Chapter 10, Problem 69P

Water at atmospheric pressure and temperature $\left(\rho =998.2{\text{kg/m}}^2\right)$ , and $\mu =1.003\times {10}^{-3}$ at free stream velosity V=0.100481 m/s flows over a two-dimensional circular cylinder of diameter d = 1.00 m. Approximate tie flow as potential flow, (a) Calculate the Reynolds number, based on cylinder diameter. Is Re large enough that potential flow should be a reasonable approximation? (3) Estimate the miiiirnum ar.c ma-ximuin speeds ??? and ??? (speed is the magnitude of velocity) and the maximum and minimum pressure difference $P-P_{\infty }$ in the flow, along with their respective locations.

To determine

(a)

The Reynolds number based on cylinder diameter.

Answer to Problem 69P

The Reynolds number is $100601$.

Explanation of Solution

Given information:

The density of fluid is $998.2\text{kg}/{\text{m}}^{\text{3}}$, Dynamic viscosity of fluid is $1.003\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$, free stream velocity is $0.100481\text{m}/\text{s},$ and diameter of the two dimensional circular cylinder is $1\text{ m}$.

Write the expression of the Reynold number.

  $\mathrm{Re}=\frac{\rho V_{\infty }d}{\mu }$   ....... (I)

Here, the density of the fluid is $\rho$, the stream velocity is $V_{\infty }$, diameter of the cylinder is $d,$ and the dynamic viscosity of the fluid is $\mu$.

Calculation:

Substitute $998.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.100481\text{m}/\text{s}$ for $V_{\infty }$, $1\text{ m}$ for $d$ and $1.003\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (I).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.100481\text{m}/\text{s}\right)\times \left(1\text{ m}\right)}{\left(1.003\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =100.601\times {10}^3\frac{\left(1\text{kg}/{\text{m}}^{\text{3}}\right)\left(\text{1 }\text{m}/\text{s}\right)\left(1\text{ m}\right)}{\left(\text{1 }\text{kg}/\text{m}\cdot \text{s}\right)}\\ =100601\end{array}$

Thus, the value of Reynolds number is very high. So, the approximation of potential flow is reasonable.

Conclusion:

Thus the Reynolds number is $100601$.

To determine

(b)

The minimum speed.

The maximum speed.

The maximum pressure difference.

The minimum pressure difference.

Answer to Problem 69P

The minimum speed is $0$.

The maximum speed is $0.200962\text{m}/\text{s}$.

The maximum pressure difference is $5.039\text{N}/{\text{m}}^{\text{2}}$.

The minimum pressure difference is $-\left(15\text{N}/{\text{m}}^{\text{2}}\right)$.

Explanation of Solution

Given information:

Write the expression of minimum speed.

  ${\left|V\right|}_{\text{min}}=0$   ....... (II)

Here, the speed at stagnation point is $0$.

Write the expression of Bernoulli's equation at the stagnation point.

  $P+\frac{1}{2}\rho V^2=P_{\infty }+\frac{1}{2}\rho V_{\infty }{}^2$   ....... (III)

Here, the pressure at starting is $P$, the density of fluid is $\rho$, the velocity at starting is $V$, the pressure at ending is $P_{\infty },$ and the free stream velocity is $V_{\infty }$.

Write the expression of maximum speed.

  ${\left|V\right|}_{\max }=2V_{\infty }$   ....... (IV)

Here, the free stream velocity is $V_{\infty }$.

Calculation:

The speed at stagnation point is $0$ from Equation (II).

Therefore, the minimum speed is $0$.

Substitute $0$ for $V$, $998.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.100481\text{m}/\text{s}$ for $V_{\infty }$ in Equation (III).

  $\begin{array}{l}P+\frac{1}{2}\left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(0\right)}^2=P_{\infty }+\frac{1}{2}\left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(0.100481\text{m}/\text{s}\right)}^2\\ P+0=P_{\infty }+\frac{1}{2}\left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(0.100481\text{m}/\text{s}\right)}^2\\ P-P_{\infty }=\frac{1}{2}\times \left(10.07\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{\text{1 kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ P-P_{\infty }=5.039\text{N}/{\text{m}}^{\text{2}}\end{array}$

So, the maximum pressure difference is $5.039\text{N}/{\text{m}}^{\text{2}}$.

Substitute $0.100481\text{m}/\text{s}$ for $V_{\infty }$ in Equation (IV).

  $\begin{array}{c}{\left|V\right|}_{\max }=2\times 0.100481\text{m}/\text{s}\\ =0.200962\text{m}/\text{s}\end{array}$

So, the maximum speed is $0.200962\text{m}/\text{s}$.

Substitute $0.200962\text{m}/\text{s}$ for $V$, $998.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.100481\text{m}/\text{s}$ for $V_{\infty }$ in Equation (III).

  $\begin{array}{l}P+\frac{1}{2}\times \left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(0.200962\text{m}/\text{s}\right)}^2=P_{\infty }+\frac{1}{2}\times \left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(0.100481\text{m}/\text{s}\right)}^2\\ P+\frac{1}{2}\times \left(\text{40}\text{.313 }\text{kg}/\text{m}\cdot {\text{s}}^2\right)=P_{\infty }+\frac{1}{2}\times \left(\text{10}\text{.078 }\text{kg}/\text{m}\cdot {\text{s}}^2\right)\\ P-P_{\infty }=\left(\left(5.039\text{kg}/\text{m}\cdot {\text{s}}^2\right)-\left(20.157\text{kg}/\text{m}\cdot {\text{s}}^2\right)\right)\left(\frac{1\text{N}}{\text{1 kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ P-P_{\infty }=-\left(15\text{N}/{\text{m}}^{\text{2}}\right)\end{array}$

So, the minimum pressure difference is $-\left(15\text{N}/{\text{m}}^{\text{2}}\right)$

Conclusion:

The minimum speed is $0$.

The maximum speed is $0.200962\text{m}/\text{s}$.

The maximum pressure difference is $5.039\text{N}/{\text{m}}^{\text{2}}$.

The minimum pressure difference is $-\left(15\text{N}/{\text{m}}^{\text{2}}\right)$.