Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress 
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress 
Making thickness t the subject; we have
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
Answer:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine (
), no unit, is defined by this formula:
(1)
Where:
- Heat input, in watts.
- Power output, in watts.
If we know that 
and 
, then the heat input from the combustion phase is:
The heat input from the combustion phase is 2000 watts.
Answer:
a) 0.3
b) 3.6 mm
Explanation:
Given
Length of the pads, l = 200 mm = 0.2 m
Width of the pads, b = 150 mm = 0.15 m
Thickness of the pads, t = 12 mm = 0.012 m
Force on the rubber, P = 15 kN
Shear modulus on the rubber, G = 830 GPa
The average shear strain can be gotten by
τ(average) = (P/2) / bl
τ(average) = (15/2) / (0.15 * 0.2)
τ(average) = 7.5 / 0.03
τ(average) = 250 kPa
γ(average) = τ(average) / G
γ(average) = 250 kPa / 830 kPa
γ(average) = 0.3
horizontal displacement,
δ = γ(average) * t
δ = 0.3 * 12
δ = 3.6 mm
Answer:
as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.
Explanation:
4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum
Answer:
Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.
Explanation:
Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that 
represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:
- Wall Thickness. More thickness, more thermal resistance.
- Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
- Cross-section Area. More cross-section area, less thermal resistance.
A expression to define the thermal resistance for the wall is as follows: 
, where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.
