Thanks
@haruspex ! In the course of transcribing my scribbled calcs into LaTex, I corrected an error and obtained the same result as the OP's simpler approach. For posterity (
) here is my updated calculation:
Bernoulli:
$$H = z + \frac {144~P} {\rho}~+~\frac {v^2} {2g}$$
where
H = head (feet)
z = elevation (feet)
P = gauge pressure, psig
##rho## = fluid weight density, lbf/ft^3
v = fluid velocity, ft/sec^2
g = grav acceleration, 32.2 ft/sec^2
I am setting z=0 at the 6-inch diameter exit. Then the elevation of the 4-inch section is h, and the top of the fluid in the tank is at elev (4+h).
I call the top of the fluid location "1", the 4-inch section location "2", and the exit location "3". I will determine the total head at these locations as H1, H2, H3. The density ##\rho## is constant at 62.4 lb/ft^3.
At location 1, the elevation is (4+h), the pressure is atmospheric (P=0 psig), and the velocity is assumed negligible (i.e., the tank area is >> the pipe area).
$$H_1~=~z_1+ \frac {144~P_1} {\rho}~+~\frac {v_1^2} {2g}$$
$$=~(4+h)~+~\frac {144~*~0} {\rho}~+~\frac {0^2} {2g}$$
$$=~(4+h)$$
At location 2, the elevation is h, and the pressure is -6 psig
$$H_2~=~z_2+ \frac {144~P_2} {\rho}~+~\frac {v_2^2} {2g}$$
$$=~h~+~\frac {144~*~(-6)} {\rho}~+~\frac {v_2^2} {2g}$$
$$=~h~+~\frac {-864} {\rho}~+~\frac {v_2^2} {2g}$$
At location 3, the elevation is zero, and the pressure is atmospheric (P=0 psig).
$$H_3~=~z_3+ \frac {144~P_3} {\rho}~+~\frac {v_3^2} {2g}$$
$$=~0~+~\frac {144~*~0} {\rho}~+~\frac {v_3^2} {2g}$$
$$=~\frac {v_3^2} {2g}$$
Equate H1 and H3, and solve for v3
$$=~(4+h)~=~\frac {v_3^2} {2g}$$
$$v_3^2~=~(4+h)2g$$
$$v_3~=~\sqrt {(4+h)2g}$$
Equate H2 and H3
$$h~+~\frac {-864} {\rho}~+~\frac {v_2^2} {2g}=~\frac {v_3^2} {2g}$$
by continuity,
$$A_2~v_2~=~A_3~v_3$$
$$v_2~=~ \frac {A_3} {A_2} ~v_3$$
And
$$v_2^2~=~( \frac {A_3} {A_2} )^2~v_3^2$$
Then
$$h~+~\frac {-864} {\rho}~+~\frac {( \frac {A_3} {A_2} )^2~v_3^2} {2g}=~\frac {v_3^2} {2g}$$
Solve for h
$$h~=~\frac {v_3^2} {2g} ~-~\frac {v_3^2} {2g} (\frac {A_3} {A_2})^2~+~\frac {864} {\rho}$$
$$h~=~\frac {v_3^2} {2g} ~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
Insert v3^2 from above:
$$h~=~(4~+h) ~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
$$h~=~4~(1~-~(\frac {A_3} {A_2})^2)~+~h~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
$$h~=~\frac {4~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}} {1~-~(1~-~(\frac {A_3} {A_2})^2)}$$
$$(1~-~(\frac {A_3} {A_2})^2)~=~(1~-~(\frac {6^2} {4^2})^2)~=~-4.0625$$
so
$$h~=~\frac {4~(-4.0625)~+~\frac {864} {62.4}} {1~-~(-4.0625)}$$
$$h~=~-0.4748$$
Showing that "my" approach yields the same result as the OP.
