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liubo4ka [24]
4 years ago
9

Another common unit for viscosity is the centipoise, which is 10-2 poise (just like a centimeter is 10-2 meters). (1 centipoise

= 0.001 N⋅s/m2) The unit poise is named after Jean Leonard Marie Poiseuille. The conversion between the centipoise and the Pa⋅s is 1 centipoise = 1 mPa.s. What is the viscosity of water at room temperature?
Engineering
1 answer:
Oduvanchick [21]4 years ago
4 0

Answer and Explanation

Picking room temperature to be 20°C, the viscosity of water (μ) obtained from literature is:

1 centipoise = 0.01 poise = 0.001 Pa.s = 0.001 N.s/m² (This viscosity is the dynamic viscosity of water at 20°C)

Kinematic viscosity of water, η = μ/ρ

At 20°C, μ = 0.001 Pa.s, ρ = 998.23 kg/m³

η = 0.001/998.23 = 1.0 × 10⁻⁶ m²/s

Picking the room temperature to be 25°C, the viscosityof water (μ) is

0.89 centipoise = 0.0089 poise = 0.00089 Pa.s = 8.9 × 10⁻⁴ N.s/m²

Kinematic viscosity of water, η = μ/ρ

At 25°C, μ = 0.00089 Pa.s, ρ = 997 kg/m³

η = 0.00089/997 = 8.9 × 10⁻⁷ m²/s

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4 years ago
Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the
Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen         σ_{max}                      σ_{min}

A                       +450                      -150

B                       +300                      -300

C                       +500                      -200

Solution:

Compute the mean stress

σ_{m} =  (σ_{max}  +  σ_{min})/2

σ_{mA} =  (450 + (-150)) / 2

       =  (450 - 150) / 2  

       = 300/2

σ_{mA} = 150 MPa

σ_{mB}  = (300 + (-300))/2

        = (300 - 300) / 2

        = 0/2  

σ_{mB}  = 0 MPa

 

σ_{mC}  = (500 + (-200))/2

        = (500 - 200) / 2

        = 300/2

σ_{mC}  = 150 MPa  

Compute stress amplitude:

σ_{a} =  (σ_{max}  -  σ_{min})/2    

σ_{aA} =  (450 - (-150)) / 2

       =  (450 + 150) / 2

       = 600/2

σ_{aA} = 300 MPa

σ_{aB} =  (300- (-300)) / 2

       =  (300 + 300) / 2

       = 600/2

σ_{aB}  = 300 MPa

σ_{aC}  = (500 - (-200))/2

        = (500 + 200) / 2

        = 700 / 2

σ_{aC}   = 350 MPa

From the above results it is concluded that the longest  fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

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Answer:

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Answer:

False

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