Question 1 of 6: Loss Factor of a Valve from Bench-Test Data

Question A1: Loss Factor of a Valve from Bench-Test Data (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The valve loss (resistance) coefficient $k_v$ at each of the three closure positions.

Approach. Treat the constant 1.5 m head as the head lost across the test section (pipe + valve), subtract the straight-pipe friction, and what remains is the valve loss — giving $k_v$ directly at each measured flow. The data themselves show that no additional exit velocity head can be charged to the 1.5 m.

1. Head balance on the test section. The rig holds a constant head $H=1.5$ m across the smooth pipe and the valve, so that head is consumed by wall friction plus the valve loss: $$H = 4f\frac{L}{D}\frac{u^2}{2g} + k_v\frac{u^2}{2g}\ \Longrightarrow\ k_v = \frac{H}{u^2/2g} - 4f\frac{L}{D}.$$ The friction term uses the Fanning factor with $4f(L/D)$ as the head-loss coefficient. Why no exit velocity head: if the jet's kinetic energy were also charged to the 1.5 m, even a loss-free valve ($k_v=0$) could pass at most $u=\sqrt{2gH/(1+4fL/D)}=\sqrt{2(9.81)(1.5)/2.372}=3.52$ m/s, yet the open valve measured $4.46$ m/s (step 2). The data therefore only close if the 1.5 m is the head lost across the pipe-and-valve section.
2. Velocity and Reynolds number at each flow. The pipe area is $A=\tfrac{\pi}{4}(0.020)^2=3.142\times10^{-4}\ \text{m}^2$, so $u=Q/A$ and $Re=\rho u D/\mu$. For the fully-open point, $$u_0=\frac{8.4\times10^{-2}/60}{3.142\times10^{-4}}=4.456\ \text{m/s},\qquad Re_0=\frac{1000(4.456)(0.020)}{1\times10^{-3}}=\boxed{8.91\times10^{4}}.$$ All three Reynolds numbers ($8.91,\ 6.37,\ 3.82\times10^4$) are turbulent and below the Blasius ceiling.
3. Friction factor and velocity head. The smooth-pipe Blasius correlation gives $f=0.079\,Re^{-0.25}$; at the open point $f_0=0.079(8.91\times10^4)^{-0.25}=4.57\times10^{-3}$ and $u_0^2/2g=4.456^2/(2\cdot9.81)=1.012\ \text{m}$, so $4f_0(L/D)=4(0.00457)(75)=1.372$.
4. Tabulate the loss factor across the three closures. Applying $k_v=H/(u^2/2g)-4f(L/D)$ at each point:

   Closure	$u$ (m/s)	$Re$	$u^2/2g$ (m)	$4f\,L/D$	$k_v$
   0%	4.456	$8.91\times10^4$	1.012	1.372	0.11
   25%	3.183	$6.37\times10^4$	0.516	1.493	1.41
   40%	1.910	$3.82\times10^4$	0.186	1.695	6.37

   For example at 40% closure $k_v=\dfrac{1.5}{0.186}-1.695=\boxed{6.37}$ and at 25% closure $k_v=\dfrac{1.5}{0.516}-1.493=\boxed{1.41}$.
5. Interpret the fully-open point. At 0% closure $k_v=\dfrac{1.5}{1.012}-1.372=\boxed{0.11}$ — small, as expected for a fully open valve whose bore nearly matches the pipe. Because it is the difference of two numbers near 1.4, it is sensitive to the friction correlation (a few percent in $f$ moves it by about $\pm0.05$), so it is best quoted as $k_v\approx0.1$. The coefficient then climbs roughly 13-fold by 25% closure and nearly 60-fold by 40% closure as the port area shrinks.