STEM Problem Detail

Problem Statement

Prompt

Solve problem number 7.28 step by step and give answer.

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## Problem 1: Time Required to Empty a Boiler

**Solution. Given :**
| Dia. of orifice, | $$d = 150 \text{ mm} = 0.15 \text{ m}$$ |
|---|---|
| $$\therefore$$ Area, | $$a = \frac{\pi}{4}(.15)^2 = 0.01767 \text{ m}^2$$ |
| Length, | $$L = 8.0 \text{ m}$$ |
| Dia. of boiler, | $$D = 3.0 \text{ m}$$ |
| $$\therefore$$ Radius, | $$R = 1.5 \text{ m}$$ |
| Initial height of water, | $$H_1 = 2.4 \text{ m}$$ |
| Find height of water, | $$H_2 = 0$$ |
| | $$C_d = 0.6.$$ |

For completely emptying the tank, $$T$$ is given by equation (7.16) as
\begin{align*}
T &= \frac{4L}{3C_d \times a \times \sqrt{2g}} [(2R)^{3/2} - (2R - H_1)^{3/2}] \text{} \\
&= \frac{4 \times 8.0}{3 \times 0.6 \times 0.01767 \times \sqrt{2 \times 9.81}} [(2 \times 1.5)^{3/2} - (2 \times 1.5 - 2.4)^{3/2}] \text{} \\
&= 227.14 [5.196 - 0.4647] = 1074.66 \text{ sec} \\
&= \mathbf{17 \text{ min } 54.66 \text{ sec. Ans.}}
\end{align*}

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## Problem 2: Discharge Through a Mouthpiece

**Solution. Given :**
| Dia. of mouthpiece | $$= 100 \text{ mm} = 0.1 \text{ m}$$ |
|---|---|
| $$\therefore \quad$$ Area, | $$a = \frac{\pi}{4}(0.1)^2 = 0.007854 \text{ m}^2 \text{}$$ |
| Head, | $$H = 4.0 \text{ m}$$ |
| $$C_d$$ for mouthpiece | $$= 0.855 \text{}$$ |
| $$\therefore \quad$$ Discharge | $$= C_d \times \text{Area} \times \text{Velocity} = 0.855 \times a \times \sqrt{2gH} \text{}$$ |
| | $$= 0.855 \times 0.007854 \times \sqrt{2 \times 9.81 \times 4.0} = \mathbf{0.05948 \text{ m}^3\text{/s. Ans.}}$$ |

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## Problem 3: Max Discharge and Outlet Diameter

**Solution. Given :**
| Dia. of throat, | $$d_c = 4.0 \text{ cm}$$ |
|---|---|
| $$\therefore$$ Area, | $$a_c = \frac{\pi}{4} (4)^2 = 12.566 \text{ cm}^2 \text{}$$ |
| Constant head, | $$H = 2.0 \text{ m}$$ |
| Find max. dia. at outlet, $$d_1$$ and $$Q_{\max}$$ | |
| | $$H_a = 10.3 \text{ m}$$ |
| | $$H_{\text{sep}} = 2.5 \text{ m (absolute)}$$ |

The discharge, $$Q$$ in convergent-divergent mouthpiece depends on the area at throat.
\begin{align*}
\therefore \quad Q_{\max} &= a_c \times \sqrt{2gH} = 12.566 \times \sqrt{2 \times 9.81 \times 200} = \mathbf{7871.5 \text{ cm}^3\text{/s. Ans.}}
\end{align*}

Now ratio of areas at outlet and throat is given by equation (7.17) as
\begin{align*}
\frac{a_1}{a_c} &= \sqrt{1 + \frac{H_a - H_c}{H}} = \sqrt{1 + \frac{10.3 - 2.5}{2.0}} \quad \{ \because H_c = H_{\text{sep}} = 2.5 \} \text{} \\
&= 2.2135 \text{} \\
\frac{\pi}{4} d_1^2 \Big/ \frac{\pi}{4} d_c^2 &= 2.2135 \quad \text{or} \quad \left( \frac{d_1}{d_c} \right)^2 = 2.2135 \text{} \\
\therefore \quad \frac{d_1}{d_c} &= \sqrt{2.2135} = 1.4877 \text{} \\
d_1 &= 1.4877 \times 4.0 = \mathbf{5.95 \text{ cm. Ans.}}
\end{align*}

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## Problem 4: Maximum Head of Water Over Mouthpiece

**Solution. Given :**
| Dia. at throat, | $$d_c = 5 \text{ cm}$$ |
|---|---|
| Dia. at exit, | $$d_1 = 10 \text{ cm}$$ |
| Atmospheric pressure head, | $$H_a = 10.3 \text{ m}$$ |

Pressure head at throat $$= 8 \text{ m (vacuum)}\therefore H_c = H_a - 8.0 = 10.3 - 8.0 = 2.3 \text{ m (abs.)}$$
Using area ratio equation (7.17):
\begin{align*}
\frac{\frac{\pi}{4} (d_1)^2}{\frac{\pi}{4} (d_c)^2} &= \sqrt{1 + \frac{10.3 - 2.3}{H}} \implies \frac{10^2}{5^2} = 4 = \sqrt{1 + \frac{8}{H}} \text{} \\
16 &= 1 + \frac{8}{H} \implies 15 = \frac{8}{H} \implies H = \frac{8}{15} = 0.5333 \text{ m} \\
\therefore &\mathbf{\text{Maximum head of water} = 0.533 \text{ m. Ans.}}
\end{align*}

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## Problem 5: Diameters at Throat and Outlet

**Solution. Given :**
| Constant head, | $$H = 1.5 \text{ m}$$ |
|---|---|
| Discharge, | $$Q = 5 \text{ litres} = 0.005 \text{ m}^3\text{/s}$$ |
| Head loss in divergent ($$h_L$$), | $$= 0.1 \times \text{kinetic head at outlet}$$ |
| $$H_c$$ or $$H_{\text{sep}}$$, | $$= 2.5 \text{ (abs.)}$$ |
| $$H_a$$, | $$= 10.3 \text{ m of water}$$ |

**(i) Dia. at throat ($$d_c$$):** Applying Bernoulli's equation:
\begin{align*}
H_a + 0 + H &= H_c + \frac{v_c^2}{2g} \implies \frac{v_c^2}{2g} = 10.3 + 1.5 - 2.5 = 9.3 \text{ m} \\
v_c &= \sqrt{2 \times 9.81 \times 9.3} = 13.508 \text{ m/s} \\
Q &= a_c \times v_c \implies 0.005 = \frac{\pi}{4} d_c^2 \times 13.508 \text{} \\
d_c &= \sqrt{\frac{0.005 \times 4}{\pi \times 13.508}} = 0.0217 \text{ m} = \mathbf{2.17 \text{ cm. Ans.}}
\end{align*}

**(ii) Dia. at outlet ($$d_1$$):**
\begin{align*}
H_a + 0 + H &= H_a + \frac{v_1^2}{2g} + 0 + 0.1 \times \frac{v_1^2}{2g} \implies H = 1.1 \frac{v_1^2}{2g} \text{} \\
v_1 &= \sqrt{\frac{2 \times 9.81 \times 1.5}{1.1}} = 5.1724 \text{ m/s} \\
Q &= A_1 v_1 \implies 0.005 = \frac{\pi}{4} d_1^2 \times 5.1724 \text{} \\
d_1 &= \sqrt{\frac{4 \times 0.005}{\pi \times 5.1724}} = 0.035 \text{ m} = \mathbf{3.5 \text{ cm. Ans.}}
\end{align*}

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Problem #30