Answer:
The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s
Explanation:
Q = MCp(T2 - T1)
Q (quantity of heat) = Power (P) × time (t)
Density (D) = Mass (M)/Volume (V)
M = DV
Therefore, Pt = DVCp(T2 - T1)
V/t (volume flow rate) = P/DCp(T2 - T1)
P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K
Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)
The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s
Answer:
Technician B is correct.
Explanation:
In every vehicles, coolants are very important as it keeps the engine cool and keep it running. The coolant serves the purpose of dissipating away the heat that is generated when the engine runs.
When the engine runs, the coolant passes all over the heating surfaces in the engine thus carrying away the heat along with it. With time the color of the coolant changes to rusty brown color when the coolant serves its purpose . It also washes away all the dirt particles from the engine part along with it. Thus the color changes and it is now can be flushed from the engine sump and fresh coolant can be put in to the engine.
Thus in the context, technician B is correct.
Answer:
2 yes ot will 2 would be a yes but i dont know how i would put that into a paragraph
Answer:
T=138 °C
Explanation:
Given that
m = 0.028 kg
Net work output W= 60 KJ
T₂=2T₁
As we know that efficiency of Carnot heat engine given as


η = 0.5
We know that

Qa=heat addition
W= net work output


Qa= 120 KJ
From first law
Qa= W+ Qr
Qr= 120 - 60
Qr= 60 KJ
Qr Is the heat rejection.
Heat rejection per unit mass
Qr=60 / 0.028 = 2142.85 KJ/kg
Qr= 2142.85 KJ/kg
Temperature at which latent heat of steam is 2142.85 KJ/kg will be our answer.
T=138 °C
The temperature corresponding to 2142.85 KJ/kg will be 138 °C.
T=138 °C
Answer:
25120 N
Explanation:
The external load acting on each bolted joint
= P x A / N
= (8 x 10⁶) (31.4 x 10³ x 10⁻⁶) / 10
= 25120 N