Answer:
well you could get some green goblin it disolves all the c rap in sink
Explanation:
Answer:
h = 375 KW/m^2K
Explanation:
Given:
Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm
steel thermal conductivity k = 15 W / mK
Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C
Air Temp T_∞ = 100 C
Assuming there are no other energy sources, energy balance equation is:
E_in = E_out
q"_cond = q"_conv
Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2
q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)
=15KW/m^2
Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:
q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )
h = 15000 W / (100 - 60 ) C = 375 KW/m^2K
Answer:
maneuverability
Explanation:
needless to say, I took the quiz
Answer:
23.3808 kW
20.7088 kW
Explanation:
ρ = Density of oil = 800 kg/m³
P₁ = Initial Pressure = 0.6 bar
P₂ = Final Pressure = 1.4 bar
Q = Volumetric flow rate = 0.2 m³/s
A₁ = Area of inlet = 0.06 m²
A₂ = Area of outlet = 0.03 m²
Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s
Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s
Height between inlet and outlet = z₂ - z₁ = 3m
Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation
Work done by pump
∴ Power input to the pump 23.3808 kW
Now neglecting kinetic energy
Work done by pump
∴ Power input to the pump 20.7088 kW
