12.38 •• water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m³/s. (a) how fast will it shoot out of a hole 4.50 cm in diameter? (b) at what speed will it shoot out if the diameter of the hole is three times as large?
12.39 •• a shower head has 20 circular openings, each with radius 1.0 mm. the shower head is connected to a pipe with radius 0.80 cm. if the speed of water in the pipe is 3.0 m/s, what is its speed as it exits the shower-head openings?
12.44 • a soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355-l cans per minute. at point 2 in the pipe, the gauge pressure is 152 kpa and the cross-sectional area is 8.00 cm². at point 1, 1.35 m above point 2, the cross-sectional area is 2.00 cm². find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and 2; (d) gauge pressure at point 1.
12.46 •• bio artery blockage. a medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. to do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20 × 10⁴ pa, while in the region of blockage it is 1.15 × 10⁴ pa. furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient’s blood is 1.06. what percentage of the cross-sectional area of the patient’s artery is blocked by the plaque?

Question

Accepted Answer

Let's solve these problems one by one. We'll use the principle of continuity ($A_1v_1 = A_2v_2$) and Bernoulli's equation ($P + \frac{1}{2}
ho v^2 +
ho gh = \text{constant}$) where applicable.

### Problem 12.38
#### Part (a)
We use the continuity equation $Q = Av$, where $Q$ is the volume flow rate, $A$ is the cross - sectional area, and $v$ is the speed.
- The diameter $d = 4.50\space cm=0.045\space m$, so the radius $r=\frac{d}{2}=0.0225\space m$.
- The cross - sectional area of the hole $A=\pi r^{2}=\pi(0.0225)^{2}\space m^{2}$.
- The volume flow rate $Q = 0.750\space m^{3}/s$.

From $Q = Av$, we can solve for $v$:
$v=\frac{Q}{A}=\frac{Q}{\pi r^{2}}$

Substitute $Q = 0.750\space m^{3}/s$ and $r = 0.0225\space m$:
$A=\pi(0.0225)^{2}\approx1.5904\times 10^{-3}\space m^{2}$
$v=\frac{0.750}{1.5904\times 10^{-3}}\approx471.6\space m/s$

#### Part (b)
If the diameter is three times as large, the new diameter $d' = 3\times4.50\space cm = 13.5\space cm=0.135\space m$, and the new radius $r'=\frac{d'}{2}=0.0675\space m$.
The new cross - sectional area $A'=\pi(r')^{2}=\pi(0.0675)^{2}\space m^{2}\approx1.4314\times 10^{-2}\space m^{2}$

Using $v'=\frac{Q}{A'}$ (since $Q$ remains the same), and $Q = 0.750\space m^{3}/s$
$v'=\frac{0.750}{1.4314\times 10^{-2}}\approx52.4\space m/s$

### Problem 12.39
We use the continuity equation $A_1v_1=A_2v_2$, where $A_1$ and $v_1$ are the area and speed in the pipe, and $A_2$ and $v_2$ are the total area and speed at the shower - head openings.
- Radius of pipe $r_1 = 0.80\space cm = 0.008\space m$, so $A_1=\pi r_1^{2}=\pi(0.008)^{2}\space m^{2}\approx2.0106\times 10^{-4}\space m^{2}$
- Radius of each shower opening $r_2 = 1.0\space mm = 0.001\space m$, area of one opening $a=\pi r_2^{2}=\pi(0.001)^{2}\space m^{2}$, and total area of 20 openings $A_2 = 20\times\pi(0.001)^{2}\space m^{2}\approx6.2832\times 10^{-5}\space m^{2}$
- Speed in pipe $v_1 = 3.0\space m/s$

From $A_1v_1=A_2v_2$, we solve for $v_2$:
$v_2=\frac{A_1v_1}{A_2}=\frac{\pi(0.008)^{2}\times3.0}{20\times\pi(0.001)^{2}}$

The $\pi$ cancels out:
$v_2=\frac{(0.008)^{2}\times3.0}{20\times(0.001)^{2}}=\frac{6.4\times 10^{-5}\times3.0}{20\times10^{-6}}=\frac{1.92\times 10^{-4}}{2\times 10^{-5}} = 9.6\space m/s$

### Problem 12.44
#### Part (a)
First, find the total volume of cans per minute. Each can is $0.355\space L$, and there are 220 cans.
Total volume $V=220\times0.355\space L = 78.1\space L$ per minute. Convert to $m^{3}$ per second:
$1\space L = 10^{-3}\space m^{3}$, so $V = 78.1\times10^{-3}\space m^{3}$ per minute.
$1$ minute $ = 60$ seconds, so volume flow rate $Q=\frac{78.1\times 10^{-3}}{60}\space m^{3}/s\approx1.3017\times 10^{-3}\space m^{3}/s$

The density of water $
ho = 1000\space kg/m^{3}$. Mass flow rate $\dot{m}=
ho Q$
$\dot{m}=1000\times1.3017\times 10^{-3}\space kg/s\approx1.30\space kg/s$

#### Part (b)
We already calculated the volume flow rate in part (a):
$Q=\frac{220\times0.355\times 10^{-3}}{60}\space m^{3}/s\approx1.30\times 10^{-3}\space m^{3}/s$ (or $1.30\space L/s$)

#### Part (c)
Using the continuity equation $Q = A_1v_1=A_2v_2$
- Area at point 1: $A_1 = 2.00\space cm^{2}=2.00\times 10^{-4}\space m^{2}$
- Area at point 2: $A_2 = 8.00\space cm^{2}=8.00\times 10^{-4}\space m^{2}$

For point 2: $v_2=\frac{Q}{A_2}=\frac{1.3017\times 10^{-3}}{8.00\times 10^{-4}}\space m/s\approx1.63\space m/s$

For point 1: $v_1=\frac{Q}{A_1}=\frac{1.3017\times 10^{-3}}{2.00\times 10^{-4}}\space m/s\approx6.51\space m/s$

#### Part (d)
Use Bernoulli's equation $P_1+\frac{1}{2}
ho v_1^{2}+
ho gh_1=P_2+\frac{1}{2}
ho v_2^{2}+
ho gh_2$
Let $h_2 = 0$ (reference level), then $h_1 = 1.35\space m$, $P_2 = 152\times 10^{3}\space Pa$, $
ho = 1000\space kg/m^{3}$, $v_1\approx6.51\space m/s$, $v_2\approx1.63\space m/s$

Rearrange for $P_1$:
$P_1=P_2+\frac{1}{2}
ho(v_2^{2}-v_1^{2})+
ho g(h_2 - h_1)$

Substitute the values:
$\frac{1}{2}
ho(v_2^{2}-v_1^{2})=\frac{1}{2}\times1000\times(1.63^{2}-6.51^{2})=\frac{1}{2}\times1000\times(2.6569 - 42.3801)=500\times(- 39.7232)=-19861.6\space Pa$

$
ho g(h_2 - h_1)=1000\times9.8\times(0 - 1.35)=-13230\space Pa$

$P_1=152000-19861.6 - 13230=118908.4\space Pa\approx119\space kPa$

### Problem 12.46
We use the continuity equation $A_1v_1 = A_2v_2$ and Bernoulli's equation (since blood is incompressible and we can assume steady flow). The specific gravity of blood $SG = 1.06$, so the density of blood $
ho=SG\times
ho_{water}=1.06\times1000 = 1060\space kg/m^{3}$

#### Step 1: Apply Bernoulli's equation
$P_1+\frac{1}{2}
ho v_1^{2}+
ho gh_1=P_2+\frac{1}{2}
ho v_2^{2}+
ho gh_2$
Assume $h_1 = h_2$ (same height), so $P_1 - P_2=\frac{1}{2}
ho(v_2^{2}-v_1^{2})$

Given $P_1 = 1.20\times 10^{4}\space Pa$, $P_2 = 1.15\times 10^{4}\space Pa$, $v_1 = 30.0\space cm/s = 0.300\space m/s$

$P_1 - P_2=1.20\times 10^{4}-1.15\times 10^{4}=500\space Pa$

So, $500=\frac{1}{2}\times1060\times(v_2^{2}-0.300^{2})$

$v_2^{2}-0.09=\frac{1000}{1060}$
$v_2^{2}=\frac{1000}{1060}+0.09\approx0.9434 + 0.09 = 1.0334$
$v_2=\sqrt{1.0334}\approx1.0166\space m/s = 101.66\space cm/s$

#### Step 2: Apply continuity equation
$A_1v_1=A_2v_2$, where $A_1$ is the area of the normal artery, $A_2$ is the area of the artery at the blockage. Let the blocked area be $A_{blocked}$, so the open area $A_2=A_1 - A_{blocked}$

$A_1\times0.300=(A_1 - A_{blocked})\times1.0166$

$0.300A_1=1.0166A_1-1.0166A_{blocked}$

$1.0166A_{blocked}=1.0166A_1 - 0.300A_1=0.7166A_1$

$\frac{A_{blocked}}{A_1}=\frac{0.7166}{1.0166}\approx0.705$

The percentage of the area blocked is $70.5\%$ (or approximately $71\%$)

### Final Answers

#### 12.38
(a) $\boldsymbol{\approx472\space m/s}$
(b) $\boldsymbol{\approx52.4\space m/s}$

#### 12.39
$\boldsymbol{9.6\space m/s}$

#### 12.44
(a) $\boldsymbol{\approx1.30\space kg/s}$
(b) $\boldsymbol{\approx1.30\times 10^{-3}\space m^{3}/s}$ (or $1.30\space L/s$)
(c) $v_1\boldsymbol{\approx6.51\space m/s}$, $v_2\boldsymbol{\approx1.63\space m/s}$
(d) $\boldsymbol{\approx119\space kPa}$

#### 12.46
$\boldsymbol{\approx71\%}$ (or $70.5\%$)

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QUESTION IMAGE

Question

Explanation:

Problem 12.38

Part (a)

Part (b)

Problem 12.39

Problem 12.44

Part (a)

Part (b)

Part (c)

Answer:

Problem 12.38

Part (a)

Part (b)

Problem 12.39

Problem 12.44

Part (a)

Part (b)

Part (c)

Part (d)

Problem 12.46

Step 1: Apply Bernoulli's equation

Step 2: Apply continuity equation

Final Answers

12.38

12.39

12.44

12.46