Question

The pressure loss p in a pipe is related to the flow rate Q by the equation P = KQn, where K and n are constants. If K = 190 x 10³ determine the value for n when a flow rate of Q = 0.04 m³/s causes a pressure loss of 180 Pascals.​

Answer 1

The value for
`\( n \)` is approximately
`\( 2.162 \)`.

The equation given is
`\( P = KQ^n \)`, where
`\( P \)` is the pressure loss,
`\( K \)` is a constant,
`\( Q \)` is the flow rate, and
`\( n \)` is another constant.

Given values:

-
`\( K = 190 * 10^3 \)` (Pa·m³/s)

-
`\( Q = 0.04 \)` (m³/s)

-
`\( P = 180 \)` (Pa)

We can substitute these values into the equation to solve for
`\( n \):`

`\[ P = KQ^n \]`

`\[ 180 = (190 * 10^3) * (0.04)^n \]`

To find
`\( n \)`, take the natural logarithm (ln) of both sides:

`\[ \ln(180) = \ln[(190 * 10^3) * (0.04)^n] \]`

Now, apply logarithm properties to simplify:

`\[ \ln(180) = \ln(190 * 10^3) + n * \ln(0.04) \]`

Now, solve for
`\( n \)`:

`\[ n = (\ln(180) - \ln(190 * 10^3))/(\ln(0.04)) \]`

Let's calculate this:

`\[ n = (\ln(180) - \ln(190 * 10^3))/(\ln(0.04)) \]`

`\[ n \approx (5.19295-12.15478)/-3.21888`

`\[ n \approx (-6.96183)/(-3.21888) \]`

`\[ n \approx 2.162 \]`

So, the value for
`\( n \)` is approximately
`\( 2.162 \)`.