Chemical Engineering - Fluid Mechanics - Discussion

According to me, Option B is correct.

Why?

Because the same flow rate of water is used, by Bernoulli's principle, the -ΔP of both cases will be the same(?).
If we go by this assumption, since -ΔP = -Δ(ρgh) = -Δρ(ghm) = (ρm - ρf)(ghm), for a constant -ΔP, a change in hm would mean that there would have to be a proportional adjustment in the density difference ρm - ρf. Since ρf is the same (water), then it is ρm that has to change.

(ρm - ρf)1*(hm)1 = (ρm - ρf)2*(hm)2.
(SGm - SGf)1*(hm)1 = (SGm - SGf)2*(hm)2.

Case I:
SGm - SGf = 13.6 - 1 = 12.6.
hm = 2.

Case II:
SGm - SGf = 1.6 - 1 = 0.6.
hm = ?

Plugging in the values:
12.6 * 2 = 0.6*(hm)2.
(hm)2 = 42 cm.

17 is right answer.

You don't need to require -1 (atmosphere pressure). Because you comparing a close system with a close system.
ρ1 * h1 = ρ2 * h2,
13.6 * 2 = 1.6 * h2,
So, h2 = 17 cm.

42 is the wrong answer.

(You have to do -1 when you compare closed system with the open system)
(13.6-1)*2 = (1.6-1)*H.
H = 42cm.

P = (ρ) * g * h.
So, P1 = P2.
ρ1 * g1 * h1 = ρ2 * g2 * h2.
ρ1 * h1 = ρ2 *h2.

Now, ρ of fluid = sp.gr. * density of water ρ for mercury (ρ1) = 13.6 * 1 gm/cc.
ρ1 = 13.6 gm/cc.

Same for ρ2 = 1.6 gm/cc.

Now from the above equation,
ρ1 * h1 = ρ2 * h2,
13.6 * 2 = 1.6 * h2,
So, h2 = 17 cm.

Agree, 17 is the right answer.

@Hemang.
I agree with you.
According to the formula, there should is delta P.

So answer should be 42.

I also agree that answer should be 42.

I appear in exam of linde engineering and they copy paste this question. Hope for good.

42 manometer are used for the measurement of the difference of pressure.

Manometer measures pressure P1 = P2.

Density 1*g*H1 = Density 2*g*H2.

H1 = 2 cms, g = 9.81, density 1 = Specific gravity*Density of water.

H2 = ((13.6*1)*9.81*2)/((1.6*1)*9.81.

= 17 cms.

That is when sealing of water is provided normally it is (rho) *g*h.

(13.6-1)*2 = (1.6-1)*H.

H= 42cm.