QUESTION 7 [TOTAL MARKS: 16 MARKS]
An engineer is designing a water filtration system where the flow rate of water through a pipe
varies due to the accumulation of particles. The relationship between the flow velocity
and the rate of particle accumulation is modelled by the Bernoulli’s equation:

is the flow velocity (in metres per second) and is the time taken (in seconds). Assuming
the initial velocity of water is 1 metre per second, determine the function of .
[16 marks]

Ask by Higgins Frank. in Brunei

Mar 22,2025

Question

QUESTION 7 [TOTAL MARKS: 16 MARKS]
An engineer is designing a water filtration system where the flow rate of water through a pipe
varies due to the accumulation of particles. The relationship between the flow velocity
and the rate of particle accumulation is modelled by the Bernoulli’s equation:

is the flow velocity (in metres per second) and is the time taken (in seconds). Assuming
the initial velocity of water is 1 metre per second, determine the function of .
[16 marks]

Ask by Higgins Frank. in Brunei

Mar 22,2025

UpStudy AI · Accepted Answer

**Step 1. Rewrite the Equation**

The given differential equation is

$$
\frac{dv}{dt} + \frac{v}{10} = \frac{5}{v}.
$$

Since the right‐hand side contains $$\frac{5}{v}$$, this equation is non‐linear. To simplify, multiply the equation by $$v$$:

$$
v \frac{dv}{dt} + \frac{v^2}{10} = 5.
$$

**Step 2. Substitute to Reduce the Equation**

Notice that the term $$v \frac{dv}{dt}$$ can be related to the derivative of $$v^2$$. Let

$$
u = v^2.
$$

Then, differentiating $$u$$ with respect to $$t$$, we have

$$
\frac{du}{dt} = 2v \frac{dv}{dt} \quad \Longrightarrow \quad v \frac{dv}{dt} = \frac{1}{2}\frac{du}{dt}.
$$

Substitute into the equation:

$$
\frac{1}{2}\frac{du}{dt} + \frac{u}{10} = 5.
$$

Multiply the entire equation by 2 to simplify:

$$
\frac{du}{dt} + \frac{u}{5} = 10.
$$

**Step 3. Solve the Linear Differential Equation**

The equation

$$
\frac{du}{dt} + \frac{1}{5}u = 10
$$

is a first-order linear differential equation. The integrating factor is computed as:

$$
\mu(t) = e^{\int \frac{1}{5}\,dt} = e^{t/5}.
$$

Multiply the entire equation by the integrating factor:

$$
e^{t/5}\frac{du}{dt} + \frac{1}{5}e^{t/5}u = 10e^{t/5}.
$$

Notice that the left-hand side is the derivative of $$u e^{t/5}$$:

$$
\frac{d}{dt}\left(u e^{t/5}\right)= 10e^{t/5}.
$$

Integrate both sides with respect to $$t$$:

$$
\int \frac{d}{dt}\left(u e^{t/5}\right)dt = \int 10e^{t/5}dt.
$$

This gives:

$$
u e^{t/5} = 10 \cdot \int e^{t/5}dt.
$$

To compute the integral on the right, use the substitution $$w = t/5$$ so that $$dt = 5dw$$:

$$
\int e^{t/5}dt = 5\int e^w\,dw = 5e^w = 5e^{t/5}.
$$

Thus,

$$
u e^{t/5} = 10 \cdot 5e^{t/5} + C = 50e^{t/5} + C,
$$

where $$C$$ is the constant of integration. Dividing through by $$e^{t/5}$$, we obtain:

$$
u(t) = 50 + Ce^{-t/5}.
$$

**Step 4. Apply the Initial Condition**

We are given that the initial velocity is $$v(0)=1$$. Since $$u = v^2$$, we have:

$$
u(0) = v(0)^2 = 1.
$$

Substitute $$t = 0$$ into the solution:

$$
1 = 50 + C e^{0} = 50 + C.
$$

Solve for $$C$$:

$$
C = 1 - 50 = -49.
$$

Thus, the solution for $$u(t)$$ becomes:

$$
u(t) = 50 - 49e^{-t/5}.
$$

**Step 5. Express the Solution in Terms of $$v(t)$$**

Since $$u(t) = v(t)^2$$, we have:

$$
v(t)^2 = 50 - 49e^{-t/5}.
$$

Taking the positive square root (as velocity is positive), the function $$v(t)$$ is:

$$
v(t)=\sqrt{50-49e^{-t/5}}.
$$
The flow velocity $$ v(t) $$ as a function of time is: $$ v(t) = \sqrt{50 - 49e^{-t/5}} $$

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