Question

(a) For a 1.25-in.-diameter pipe, what is the maximum volumetric flow rate at which water can be pumped and the flow will remain laminar? Express your result in the dimensions of gallons per minute.

(b) What would be the maximum fl ow rate for SAE 30 oil?

Answer 1

For maintaining the flow in laminar condition, Reynolds number should be less than or equal to 2000. Maximum Reynolds number for a flow to be laminar is 2000.

 

Calculate the maximum velocity of the water;

 

The formula for Reynolds number is follows;

Re = ρwater × maximum velocity(υmax) × d/µwater

 

Rearrange the above equation for maximum velocity as follows;

υmax = Re × µwater/ρwater × d

Here, Re is the Reynolds number, µwater is the viscosity of the water, d is the diameter of the pipe and ρwater is the density of the water.

 

Substitute 2000 for Re, 1 × 10-3 for µwater, 1000 kg/m3 for ρwater and 1.25in. for d.

υmax = [2000 × {1 × 10-3 (kg/m.s)}]/{1000 kg/m3 × (1.25 in × 1m/39.3in)}

        = 0.0625 m/s

 

Calculate the maximum flow rate in pipe under laminar condition:

Vmax = Apipe × υmax

Here, Apipe is the area of the pipe and υmax is the maximum velocity.

 

Calculate the area of the pipe:

Apipe = π/4 × d2

Here, d is the diameter of the pipe.

 

Substitute 1.25 in. for d.

Apipe = π/4 × (1.25 in. × 1m/39.3 in.)2

        = 9.94556 × 10-4 m2

 

Substitute 0.0625 m/s for υmax and 9.94556 × 10-4 m2 for Apipe.

Vmax = (7.94556 × 10-4 m2) × 0.0625 m/s

         = 4.9659 × 10-5 m3/s

         = 4.9659 × 10-5 m3/s × 264.2 gal/1m3 × 60s/1 min

         = 0.7872 gal/min

 

Therefore, the maximum flow rate of the water is 0.7872 gal/min.

 

b.

Calculate the maximum velocity of the SAE 30 oil as follows;

Density and viscosity for SAE 30 oil is 917 kg/m3 and 0.26 kg/m∙s respectively.

 

The formula for the Reynolds number is follows.

Re = ρSAE30 × maximum velocity(υmax) × d/µSAE30

 

Rearrange the above equation for the maximum velocity.

υmax = Re × µSAE30/ρSAE30 × d

 

Here, Re is the Reynolds number, µSAE30 is the viscosity of the oil, d is the diameter of the pipe and ρSAE30 is the density of the oil.

 

Substitute 2000 for Re, (0.26 kg/m∙s) for µSAE30, 917 kg/m3 for ρSAE30 and 1.25 for d.

Re = {2000 × (0.26 kg/m∙s)}/{(917 kg/m3) × (1.25 × 1m/39.3 in.)}

      = 17.7 m/s

 

Calculate the maximum flow rate in pipe under laminar condition as follows;

Vmax = Apipe × υmax

 

Here, Apipe is the area of the pipe and υmax is the maximum velocity.

 

Calculate the area of the pipe:

Apipe = π/4 × d2

Here, d is the diameter of the pipe.

 

Substitute 1.25 in. for d.

Apipe = π/4 × (1.25 in. × 1m/39.3 in.)2

         = 7.94556 × 10-4 m2   

 

Substitute 17.7 m/s for υmax and 7.94556 × 10-4 m2 for Apipe.

Vmax = (7.94456 × 10-4) × 17.7 m/s

         = 0.014 m3/s

         = 0.014 m3/s × 264.2 gal/1m3 × 60s/1m

         = 221.93 gal/min

 

Therefore, the maximum flow rate of the SAE 30 oil is 221.93 gal/min.

a. Therefore, the maximum flow rate of the water is 0.7872 gal/min.