For laminar flow between parallel plates separated by a distance 2h, head loss varies

D i r e c t l y a s h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I n v e r s e l y a s h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Inversely as h^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Inversely as h^{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $Inversely as h^{3}$
For parallel plates

Head loss,

$h_{f} = \frac{12 μ Q L}{ρ g (2 h)^{2}}$

(Discharge remains same)

$Q = (2 h) . ¯ U$

$So h_{f} = \frac{12 μ Q L}{ρ g (2 h)^{3}}$

Suggest Corrections

Similar questions

Q. The loss of pressure head for the laminar flow through pipes varies

Q. Laminar flows takes place between two parallel plates. The distance between the plates is

B

. The minimum depth of flow from the bottom plate where local velocity is equal to

3 / 4^{t h}

the average velocity of flow is kB. The value of k (correct upto two decimal place) is ____.

Q. Energy stored in between the plates of parallel plate capacitor of area

A

, separated by distance

d

is:

A parallel plate capacitor has rectangular plates each of area $20 c m^{2}$ is separated by a distance D the potential difference between the two plates?

Q. A parallel plate capacitor has plates of area

A

separated by distance

d

between them. It is filled with a dielectric which has a dielectric constant which varies as

k (x) = k (1 + α x)

, where `

x

' is the distance measured from one of the plates. If

(α d ≪ 1)

, the total capacitance of the system is
best given by the expression:

Join BYJU'S Learning Program

For laminar flow between parallel plates separated by a distance 2h, head loss varies

The correct option is D Inversely as h3 For parallel plates Head loss, hf=12μQLρg(2h)2 (Discharge remains same) Q=(2h).¯U So hf=12μQLρg(2h)3

The correct option is D $Inversely as h^{3}$
For parallel plates

Head loss,

$h_{f} = \frac{12 μ Q L}{ρ g (2 h)^{2}}$

(Discharge remains same)

$Q = (2 h) . ¯ U$

$So h_{f} = \frac{12 μ Q L}{ρ g (2 h)^{3}}$