Answer:
The answer to this question is 1273885.3 ∅
Explanation:
<em>The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.</em>
<em>Given that,</em>
<em>The distance = 50mm</em>
<em>The time t =10 seconds</em>
<em>The force F = 10kN</em>
<em>The piston diameter is = 100mm</em>
<em>The pressure = F/A</em>
<em> 10 * 10^3/Δ/Δ </em>
<em> P = 1273885.3503 pa</em>
<em>Then</em>
<em>Power = work/time = Force * distance /time</em>
<em> = 10 * 1000 * 0.050/10</em>
<em>which is =50 watt</em>
<em>Power =∅ΔP</em>
<em>50 = 1273885.3 ∅</em>
Answer:
The Poisson's Ratio of the bar is 0.247
Explanation:
The Poisson's ratio is got by using the formula
Lateral strain / longitudinal strain
Lateral strain = elongation / original width (since we are given the change in width as a result of compession)
Lateral strain = 0.15mm / 40 mm =0.00375
Please note that strain is a dimensionless quantity, hence it has no unit.
The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.
Longitudinal strain = 4.1 mm / 270 mm = 0.015185
Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247
The Poisson's Ratio of the bar is 0.247
Please note also that this quantity also does not have a dimension
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.