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:
Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
<u>Determine the design flow ins ft^3/s </u>
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
Answer:
Explanation:C7H6O2 + (15/2) O2 = > 3H2O + 7CO2 delH = Σ stoichiometric coefficient* enthalpy of formation delH = (7*-393.509 kJ/mol) + (3*-285.83 kJ/mol) - (15/2 * 0) -(1*-385.2 kJ/mol) delH = -3226.853 kJ/mol benzoic acid ...
Answer:
work done = 665.12 kJ/k.mol of nitrogen
dQ = 685.905 kJ/k.mol of nitrogen
Explanation:
given data
pressure = 80 bar
temperature t1 = 220 K
temperature t2 = 300 K
solution
first we get here work done as considering ideal gas condition
work done = P (v2-v1 ) = nR (t2-t1)
put here value
work done = 1 × 8.314 ×( 300 - 220 )
work done = 665.12 kJ/k.mol of nitrogen
and
now we get heat transfer by 1st law of thermodynamics that is
heat transfer dQ = dv + dw
dQ = Cv dT + dw
put here value and we get
dQ =
× (t2-t1) + 665.12
dQ =
× (300-220) + 665.12
dQ = 685.905 kJ/k.mol of nitrogen