Answer:
a)A constant volume process is called isochoric process.
b)Yes
c)Work =0
Explanation:
Isochoric process:
A constant volume process is called isochoric process.
In constant volume process work done on the system or work done by the system will remain zero .Because we know that work done give as
work = PΔV
Where P is pressure and ΔV is the change in volume.
For constant volume process ΔV = 0⇒ Work =0
Yes heat transfer can be take place in isochoric process.Because we know that temperature difference leads to transfer of heat.
Given that
Initial P=10 MPa
Final pressure =15 MPa
Volume = 100 L
Here volume of gas is constant so the work work done will be zero.
Answer:

Solution:
Note: Refer the diagram


Drag coefficient data for selected objects table at
Hemisphere (open end facing flow), 
Substituting all parameters,

Then,
![\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3DV_%7Bw%7D-%5Cleft%5B%5Cfrac%7B2%20F_%7BR%7D%7D%7B%5Crho%5Cleft%28C_%7BD%2C%20w%7D%20A_%7Bw%7D%2BC_%7BD%2C%20B%7D%20A_%7Bb%7D%5Cright%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cdots%5C%5C%26V_%7Bw%7D%3D24%20%5Ctimes%201000%20%5Ctimes%20%5Cfrac%7B1%7D%7B3600%7D%5C%5C%26V_%7Bw%7D%3D6.67%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
And the equation becomes,
![\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3D6.67-%5Cleft%5B%5Cfrac%7B2%20%5Ctimes%205.52%7D%7B1.23%281.42%20%5Ctimes%201.17%2B1.2%20%5Ctimes%200.3%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%26V_%7Bb%7D%3D6.67-2.11%5C%5C%26V_%7Bb%7D%3D4.56%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
Thus the floyds travels at
wind speed.
The friction force f = 10000 N
The heat transfer Q = 1.7936 KJ
<u>Explanation:</u>
Given data:
Surface area of Piston = 1 
Volume of saturated water vapor = 100 K Pa
Steam volume = 0.05 
Using the table of steam at 100 K pa
Steam density = 0.590 Kg/
Specific heat
= 2.0267 KJ/Kg K
Mass of vapor = S × V
m = 0.590 × 0.05
m = 0.0295 Kg
Solution:
a) The friction force is calculated
Friction force = In the given situation, the force need to stuck the piston.
= pressure inside the cylinder × piston area
= 100 ×
× 0.1
f = 10000 N
b) To calculate heat transfer.
Heat transfer = Heat needs drop temperatures 30°C.

Q = 0.0295 × 2.0267 ×
× 30
Q = 1.7936 KJ
Answer:
c
Explanation:
You never want short system terminals
Watts I believe is the answer