energy of the process. The fan needs to maintain a pressure difference $p=50\text{ Pa}$. With this information and the given power of the fan $P=60\text{ W}$ we can calculate the mass flow rate $\dot{m}$ and from that the needed velocity $V$.

We first rearrange the power formula to get the mass flow rate:

\begin{align*} P&=\dot{m} \cdot \frac{p}{\rho} \\ \dot{m}&= \frac{P \cdot \rho}{p} \end{align*}

Before we calculate the velocity $V$ we first need to calculate the surface area of the air duct $A$ using the given diameter $D=30\text{ cm}$.

\begin{align*} D&=30\text{ cm} \cdot \frac{1 \text{ m}}{100\text{ cm}} =0.3\text{ m} \\ A&=D^2\cdot \pi \\ A&=\left(0.3\text{ m}\right) ^2 \cdot 3.14 \\ A&=\color{#4257b2}0.283\text{ m}^2 \\ \end{align*}

Now we can solve the mass flow rate for the highest possible velocity of the air $V$.

\begin{align*} \dot{m}&=\rho \cdot A \cdot V \\ V&=\frac{\dot{m}}{A \cdot \rho} \\ V&=\frac{\dfrac{P\cdot \cancel{\rho}}{p}}{A \cdot \cancel{\rho}} \\ V&=\frac{ \dfrac{ 60\text{ W} \cdot } {50\text{ Pa}} } {0.283\text{ m}^2} \\ V&=\boxed{\color{#c34632}4.24 \,\frac{\text{m}}{\text{s}} } \end{align*}

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