What are the Basic requirements of drinking Water ?

A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o

Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0

3 years ago

Suppose a large amount of power is required. Which engine would you choose between Otto and Diesel? Why?

Firdavs [7]

Answer:

Otto engine

Explanation:

As we know that

Power = Torque x speed

So we can say that when speed of engine then power of engine also will increases.

The speed of Otto engine is more as compare to Diesel engine so the power of Otto engine is more.But on the other hand torque of Diesel engine is more as compare to Otto engine but the speed is low so the product of speed and torque is more for Otto engine .It means that when requires large amount of power then Otto engine should be use.

6 0

3 years ago

An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a speed of 10 ft/s thr

Sladkaya [172]

Answer:

Part A

The mass plow rate, is approximately 97.0 lbm/s

Part B

The power used to overcome friction, is approximately 1.9 hp

Explanation:

The efficiency of the pump, η = 80%

The power input to the pump, P = 20 hp

The speed of the water through the pipe, v = 10 ft./s

The diameter of the pipe, d = 5.2 inches = 13/30 ft.

The free surface of the pool above the lake, h = 80 ft.

The density of the water, ρ = 62.4 lbm/ft.³

Part A

The mass plow rate, $\dot m$ = Q × ρ

Where;

ρ = 62.4 lbm/ft³

Q = A × v

A = The cross-sectional area of the pipe

∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s

∴ The mass plow rate, $\dot m$ ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s

The mass plow rate, $\dot m$ ≈ 97.0 lbm/s

Part B

The power to pump the water at the given rate, $P_w$ = $\dot m$ ·g·h

∴ $P_w$ = 97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp

$P_w$ ≈ 14.1130725 Hp

The power output of the pump, $P_{out}$ = 0.8 × 20 hp = 16 hp

Therefore, the power used to overcome friction, $P_f$ = $P_{out}$ - $P_w$

∴ $P_f$ ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp

The power used to overcome friction, $P_f$ ≈ 1.9 hp

7 0

2 years ago

Jylan enjoys being in and around water. Based on his interests, which of the following careers might be enjoyable for him? dairy

Tasya [4]

Answer: the answer is a marine biologist

Explanation: the answer is marine biologist because Jylan likes to be in and out the water which means he would be perfect as marine biologist.

6 0

3 years ago

Pressurized steam at 400 K flows through a long, thin-walled pipe of 0.6-m diameter. The pipe is enclosed in a concrete (stone m

iris [78.8K]

Answer:

Rate of heat loss per unit length of pipe, q' = 767.01 W/m

Explanation:

Let q' be the Rate of heat loss per unit length

Let q be the Rate of heat loss

q' = q/L

Where L is the length of the pipe

Diameter, D= 0.6m

The rate of heat loss q is given by the formula: q = Sk(T₂ - T₁)

Where k is the thermal conductivity of the concrete at 300 K

k = 1.4 Wb/m-K (at 300K)

And S is the shape factor given by the formula:

S = 2πL/ ln(1.08w/D)

S = (2π*L) / ln(1.08*1.75/0.6))

S = (2π*L) / 1.147

S = 5.48 L

q = 5.48L*1.4(400-300)

q = 767.01 L

q' = q/L

q' = 767.01L/L

q' = 767.01 W/m

4 0

3 years ago