University of South Florida Dynamic Loss Associated with Flow from A Room Question

User Generated

zozo

Engineering

University of South Florida

Description

A duct is made of drawn tubing that has an inside diameter of 8 inches. Air flows through the duct at a rate of 500 cfm. The duct is X ft long and lays horizontally. At the duct inlet, the dynamic loss associated with flow from the room into the duct is K = 0.35. At the duct outlet, the dynamic loss associated with an attached vent is K = 7 * Y - 26.

The density and kinematic viscosity of the air are 0.0753 lbm/ft^3 and 1.62E-4 ft^2/s. The surface roughness of the drawn tubing is 0.0000015 ft.

  1. Determine the pressure drop (inches of air) along the duct length.
  2. Plot the pressure drop (inches of air) as a function of volumetric flow rate (ft^3/min), from 100 to 1000 cfm, in increments of 100 cfm.
  3. Plot the frictional head loss (inches of air) as a function of volumetric flow rate (ft^3/min), from 100 to 1000 cfm, in increments of 100 cfm.
  4. Plot the dynamic head loss (inches of air) as a function of volumetric flow rate (ft^3/min), from 100 to 1000 cfm, in increments of 100 cfm.

*All plots must be computer-generated*

The number of letters in your first name multiplied by 10 = X

The number of letters in your last name = Y

My name if 5 letters

X = 50

y - 5

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Explanation & Answer

Attached. Please let me know if you have any questions or need revisions.

A duct is made of drawn tubing that has an inside diameter of 8 inches. Air flows through the duct
at a rate of 500 cfm. The duct is X ft long and lays horizontally. At the duct inlet, the dynamic loss
associated with flow from the room into the duct is K = 0.35. At the duct outlet, the dynamic loss
associated with an attached vent is K = 7 * Y - 26.
The density and kinematic viscosity of the air are 0.0753 lbm/ft^3 and 1.62E-4 ft^2/s. The surface
roughness of the drawn tubing is 0.0000015 ft.
1. Determine the pressure drop (inches of air) along the duct length.
2. Plot the pressure drop (inches of air) as a function of volumetric flow rate (ft^3/min), from
100 to 1000 cfm, in increments of 100 cfm.
3. Plot the frictional head loss (inches of air) as a function of volumetric flow rate (ft^3/min),
from 100 to 1000 cfm, in increments of 100 cfm.
4. Plot the dynamic head loss (inches of air) as a function of volumetric flow rate (ft^3/min),
from 100 to 1000 cfm, in increments of 100 cfm.
*All plots must be computer-generated*
The number of letters in your first name multiplied by 10 = X
The number of letters in your last name = Y

Given:
𝐷 = 8 𝑖𝑛 =

2
3

𝑓𝑑

𝜈 = 1.62 π‘₯ 10βˆ’4

π‘ž = 500
𝑓𝑑 2
𝑠

π‘˜1 = 0.35
𝐿 = 𝑋 𝑓𝑑 = 50 𝑓𝑑
𝐾2 = 7π‘Œ βˆ’ 26 = 30

𝑓𝑑 3
π‘šπ‘–π‘›

=

25 𝑓𝑑 3
3 𝑠

𝑙𝑏

𝜌 = 0.0753 𝑓𝑑 3
∈= 0.0000015 𝑓𝑑 = 1.5 π‘₯ 10βˆ’6 𝑓𝑑
(𝑆𝑖𝑛𝑐𝑒 𝐼 β„Žπ‘Žπ‘£π‘’ 5 π‘™π‘’π‘‘π‘‘π‘’π‘Ÿπ‘  𝑖𝑛 π‘šπ‘¦ π‘›π‘Žπ‘šπ‘’, π‘ π‘œ 𝑋 = 10(5) = 50 𝑓𝑑)
(𝑆𝑖𝑛𝑐𝑒 𝐼 β„Žπ‘Žπ‘£π‘’ 8 π‘™π‘’π‘‘π‘‘π‘’π‘Ÿπ‘  𝑖𝑛 π‘šπ‘¦ π‘ π‘’π‘Ÿπ‘›π‘Žπ‘šπ‘’, π‘ π‘œ 𝐾2 = 7π‘Œ βˆ’ 26 = 7(8) βˆ’ 26 = 30)

Solution:
1. Determine the pressure drop (inches of air) along the duct length.
From mechanical energy balance,
π‘Šβ€² =

βˆ†π‘ƒ
+ βˆ†πΎπΈ + βˆ†π‘ƒπΈ + βˆ‘ πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘™π‘œπ‘ π‘ 
𝜌

Since there is no pump involved and there is no change in elevation while the diameter of the pipe
is uniform throughout, change in potential energy can be ignored and total work is zero. Therefore,
the equation will be:
βˆ†π‘ƒ
+ βˆ†πΎπΈ + βˆ‘ πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘™π‘œπ‘ π‘  = 0
𝜌

The pressure drop can be calculated as:
βˆ’βˆ†π‘ƒ = 𝜌 (βˆ†πΎπΈ + βˆ‘ πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘™π‘œπ‘ π‘  )
where βˆ‘ πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘Žπ‘›π‘Žπ‘™ π‘™π‘œπ‘ π‘  = πΉπ‘™π‘œπ‘ π‘  𝑑𝑒𝑒 π‘‘π‘œ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› + πΉπ‘™π‘œπ‘ π‘  𝑑𝑒𝑒 π‘‘π‘œ π‘π‘œπ‘›π‘‘π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› + πΉβ„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘  +
πΉπ‘™π‘œπ‘ π‘  𝑑𝑒𝑒 π‘‘π‘œ π‘£π‘Žπ‘™π‘£π‘’π‘ /𝑓𝑖𝑑𝑑𝑖𝑛𝑔𝑠 + πΉπ‘™π‘œπ‘ π‘  𝑑𝑒𝑒 π‘‘π‘œ π‘šπ‘’π‘‘π‘’π‘Ÿπ‘–π‘›π‘” 𝑑𝑒𝑣𝑖𝑐𝑒

Since only head loss is present,
βˆ’βˆ†π‘ƒ = 𝜌(πΉβ„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘  )
π‘€β„Žπ‘’π‘Ÿπ‘’, πΉβ„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘  = πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ β„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘  + πΉπ‘‘π‘¦π‘›π‘Žπ‘šπ‘–π‘ β„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘ 

Therefore, the final equation will be:
βˆ’βˆ†π‘ƒ = 𝜌(βˆ†πΎπΈ + πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ β„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘  + πΉπ‘‘π‘¦π‘›π‘Žπ‘šπ‘–π‘ β„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘  )

Solving f...


Anonymous
Really great stuff, couldn't ask for more.

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