Answer:
The circulation around the cylinder is 0.163 
Explanation:
Given :
Velocity of spinning cylinder

Sea level density

Sea level span

Lift per unit circulation is given by,

Where
circulation around cylinder



Therefore, the circulation around the cylinder is 0.163 
Answer:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 -
............1
efficiency ( reversible ) η1 = 1 -
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency =
second-law efficiency =
second-law efficiency = 62.42 %
Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.
One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.
Answer:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm
Required:
•). The magnitude of current= ?
•••••••••••••SOLUTION•••••••••••••
We can find the relation ship between current, voltage and resistance with the help of Ohms law.
According to ohms law;
V= IR.
Rearranging the above equation;
I= V/ R
Putt the values in the above equation; we get
I= 15V/ 1000ohm
I = 0.015 A( ampere)
••••••••••••••• CONCLUSION•••••••
The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.