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liq [111]
3 years ago
8

In a flow over a flat plate, the Stanton number is 0.005: What is the approximate friction factor for this flow a)- 0.01 b)- 0.0

2 c)- 0.001 d)- 0.1
Engineering
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

(a) .01

Explanation:

  • stanton number is a dimensionless quantity
  • stanton is expressed as \frac{heat transer}{thermal capacity}
  • stanton number is discovered by Thomas edward stanton

there is relation between friction factor and stanton number and friction factor that is stanton number is half of friction factor

stanton number =\frac{friction factor}{2}

.005=\frac{friction factor}{2}

friction factor =2×.005

friction factor=.01

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Firdavs [7]

Answer:

c) site preparation

Explanation:

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7 0
3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

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