measure small pressure differences. When the reading is zero, the level in the t

Question

measure small pressure differences. When the reading is zero, the
level in the two reservoirs are equal. Assume that fluid A is methane
at atmospheric pressure and 60?F, that liquid B in the reservoirs is
kerosine (specific gravity = 0.815), and that liquid C in the U tube is
water. The inside diameters of the reservoirs and U tube are 2.0 and
0.25 in., respectively. If the reading of the manometer is 5.72 in., what
is the pressure difference (|Pa-Pb|) in inches of water (a) when the
change in levels in the reservoirs is neglected, (b) when the change in
levels in the reservoirs is taken into account? (c) What is the
percentage error in the answer to part (a)?

Explanation / Answer

given,

density of water rho_w=1000 kg/m^3 ------>in C tube

density of methane rho_m=815 kg/m^3 -----> in A tube

density of kerosene rho_K=1000 kg/m^3 ---> in B tube

let

h=5.72 inches =0.145 m

d1=2 inches=0.051 m

d2=0.25 inches=0.0063 m

now,

A)

when change in level in the reservoir(B) is neglected

p1+(rho_m*g*h)=P2+(rho_w*g*h)

===>

change in pressure,

p1-p2=(rho_w-rho_m)*g*h

=(1000-815)*9.8*0.145

=262.88 Pa

B)

when change in level in the reservoir(B) is taken,

let change in level is x,

now,

p1+(rho_m*g*h)=P2+(rho_w*g*h)+(rho_m*g*x)

p1-p2=(rho_w-rho_m)*g*h+(rho_m*g*x)

here

volume increased ---> pi*(d2/2)^2*x=pi*(d1/2)^2*h

====>

x=(d1/d2)^2*h

now,

p1-p2=(rho_w-rho_m)*g*h+(rho_m*g*x)

p1-p2=(rho_w-rho_m)*g*h+(rho_m*g*(d1/d2)^2*h)

=((1000-815)*9.8*0.145)+(815*9.8*(0.0063/0.051)^2*0.145)

=280.56 Pa

C)

error=(280.56-262.88)/280.56

=0.063

error in % is =6.3 %

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