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A commonly used strategy to improve the efficiency of a power plant is to withdraw steam from the turbine before it fully expand
s, reheat it in the boiler and send it to another turbine. Consider a case where 1 kg/s steam enters a turbine at 5MPa, 500 °C, and is expanded to 1 MPa pressure. At this point, the steam is withdrawn from the turbine, reheated back to 500 °C at constant pressure, and then expanded in another turbine to 50 kPa pressure. Calculate (a) The temperature at which steam was sent back for reheating, (b) Heat added during the reheati
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Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.
<h3>What is clockwise torque?</h3>The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.
A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.
Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.
The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.
When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.
You can calculate the torque's magnitude using
To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.
Clockwise torque due to 100 g ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm
Clockwise torque due to 200 g ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm
Clockwise torque due to stick mass ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm
Counter-clockwise torque due to normal force ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm
Learn more about torque
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Answer:
This question is comprising many parts (a to r). That is impossible to answer in one sheet. Following are attached images having answers to most of the parts.
I hope it will help you a lot.
Explanation:
Answer:
the pressure gradient in the x direction = -15.48Pa/m
Explanation:
- The concept of partial differentiation was used in the determination of the expression for u and v.
- each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.
