The exam is closed book and closed notes.

---

The exam is closed book and closed notes.

1. Air discharge from a 2 in diameter nozzle and strikes a curved vane, which is in a vertical plane as shown in below figure. A stagnation tube connected to a water U-tube manometer is located in the free air jet. Determine the horizontal component of the force that the air jet exerts on the vane.

Neglect the weight of the air and all friction.

𝛾

: 62.4 𝑙𝑏/𝑓𝑡

∑ 𝐹𝑥 = 𝑚̇𝑉 ⃗⃗⃗ − 𝑚̇𝑉

⃗⃗⃗

2. The power P developed by a wind turbine is a function of turbine’s diameter D, air density ρ,

wind speed V, and rotation rate ω. Namely it can be written as P = f(D, ρ, V, ω). Viscosity effects

are negligible. Rewrite this relationship in dimensionless form.

---

3. The viscous, incompressible flow between the parallel plates shown in Figure is caused by the motion of the bottom plate. Assuming pressure gradient along the 𝑥 direction is a constant and not a zero(

𝜕𝑥

≠ 0), steady, 2D, and parallel flow and using differential analysis: (a) Show that the flow is fully developed using continuity equation; (b) Find the velocity profile 𝑢(𝑦) using Navier- Stokes equations with appropriate boundary conditions; (c) Find wall shear stress at bottom wall;

and (d) Find the flow rate (hint: 𝑄 = ∫ 𝑉 ⋅ 𝑛 𝑑𝐴 and assume constant width w). Explicitly state all assumptions.

Incompressible Continuity Equation:

𝜕𝑢

𝜕𝑥+^𝜕𝑣

𝜕𝑦+^𝜕𝑤

𝜕𝑧 = 0

Incompressible Navier-Stokes Equations in Cartesian Coordinates:

𝜌𝑔𝑥−^𝜕𝑝

𝜕𝑥+ 𝜇 (^𝜕2𝑢

𝜕𝑥²+^𝜕2𝑢

𝜕𝑦²+^𝜕2𝑢

𝜕𝑧²) = 𝜌 (^𝜕𝑢

𝜕𝑡+ 𝑢^𝜕𝑢

𝜕𝑥+ 𝑣^𝜕𝑢

𝜕𝑦+ 𝑤^𝜕𝑢

𝜕𝑧) 𝜌𝑔𝑦−^𝜕𝑝

𝜕𝑦+ 𝜇 (^𝜕2𝑣

𝜕𝑥²+^𝜕2𝑣

𝜕𝑦²+^𝜕2𝑣

𝜕𝑧²) = 𝜌 (^𝜕𝑣

𝜕𝑡+ 𝑢^𝜕𝑣

𝜕𝑥+ 𝑣^𝜕𝑣

𝜕𝑦+ 𝑤^𝜕𝑣

𝜕𝑧) 𝜌𝑔_𝑧−^𝜕𝑝

𝜕𝑧+ 𝜇 (^𝜕2𝑤

𝜕𝑥²+^𝜕2𝑤

𝜕𝑦²+^𝜕2𝑤

𝜕𝑧²) = 𝜌 (^𝜕𝑤

𝜕𝑡+ 𝑢^𝜕𝑤

𝜕𝑥+ 𝑣^𝜕𝑤

𝜕𝑦+ 𝑤^𝜕𝑤

𝜕𝑧) Shear stress (At free surface, shear stress is zero)

𝜏_𝑥𝑦= 𝜏_𝑦𝑥= 𝜇 (𝜕𝑢

𝜕𝑦+𝜕𝑣

𝜕𝑥) 𝜏𝑥𝑦= 𝜏𝑦𝑥= 𝜇 (𝜕𝑢

𝜕𝑦+𝜕𝑣

𝜕𝑥) 𝜏_𝑥𝑦= 𝜏_𝑦𝑥= 𝜇 (𝜕𝑢

𝜕𝑦+𝜕𝑣

𝜕𝑥)

---

Solution 1:

KNOWN: Inlet and outlet diameter, Outlet flow angle, manometer height.

FIND:

- Force 𝑅

that requires to hold the plate stationary ASSUMPTIONS:

Neglect weight of air Neglect friction loss

∑ 𝐹𝑥 = −𝑅

= 𝑚̇𝑉

− 𝑚̇𝑉

𝑅

= 𝑚̇𝑉

− 𝑚̇𝑉

Diameter of section

and

is same

∴ |𝑉

| = |𝑉

|

𝑉

Direction is negative 𝑥 and has cos (30

)

∴ 𝑅

= 𝑚̇𝑉

− 𝑚̇(−𝑉

𝑐𝑜𝑠30

) 𝑅

= 𝑚̇𝑉

(1 + 𝑐𝑜𝑠30

)

Calculate 𝑉

using Bernoulli equation (Apply stagnation point and section

) 𝑃

𝜌

+ 𝑉

2 = 𝑃

𝜌

+ 𝑉

2

𝑃

𝜌

= 𝑃

𝜌

+ 𝑉

2 From the manometer,

𝑃

= 𝑃

+ ℎ𝛾

𝑃

= 𝑃

𝑃

+ ℎ𝛾

𝜌

= 𝑃

𝜌

+ 𝑉

2

∴ 𝑉

2 = ℎ𝛾

𝜌

, 𝑉

= √2ℎ 𝛾

𝜌

(1)

(1)

(2)

(2)

(2)

---

Plug in 𝑉

to the 𝑅

𝑅

= 𝑚̇𝑉

(1 + 𝑐𝑜𝑠30

) = 𝜌

𝜋 4 ( 2

12 )

2

𝑉

(1 + 𝑐𝑜𝑠30

)

𝑅

= 𝜌

𝜋 4 ( 2

12 )

2

2ℎ 𝛾

𝜌

(1 + 𝑐𝑜𝑠30

) 𝑅

= 𝜋ℎ

2 ( 2 12 )

2

𝛾

(1 + 𝑐𝑜𝑠30

)

𝑅

= 𝜋(7/12) 2 ( 2

12 )

2

62.4(1 + 𝑐𝑜𝑠30

)

𝑅

= 2.963 𝑙𝑏

---

Solution 2:

ASSUMPTIONS: the problem is only a function of the above dimensional variables ANALYSIS:

P = f(D, ρ, V, ω).

𝑛 = 5

𝑇³ } , 𝐷 = {𝐿}, 𝜌 = {𝑀

𝐿³} , 𝑉 = {𝐿

𝑇} ,

𝜔

𝑗 = 3 → 𝑘 = 𝑛 − 𝑗 = 5 − 3 = 2 The repeating variables are 𝐷, 𝜌, 𝑉

Calculate first non-dimension variable

Π₀ = 𝑃𝐷^𝑎𝜌^𝑏𝑉^𝑐

𝑢𝑛𝑖𝑡 [Π₀] → {𝑀𝐿²

𝑇³ } {𝐿^𝑎} {𝑀^𝑏 𝐿^3𝑏} {𝐿^𝑐

𝑇^𝑐}

Above one should be non-dimension

[𝑀] → 1 + 𝑏 = 0 [𝐿] → 2 + 𝑎 − 3𝑏 + 𝑐 = 0

[𝑇] → −3 − 𝑐 = 0

∴ 𝑏 = −1, 𝑐 = −3, 𝑎 = −2

∴ Π₀= 𝑃𝐷⁻²𝜌⁻¹𝑉⁻³= 𝑃 𝜌𝐷²𝑉³

Calculate second non-dimension variable

Π₁=

𝜔

𝑇} {𝐿^𝑎} {𝑀^𝑏 𝐿^3𝑏} {𝐿^𝑐

𝑇^𝑐} [𝑀] → 𝑏 = 0

[𝐿] → 𝑎 − 3𝑏 + 𝑐 = 0 [𝑇] → −1 − 𝑐 = 0

∴ 𝑏 = 0, 𝑐 = −1, 𝑎 = 1

∴ Π₁=

𝜔

𝜔𝐷

Π0= 𝑓𝑛𝑐(Π1) → 𝑃

𝜌𝐷²𝑉³= 𝑓𝑛𝑐 (

𝜔𝐷

(4)

(1) (1)

(2) (2)

---

Solution 3:

KNOWN: Flow condition, Boundary condition FIND:

- Show that the flow is fully developed - velocity profile 𝑢(𝑦)

- Find wall shear stress at bottom wall - Find the flow rate

ASSUMPTIONS:

- steady:

𝜕𝑡

= 0 - 2D flow: 𝑤 = 0;

𝜕𝑧

= 0 - Parallel flow: 𝑣 = 𝑤 = 0 -

𝜕𝑥

= constant

ANALYSIS:

(a) Continuity

𝜕𝑢

𝜕𝑥 + 𝜕𝑣

𝜕𝑦 + 𝜕𝑤

𝜕𝑧 = 0

𝜕𝑢

𝜕𝑥 + 0 + 0 = 0 → 𝜕𝑢

𝜕𝑥 = 0 → 𝑓𝑢𝑙𝑙𝑦 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑

(b) x-momentum equation 𝜌 ( 𝜕𝑢

𝜕𝑡 + 𝑢 𝜕𝑢

𝜕𝑥 + 𝑣 𝜕𝑢

𝜕𝑦 + 𝑤 𝜕𝑢

𝜕𝑧 ) = 𝜌𝑔

− 𝜕𝑝

𝜕𝑥 + 𝜇 ( 𝜕

𝑢

𝜕𝑥

+ 𝜕

𝑢

𝜕𝑦

+ 𝜕

𝑢

𝜕𝑧

)

0 = − 𝜕𝑃

𝜕𝑥 + 𝜇 𝜕

𝑢

𝜕𝑦

𝜕

𝑢

𝜕𝑦

= 1 𝜇

𝜕𝑝

𝜕𝑥

Integrate twice

𝑢(𝑦) = 1 2𝜇

𝜕𝑝

𝜕𝑥 𝑦

+ 𝑐

𝑦 + 𝑐

(1)

(1.5)

(1.5)

---

Apply boundary conditions

𝑦 = 0 → 𝑢 = 𝑈 → 𝑐

= 𝑈

𝑦 = 𝑏 → 𝜏 = 0 → 𝜇 ( 𝜕𝑢

𝜕𝑦 ) = 0 → 𝜇 ( 1 𝜇 ( 𝜕𝑝

𝜕𝑥 ) 𝑏 + 𝑐

) = 0

𝑐

= − 1 𝜇 ( 𝜕𝑝

𝜕𝑥 ) 𝑏

Therefore,

𝑢(𝑦) = 1 2𝜇

𝜕𝑝

𝜕𝑥 𝑦

− 𝑏 𝜇 ( 𝜕𝑝

𝜕𝑥 ) 𝑦 + 𝑈

𝑢(𝑦) = 1 2𝜇 ( 𝜕𝑝

𝜕𝑥 ) (𝑦

− 2𝑏𝑦) + 𝑈

(c) Shear stress at bottom wall 𝜏 = 𝜇 𝑑𝑢

𝑑𝑦 = 𝜇 { 1 2𝜇 ( 𝜕𝑝

𝜕𝑥 ) (2𝑦 − 2𝑏)} = ( 𝜕𝑝

𝜕𝑥 ) (𝑦 − 𝑏)

𝜏

= 𝜏(0) = −𝑏 ( 𝜕𝑝

𝜕𝑥 )

(d) Flow rate

𝑄 = ∫ 𝑉 ⋅ 𝑛 𝑑𝐴 = 𝑤 ∫ 𝑢(𝑦)𝑑𝑦

𝑏 0

= 𝑤 2𝜇 ( 𝜕𝑝

𝜕𝑥 ) ( 1

3 𝑦

− 𝑏𝑦

)

0 𝑏

+ 𝑤𝑈(𝑦)

𝑄 = 𝑤 2𝜇 ( 𝜕𝑝

𝜕𝑥 ) ( 1

3 𝑏

− 𝑏

) + 𝑤𝑈𝑏 = − 𝑤𝑏

3𝜇 ( 𝜕𝑝

𝜕𝑥 ) + 𝑤𝑈𝑏

(2)

(1.5)

(1.5)