Background In this experiment we examine the dynamic response of the fluid system represented schematically in the figure. h Rive The open tank acts as a fluid capacitance, for which the elemental equation is dp Q=cª dt (1) where is the net volumetric flow rate of fluid into the tank and P is the pressure at the bottom of the tank. The capacitance constant is Atank pg C= (2) where Atank is the cross-sectional area of the tank, p is the fluid mass density, and g is the local acceleration due to gravity. Pre-Lab Calculations 1. The area of the tank in our experiment is Atank = 0.0314 m². Calculate the capacitance of the tank C. (Eq 2) (mks units).

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Background
In this experiment we examine the dynamic response of
the fluid system represented schematically in the figure.
h
The open tank acts as a fluid capacitance, for which
the elemental equation is
Q=cª
(1)
where is the net volumetric flow rate of fluid into the
tank and P is the pressure at the bottom of the tank. The
capacitance constant is
dp
dt
C =
(2)
where Atank is the cross-sectional area of the tank, p
is the fluid mass density, and g is the local acceleration
due to gravity.
The orifice at the bottom of the tank acts as a fluid
resistance. Recall that the pressure drop across an orifice
resistor is.
Qout
Atank
pg
PR - (2013) Qu
20²
Rodp
dt
RC
where Ao is the area of the orifice, C is the discharge
coefficient (Cd = 0.6), and Q is the volumetric flow
through the orifice.
When the volumetric inflow rate Qin is a constant Qino
the pressure at the bottom of the tank will settle to a
steady-state value Po, and the flow rate through the ori-
fice will be QRO = Qino. Consider small perturbations in
resistor pressure and flow away from steady-state values:
P = Pr - Po, QR = QR-QRO, For these conditions,
the orifice can be described by the linear relationship
PR=RQR
where R=
,dh*
dt
QRO
CA²
Then a linear system equation for the pressure can be
derived.
Or, stated in terms of fluid level,
R
+h* = -Qin
pg
where h*=h-ho and Qin = Qin - ino
(3)
+ P* = RQin*
(5)
(6)
Pre-Lab Calculations
1. The area of the tank in our experiment is Atank =
0.0314 m². Calculate the capacitance of the tank
C. (Eq 2) (mks units).
7
Transcribed Image Text:Background In this experiment we examine the dynamic response of the fluid system represented schematically in the figure. h The open tank acts as a fluid capacitance, for which the elemental equation is Q=cª (1) where is the net volumetric flow rate of fluid into the tank and P is the pressure at the bottom of the tank. The capacitance constant is dp dt C = (2) where Atank is the cross-sectional area of the tank, p is the fluid mass density, and g is the local acceleration due to gravity. The orifice at the bottom of the tank acts as a fluid resistance. Recall that the pressure drop across an orifice resistor is. Qout Atank pg PR - (2013) Qu 20² Rodp dt RC where Ao is the area of the orifice, C is the discharge coefficient (Cd = 0.6), and Q is the volumetric flow through the orifice. When the volumetric inflow rate Qin is a constant Qino the pressure at the bottom of the tank will settle to a steady-state value Po, and the flow rate through the ori- fice will be QRO = Qino. Consider small perturbations in resistor pressure and flow away from steady-state values: P = Pr - Po, QR = QR-QRO, For these conditions, the orifice can be described by the linear relationship PR=RQR where R= ,dh* dt QRO CA² Then a linear system equation for the pressure can be derived. Or, stated in terms of fluid level, R +h* = -Qin pg where h*=h-ho and Qin = Qin - ino (3) + P* = RQin* (5) (6) Pre-Lab Calculations 1. The area of the tank in our experiment is Atank = 0.0314 m². Calculate the capacitance of the tank C. (Eq 2) (mks units). 7
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