Question

- Motor oil at T=-30 deg F has a kinematic viscosity of
ν=0.11ft
^{2}/s and a specific gravity of SG=0.88. (The rest of this is made up numbers and may not be indicative of actual car engine behavior) Assume that the oil flows at a velocity of V=2ft/s through a cast iron conduit within the engine block. Treat the conduit as a “best case roughness” (use smallest number for ε if needed) cast iron pipe with a diameter of 0.5in and a length of 5ft. Assume oil height is raised by 1ft.- What is the Reynolds number?
- Is this laminar or turbulent flow?
- What is the friction factor f? Show how you determined this by calculation or moody diagram as applicable.
- What is the head loss in ft?

Answer #1

Given,

Kinematic viscosity , v = 0.11 ft2/s

Specific Gravity , G = 0.88

Velocity , V = 2 ft/s

Diameter , d = 0.5 inches = 0.04167 ft

Length of Conduit , L = 55ft

**a)**

**Reynolds Number, Re =
**

**= **

**=** **1000**

**b,**

- If Re < 2000, the flow is called Laminar
- If Re > 4000, the flow is called turbulent
- If 2000 < Re < 4000, the flow is called transition

therefore the flow is **Laminar**

**c,**

**Friction factor , f =
( for laminar )**

**=
**

**= 0.064**

Due to rough regime pipe it is essential to employ yet another
dimensionless parameter of the flow, the *roughness Reynolds
number*

( were ε = 1 )

= 2166

**d,**

**Head loss =
**

**= 0.064*1322.115*0.2**

**= 16.92 ft**

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