Answer:
Work = - 4175.23 J
Heat Transfer = -4175.23 J
Explanation:
The work done in an isothermal process, is given by the following formula:
W = RT ln (P1/P2)
where,
W = Work done
R = Universal Gas Constant = 8.314 J/mol.k
T = Constant Temperature = 300 K
P1 = initial pressure = 150 KPa
P2 = final pressure = 800 KPa
using these values, we get:
W = (8.314 J/mol.K)(300 k) ln (150/800)
<u>W = - 4175.23 J</u>
Here, negative sign shows that work is done on the system.
In isothermal process, from 1st law of thermodynamics:
Heat Transfer = Q = W (Since change in internal energy is zero for isothermal processes)
<u>W = - 4175.23 J</u>
Here, negative sign shows that heat is transferred from the system to surrounding.
Answer:
Process capability index = (USL - LSL)/6Sigma = 1/6(0.3)= 0.56
Answer (a) 0.56
Explanation:
The net resultant direct force and angle on the vane is created when the water jet exits the vane at position 2 with 92% of its initial velocity.
<h3>What is mean by velocity?</h3>
- The speed at which a body or object is moving determines its direction of motion. A scalar quantity, speed is primarily. As a matter of fact, velocity is a vector quantity.
- The rate at which distance changes is what it is. It measures the displacement's rate of change. A body's velocity is defined as its speed in a particular direction.
- Velocity is a measure of how quickly a distance changes in relation to time. Having both magnitude and direction, velocity is a vector quantity.
- The rate of change in a body's displacement with respect to time is referred to as velocity. In the SI, m/s is its unit.
Given,
External angle of Curved Vane = 158°
mean velocity at 1 = 12 m/s
Volumetric flow rate = 
mean velocity at 
i) Force exerted in x - friction A C 1 = Volume
![F_{S_x} &=\rho A C_1\left[C_2 \cos \theta-C_1\right] \\](https://tex.z-dn.net/?f=F_%7BS_x%7D%20%26%3D%5Crho%20A%20C_1%5Cleft%5BC_2%20%5Ccos%20%5Ctheta-C_1%5Cright%5D%20%5C%5C)
![&=1000 \times \frac{55}{3600}\left[\left(11.04 \cos 158^{\circ}\right)-12\right]](https://tex.z-dn.net/?f=%26%3D1000%20%5Ctimes%20%5Cfrac%7B55%7D%7B3600%7D%5Cleft%5B%5Cleft%2811.04%20%5Ccos%20158%5E%7B%5Ccirc%7D%5Cright%29-12%5Cright%5D)
i![\rangle F_{\text {sc }}=\supseteq A c_1\left[C_2 \sin \theta\right] \\](https://tex.z-dn.net/?f=%5Crangle%20F_%7B%5Ctext%20%7Bsc%20%7D%7D%3D%5Csupseteq%20A%20c_1%5Cleft%5BC_2%20%5Csin%20%5Ctheta%5Cright%5D%20%5C%5C)






The complete question is:
A horizontal jet of water strikes a curved vane as shown in Figure C.1. The external angle of the curved vane is 158°.The mean velocity and volumetric flow rate of the water jet at position 1 are 12 m/s and 55 m³/h respectively. Due to friction, the water jet leaves the vane at position 2 with 92 % its original velocity.
(i) Direct force exerted by the water jet on the vane in the x - direction.
(ii) Direct force exerted by the water jet on the vane in the y - direction.
(ii) Net resultant direct force and angle on the vane.
To learn more about velocity, refer to:
brainly.com/question/24681896
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Answer:
F=1.47 KN
Explanation:
Given that
Diameter of plate = 25 cm
Height of pool h = 3 m
We know that force can be given as
F= P x A
P=ρ x g x h
Now by putting the values
P=1000 x 10 x 3
P= 30 KPa


F= 30 x 0.049 KN
F=1.47 KN
So the force on the plate will be 1.47 KN.