Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
2.5 is the required details
Answer:
0.023 Pa*s
Explanation:
The surface area of the side of the inner cylinder is:
A = π*d*l
A = π*0.15*0.75 = 0.35 m^2
At 200 rpm the inner cylinder has a tangential speed of:
u = w * r
u = w * d/2
w = 200 rpm * 2π / 60 = 20.9 rad/s
u = 20.9 * 0.15 / 2 = 1.57 m/s
The torque is of 0.8 N*m, this means that the force is:
T = F * r
F = T / r
F = 2*T / d
For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:
F = μ*A*u / y
Where
μ: viscocity
y: separation between pates
A: surface area of the plates
Then:
2*T / d = μ*A*u/y
Rearranging:
μ = 2*T*y / (d*A*u)
μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s
Answer:
Explanation:
The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.
The velocity of water delivered by the fire hose is:
The maximum height is cleared in the Bernoulli's equation:
