Question

The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS

Answer 1

a.

• i. PU is greater than PQ.

• ii. VU is Greater than 2VT

b.

• i. PR is Equal to PU.

• ii. VR is Equal to VS

c.

• i. VQ is Equal to VU

• ii. PR is Greater than PS.

What is pressure?

Pressure is the force per unit area on a surface.

What is speed?

Speed is the distance moved per unit time.

Pressure

Since pressure, P = hρg where

• h = depth,
• ρ = density of liquid and
• g = acceleration due to gravity.

Since ρ and g are constant

P ∝ h

So, we see that pressure is directly proportional to depth.

a. i. Pressure between R and U

Since U is lower than Q, Pressure at U is greater than pressure at Q.

So,PU is greater than PQ.

ii. Speed  between U and T

Using the continuity equation

VUAU = VTAT where

• VU = speed at U,
• AU = cross-sectional area at U = π(dU)² where
• dU = diameter at U = 1.0 mm
• VUT= speed at T,
• AT = cross-sectional area at T = π(dT)² where
• dT = diameter at T = 2.0 mm

So, VUAU = VTAT

VUπ(dU)² = VTπ(dT)²

VU = VT(dT)²/(dU)²

VU = VT(2.0)²/(1.0)²

VU = VT(4)

VU = 4VT

Since VU = 4VT,VU is Greater than 2VT

b i. Pressure between R and U

Since R is at the same depth as U, Pressure at R is equal to pressure at U.

So,PR is Equal to PU.

ii. Speed between R and S

Using the continuity equation

VRAR = VSAS where

• VR = speed at R,
• AR = cross-sectional area at R = π(dR)² where
• dR = diameter at R = 2.0 mm
• VS= speed at S,
• AS = cross-sectional area at S = π(dS)² where
• dS = diameter at S = 2.0 mm

So, VRAR = VSAS

VRπ(dR)² = VSπ(dS)²

VR = VS(dS)²/(dS)²

VR = VS(2.0)²/(2.0)²

VR = VS(1)

VR = VS

Since VR = VS,VR is Equal to VS

c. i. Speed between Q and U

Using the continuity equation

VQAQ = VUAU where

• VQ = speed at Q,
• AQ = cross-sectional area at Q = π(dQ)² where
• dQ = diameter at Q = 1.0 mm
• VU = speed at U,
• AU = cross-sectional area at U = π(dU)² where
• dU = diameter at U = 1.0 mm

So, VQAQ = VUAU

VQπ(dQ)² = VUπ(dU)²

VQ = VU(dU)²/(dQ)²

VQ = VU(1.0)²/(1.0)²

VQ = VU(1)

VQ = VU

Since VQ = VU, VQ is Equal to VU

Ii. Pressure between R and S

Since R is lower than S, Pressure at R is greater than pressure at S.

So,PR is Greater than PS.

Learn more about pressure here:

https://brainly.com/question/27945027

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