Answer:
D
Explanation:
To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete. You have to multiply them to get the total cost of the project.
![\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D%26Cost%5C%20per%5C%20day%5C%20%28%5C%24%29%26Time%5C%20to%5C%20complete%5C%20%28days%29%26Total%5C%20cost%5C%20%28%5C%24%29%5C%5CZoe%26500%268%264000%5C%5CGreg%26650%2610%266500%5C%5COrion%26400%2612%264800%5C%5CJin%26700%265%263500%5Cend%7Barray%7D%5Cright%5D)
As you can see, Greg is the least cost-effective because he charges the most for the project.
Answer:

Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ =
= 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than
and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

Now, we will find
(enthalpy at the outlet for the isentropic process):

Now, the isentropic efficiency of the turbine can be given as follows:

It’s in Wolfsburg Germany
Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as

Putting all the value to get the velocity of the flow


V = 0.5 m/s