How to determine effect of obstruction on pressure? I am struggling to figure out what theory explains a situation I observed. A steam boiler had a pressure control which would turn off the boiler if it sensed 2 psi. The boiler would frequently switch off due to the pressure control sensing pressures exceeding 2 psi. When the pressure control was removed a blockage was found in the curled tube that lead into the pressure control. After the blockage was removed the steam boiler stopped sensing pressures in excess of 2 psi. What explains the effect of the blockage on the pressure sensor.
Things I have thought of:

*

*Ideal gas law: $PV = nRT$, I think this would have no affect. Even though the volume of the system is decreased slightly due to the a narrower tube it is such a small decrease in volume relative to the whole system.


*Bernoulli's Principle: If we measure pressure at a blockage the pressure should be lower than the overall pressure of the system because fluids flow faster and at lower pressure pressure through narrower tubes. We aren't measuring pressure at the blockage though so I assume that this does not apply here.


*Poiseuille Equation: Change in Pressure = Flow Rate $×$ Resistance. I think the equation is a little more complicated with a gas, but the general idea should still apply. If we narrow the tube then the resistance will increase and we should see an increase in pressure at the control relatively to pressure in the rest of the system. I am confused by this idea because it seems to say the opposite of what Bernoulli's principle says, but seems like it is the most likely explanation here.


*The pressure control is broken.


*There is another principle at work that I don't know about.
Heating System Diagram:

Obstruction:

 A: What you are looking at is compressible flow, which is governed by conservation of mass and momentum.  If you combine these equations, you can find that:
$$
dP \left( 1 - M^{2} \right) = \rho \ v^{2} \ \frac{ dA }{ A } \tag{0}
$$
where $dP$ is the differential change in pressure, $M$ is the Mach number, $\rho$ is the mass density of the fluid/gas, $v$ is the flow velocity, $A$ is the area of the pipe, and $dA$ is the differential change in area of the pipe.
We can use the continuity equation to show that:
$$
\rho \ A \ v = \text{constant} \tag{1}
$$
We can differentiate Equation 1 to find:
$$
\frac{ dv }{ v } + \frac{ dA }{ A } + \frac{ d\rho }{ \rho } = 0 \tag{2}
$$
The conservation of energy, assuming isentropic flow, is given as:
$$
\frac{ 1 }{ 2 } v^{2} + \frac{ \gamma }{ \gamma - 1 } \frac{ P }{ \rho } = \text{constant} \tag{3}
$$
where $\gamma$ is the ratio of specific heats (sometimes called the polytrope index) and it does not change, i.e., it's a constant.  Note that in an isentropic process, one can assume the pressure of an ideal gas follows:
$$
P \ \rho^{-\gamma} = \text{constant} \tag{4a}
$$
which allows us to see that:
$$
dP = \frac{ \gamma P }{ \rho } d\rho \tag{4b}
$$
We can then differentiate Equation 3 and use Equations 2 and 4b to find:
$$
\begin{align}
  v \ dv + \left( \frac{ \gamma }{ \gamma - 1 } \right) \left( \frac{ \rho \ dP - P \ d\rho }{ \rho^{2} } \right) & = 0 \tag{5a} \\
  v \ dv + \left( \frac{ \gamma }{ \gamma - 1 } \right) \left[ \frac{ \rho \left( \frac{ \gamma P }{ \rho } \right) d\rho - P d\rho }{ \rho^{2} } \right] & = 0 \tag{5b} \\
  v \ dv + \left( \frac{ \gamma \ P }{ \rho^{2} } \right) d\rho & = 0 \tag{5c} \\
  v \ dv - \left( \frac{ \gamma \ P }{ \rho } \right) \left( \frac{ dv }{ v } + \frac{ dA }{ A } \right) & = 0 \tag{5d} \\
  \frac{ dv }{ v } \left( 1 - \frac{ \gamma \ P }{ \rho \ v^{2} } \right) - \left( \frac{ \gamma \ P }{ \rho \ v^{2} } \right) \frac{ dA }{ A } & = 0 \tag{5e} \\
  \frac{ dv }{ v } \left( \frac{ \rho \ v^{2} }{ \gamma \ P } - 1 \right) & = \frac{ dA }{ A } \tag{5f} \\
  \frac{ dv }{ v } \left( M^{2} - 1 \right) & = \frac{ dA }{ A } \tag{5g}
\end{align}
$$
Therefore, for subsonic flow rates (i.e., $M$ < 1), we see that a converging channel ($dA$ < 0) will cause an increase in velocity ($dv$ > 0) while a diverging channel ($dA$ > 0) will cause a decrease in velocity ($dv$ < 0).  That is, a converging channel ($dA$ < 0) will cause a decrease in pressure ($dP$ < 0) while a diverging channel ($dA$ > 0) will cause an increase in pressure ($dP$ > 0).
So the the air speed increases as it enters the constriction then decreases on the other side but the pressure will increase.  In fact, depending on the shape of the blockage and the temperatures/speeds involved, the gas can actually accelerate, relative to the pre-blockage velocity, on the pressure gauge side (e.g., look up articles on things like the de Laval nozzle).
Side Note:  This all assumes that flow actually makes it past the blockage area and your gauge is not measuring a negative pressure (relative to the pre-blockage pressure or atmospheric pressure, whatever it's calibrated to).
A: It can be assumed that as a result of obstruction, a closed cavity is formed containing steam and water.For a vapor in a closed volume, such a scenario of self-oscillations is possible: 
1) the temperature of the tube rises, the pressure in the cavity rises and reaches 2 psi; 
2) the sensor worked, the fuel supply stopped, the boiler cools down, the temperature of the tube decreases, the pressure drops and reaches 2 psi. Cycle repeats.
The temperature in the cavity depends on the temperature of the tube, which depends on the conditions of heat transfer. With a significant decrease in temperature in the cavity, steam condensation is possible, and when heated, water evaporates. Then the pressure in the cavity depends on the temperature and on the mass of the vapor. In such a system, self-oscillations are possible due to heating-cooling and evaporation-condensation cycles. All this can be formulated in the form of a kinetic model containing control parameters:
$$p'(t) = m'(t) T(t) + m(t) T'(t)$$
$$ h m'(t) = T(t) - T_c(p)$$
$$T'(t) = k (T_s - T) - h m'(t)$$
$$T_s'(t) = k_1 (T_{out} - T_s) + k_2 (T_{in} - T_s)$$
$$T_{in}'(t) = k_3 (T_{out} - T_{in}) + q(p)$$
Here $p,m,T$ are the pressure, mass of vapor and temperature in the cavity; $T_s$ is a temperature of the tube,$T_{out}$ is an ambient temperature, $T_{in}$ is the temperature inside the boiler, $h, T_c$ are the heat of evaporation and boiling point; $k_i$ are heat transfer parameters, and $q$ is heat release rate in the boiler.  I used the following parameters and initial data
$$T_c = T_0 + a (p - p0) + b (p - p0)^2$$ 
with $T_0=4/3, p_0=1, a=0.1, b=0.1$.
$$q=0,  p>2$$ $$q=1,  p\le 2$$
$T_{out} = 1; h = 2; k_2 = 1; k_1 = 0.1; k = 0.0571;  k3 = 0.1$, and
$$p(0) = 2.208290783814135, m(0) = 1.4922705216048369, 
T(0) = 1.6138432697549574, T_s(0) = 1.867229822708354, 
T_{in}(0) = 1.4922705216048369$$
Using these data, we find the following solution describing the pressure oscillations, which explains the question

