The Role of Fuel Cells in Renewable Energy Solutions

The statement that is NOT true about the concept of a boundary layer on an object is: a. the Reynolds number is greater than uni

Ilya [14]

Answer:

Option E

Explanation:

All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.

5 0

3 years ago

Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming

kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

The concept of partial differentiation was used in the determination of the expression for u and v.
each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.

4 0

3 years ago

DRIVERS ED

forsale [732]

Answer:

b

Explanation:

only if there signal is turned on

8 0

3 years ago

Read 2 more answers

A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find

Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

$\sigma=\frac{\textup{PD}}{\textup{2t}}$

where, t is the thickness

on substituting the respective values, we get

$100=\frac{\textup{2\times800}}{\textup{2t}}$

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0

3 years ago

A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th

SCORPION-xisa [38]

Answer:

$F_{thrust}$ ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

$F_{drag}$ = $\frac{1}{2}$ × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, $F_{drag}-F_{thrust} = 0$

We can as well say:

$F_{drag}= F_{thrust}$

We can now replace $F_{thrust} with F_{drag}$ in the above equation.

Therefore, $F_{thrust}$ = $\frac{1}{2}$ × CρAv²

The A which stands as the area of the jet is given by the formula:

$A=\frac{\pi d^2}{4}$

We can now have a new equation after substituting our A into the previous equation as:

$F_{thrust}$ = $\frac{1}{2}$ × Cρ $(\frac{\pi d^2}{4})v^2$

Substituting our data from above; we have:

$F_{thrust}$ = $\frac{1}{2}$ × $(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2$

$F_{thrust}$ = $\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2$

$F_{thrust}$ = 110,990N

$F_{thrust}$ in N (newton) to KN (kilo-newton) will be:

$F_{thrust}$ = $(110,990N)*\frac{1KN}{1,000N}$

$F_{thrust}$ = 110.990 KN

$F_{thrust}$ ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0

3 years ago