The discharge ($Q$) from a drip emitter is related to the operating pressure ($P$) and the emitter constant ($K$) by the general formula:
$Q = K \cdot P^x$
The exponent '$x$' indicates the flow regime:
Given:
Substitute the values into the formula:
$4 = 0.3 \cdot (110)^x$
Rearrange to solve for '$x$':
1. Isolate the pressure term:
$ \frac{4}{0.3} = (110)^x $
$ 13.333... = (110)^x $
2. Take the logarithm of both sides (using natural logarithm, ln, for calculation):
$ \ln(13.333...) = \ln((110)^x) $
3. Use the logarithm power rule: $\ln(a^b) = b \cdot \ln(a)$:
$ \ln(13.333...) = x \cdot \ln(110) $
4. Solve for '$x$':
$ x = \frac{\ln(13.333...)}{\ln(110)} $
$ x \approx \frac{2.590}{4.700} \approx 0.55 $
The calculated exponent '$x$' is approximately 0.55. This value is close to 0.5, which characterises turbulent flow.
Turbulent flow is typical for simple orifice emitters where the water flows through a short opening.
Therefore, based on the calculated exponent, the emitter is most likely an orifice type.