As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .
also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)
Mechanical and electrical
Answer:
The work flow required by the compressor = 100.67Kj/kg
Explanation:
The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .
The work flow can be determined using the equation:
M1h1 + W = Mh2
U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2
Workflow = P2alpha2 - P1alpha1
Workflow = (h2 -U2) - (h1 - U1)
Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)
Workflow = ( 193.191 - 92.519)Kj/kg
Workflow = 100.672Kj/kg
Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A =
= 6.77 ×
m²
Now Area of cylinder is :
A =
d²
solving for d:
d = 
d = 9.28 cm