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3 years ago

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Fluid originally flows through a tube at a rate of 100 cm³/s. To illustrate the sensitivity of flow rate to various factors, cal

culate the new flow rate for the following changes with all other factors remaining the same as in the original conditions. (a) Pressure difference increases by a factor of 1.50. (b) A new fluid with 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube is used with a radius 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half the length, and the pressure difference is increased by a factor of 1.50.

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:

a) Q₂ = 150 cm³/s

b) Q₂ = 33.33 cm³/s

c) Q₂ =25.0 cm³/s

d) Q₂ = 0.01 cm³/s.

e) Q₂ = 0.03 cm³/s

Explanation:

a) Original flow rate, Q₁ = 100 cm³/s

Based on the Poiseulle's equation

ΔP = (128цLQ/πD⁴)

ΔP₂/ΔP₁ = 1.50

ΔP ∝ Q

ΔP₂/ΔP₁ = Q₂/Q₁

1.5 = Q₂/Q₁

Q₂ = 1.5 Q₁

Q₂ = 1.5 * 100

Q₂ = 150 cm³/s

b)ΔP = (128цLQ/πD⁴)

Q ∝1/ц

Q = k/ц

Q₁ = k/ц₁

Q₂ = k/ц₂

Q₂/Q₁ = ц₁/ц₂

ц₂/ц₁ = 3

ц₁/ц₂ =1/3

Q₂/100 = 1/3

Q₂ = 100/3

Q₂ = 33.33 cm³/s

c)Q ∝1/L

Q = k/L

Q₁ = k/L₁

Q₂ = k/L₂

Q₂/Q₁ = L₁/L₂

L₂/L₁ = 4

L₁/L₂ =1/4

Q₂/100 = 1/4

Q₂ = 100/4

Q₂ =25.0 cm³/s

d) Q ∝ D⁴

Q₂/Q₁ = (D₂/D₁)⁴

2R₂ = D₂

2R₁ = D₁

D₂/D₁ = R₂/R₁ = 0.1

Q₂/Q₁ = 0.1⁴

Q₂/Q₁ = 0.0001

Q₂ = 0.0001Q₁

Q₂ = 0.0001 * 100

Q₂ = 0.01 cm³/s.

e) Q ∝ D⁴∝ΔP∝1/L

Q = k D⁴ΔP/L

Q₂/Q₁ = (D₂/D₁)⁴(ΔP₂/ΔP₁)(L₁/L₂)

D₂/D₁ = R₂/R₁ = 0.1

L₂/L₁ = 1/2

L₁/L₂ = 2

ΔP₂/ΔP₁ = 1.5

Q₂/Q₁ = (0.1)⁴*(1.5)*(2)

Q₂/Q₁ = 0.0003

Q₂ = 0.0003 * 100

Q₂ = 0.03 cm³/s

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