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777dan777 [17]
4 years ago
7

A 35-pound force is applied to a 3-inch cylinder with a 6-inch stroke. What is the power produced by the cylinder that moves its

full stroke in 0.7 s?
Engineering
1 answer:
vampirchik [111]4 years ago
7 0

Explanation:

Power = work / time

Power = force × distance / time

P = (35 lbf) (6 in) / (0.7 s)

P = 300 lbf in/s

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What word is typically written at the bottom of a cover letter to indicate an
Licemer1 [7]

Answer:

i my have been taught differntly but it may be enclosure

Explanation:

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3 years ago
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In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approxim
diamong [38]

Answer:

Explanation:

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:

[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )

where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.

3 0
4 years ago
A piston–cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m3. The air undergoes a process to a state
Harman [31]

Work, W = 277.269kJ

Internal energy, Q = 277.269kJ

<u>Explanation:</u>

Given-

Pressure, P1 = 2 bar

Temperature, T1 = 300K

Volume, V1 = 2m³

P2 = 1 bar

PV = constant

Let,

mass in kg be m

Work in kJ be W

Heat transfer in kJ be Q

R' = 8.314 kJ/kmolK

Mass of air, Mair = 28.97 kg/kmol

R = 0.289 kJ/kgK

We know,

PV = mRT

m = \frac{P_1V_1}{RT_1}

m = 5.65kg

To calculate V₂:

PV = constant = P₁V₁ = P₂V₂

P₁V₁ = P₂V₂

V_2 = \frac{P_1V_1}{P_2}

V₂ = 4m³

To calculate the work:

P₁V₁ = C

P₁ = C/ V₁

W = \int\limits^V_V {pdV} \,

where limit is V₁ to V₂

W =  \int\limits^V_V {\frac{c}{v} } \, dV \\\\W = C\int\limits^V_V {v^-^1} \, dV\\ \\W = P_1V_1 (ln\frac{V_2}{V_1} ) \\\\W = (2 bar) (2m^3) (ln\frac{4m^3}{2m^3}) (\frac{10^5 N/m^2}{1 bar}) \\\\W = 277.259kJ

To calculate heat transfer:

Q - W = Δu

Q - W = m (u₂ - u₁)

Q = W + m (u₂ - u₁)

Q = W + m X cv X (T₂ - T₁)

Since, T₁ ≈ T₂

There is no change of internal energy.

W = Q

Q = 277.269kJ

6 0
3 years ago
A heat engine does 210 J of work per cycle while exhausting 440 J of waste heat. Part A What is the engine's thermal efficiency?
Oxana [17]

Answer:

The engine's thermal efficiency is 0.32

Explanation:

Thermal efficiency = work done ÷ quantity of heat supplied

Work done = 210 J

Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J

Thermal efficiency = 210 ÷ 650 = 0.32

6 0
3 years ago
There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time
Aleksandr-060686 [28]

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

4 0
3 years ago
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