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Paul [167]
3 years ago
6

According to information found in an old hydraulies book, the energy loss per unit weight of fluid flowing through a nozzle conn

ected to a hose can be estimated by the formulah = (0.04 to 0.09)(D/d)^4*V^2/2*g where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. do you think this equation is a valid in any system units? Explain.
Engineering
1 answer:
nekit [7.7K]3 years ago
3 0

Answer:

Yes equation is valid.

Explanation:

Given:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

Using SI units to assign dimensions to every quantity as follows:

Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]

Hose diameter D = [m]

Nozzle tip diameter d = [m]

Fluid velocity in the hose V = [ m s^-1 ]

Acceleration of gravity g = [ m s^-2 ]

Using the Given Equation and plug the SI units of respective quantities:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]

Simplify the equation above:

[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]

[m] = [m]

Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.

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. A 10W light bulb connected to a series of batteries may produce a brighter lightthan a 250W light bulb connected to the same b
Dvinal [7]

Answer:

Explanation:

From the equation:

Power dissipated= square of voltage supplied by battery ÷ Resistance of the load

i.e P= V^2/R

It means that at constant voltage, the the power consumed is inversely related to the resistance. Therefore the 10W bulb which has a higher resistance will consume less power using the sufficiently excess power dissipated to glow brighter than the 250W bulb which has a low resistance. The power dissipated will partly be used to overcome this low resistance making less power available for heating up the 250W bulb .

3 0
3 years ago
Read 2 more answers
Block A has a weight of 8 lb. and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic fric
kkurt [141]

Answer:

For block A, a = 9.66 ft/s²

For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

Mass of block B = m₂ = 6 lb

Coefficient of kinetic friction = μ

Normal reaction on the blocks = N

Spring stiffness of the spring btw block A and B = k = 20 lb/ft

Compression of the spring = 0.2 ft

Analysing Block A first

The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 8 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 8/32.2 = 0.248 lbm

(20×0.2) - (0.2 × 8) = (0.248) aₓ

aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 6 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 6/32.2 = 0.186 lbm

(20×0.2) - (0.2 × 6) = (0.186) aₓ

aₓ = 15 ft/s²

4 0
3 years ago
Select the best answer to the questo
Norma-Jean [14]

Answer:

C

Explanation:

7 0
3 years ago
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50.38
klemol [59]

Answer:

International Building Code (IBC)

Explanation:

6 0
2 years ago
A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s
sergiy2304 [10]

Answer:

D_o=11.9inch

Explanation:

From the question we are told that:

Thickness T=0.5

Internal PressureP=2.2Ksi

Shear stress \sigma=12ksi

Elastic modulus \gamma= 35000

Generally the equation for shear stress is mathematically given by

 \sigma=\frac{P*r_1}{2*t}

Where

r_i=internal Radius

Therefore

 12=\frac{2.2*r_1}{2*0.5}

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Generally

 r_o=r_1+t

 r_o=5.45+0.5

 r_o=5.95

Generally the equation for outer diameter is mathematically given by

 D_o=2r_o

 D_o=11.9inch

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

 D_o=11.9inch

7 0
3 years ago
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