Answer:
Detailed solution is given below:
Matthew wants to manufacture a large quantity of products with standardized products having less variety. Which type of producti
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Answer:
Magnitude of force P = 25715.1517 N
Explanation:
Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.
To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Proof -
Given that,
Diameter = 12 mm = 0.012 m
Length = 0.6 m
= 0.015°
Youngs modulus of elasticity of 34 stainless steel is 193 GPa
Now,
By applying the conditions of equilibrium, we have
∑fₓ = 0, ∑ = 0, ∑M = 0
If ∑ = 0
⇒×0.9 - P × 0.6 = 0
⇒×3 - P × 2 = 0
⇒ =
If ∑ = 0
⇒×0.9 = P × 0.3
⇒ ×3 = P
⇒ =
Now,
Area, A = = 1.3097 × 10⁻⁴ m²
We know that,
Change in Length , =
Now,
= 9.1626 × 10⁻⁹ P
= 1.83253 × 10⁻⁸ P
Given that,
= 0.015°
⇒ = 2.618 × 10⁻⁴ rad
So,
⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9
⇒P = 25715.1517 N
∴ we get
Magnitude of force P = 25715.1517 N
Answer:
TBC thickness of 4 mm is insufficient to prevent fire hazard
Explanation:
Given:
- Temperature of hot-fluid inner surface T_i = 333°C
- The convection coefficient hot-fluid h_i = 7 W/m^2K
- The thermal conductivity of engine cover k_1 = 14 W/mK
- The thickness of engine cover L_1 = 0.01 m
- The thermal conductivity of TBC layer k_2 = 1.1 W/mK ... (Typing error)
- The thickness of TBC layer L_2 = 0.004 m
- Temperature of ambient air outer surface T_o = 69°C
- The convection coefficient ambient air h_o = 7 W/m^2K
Find:
Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?
Solution:
- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.
The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.
- We will calculate the total heat flux for the entire system q:
q = ( T_i - T_o ) / R_total
- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:
R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o
- Plug in the values and compute:
R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7
R_total = 0.2900649351 T-m^2 / W
- Calculate the Total heat flux q:
q = ( 333 - 69 ) / 0.2900649351
q = 910.141 W / m^2
- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:
q = ( T_i - T_2 ) / R_i2
Where,
R_i2 = 1 / h_i + L_1 / k_1
R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W
Hence,
( T_i - T_2 ) = q*R_i2
T_2 = T_i - q*R_i2
Plug the values in:
T_2 = 333 - 910.141*0.14357
T_2 = 202.33°C
- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard
