Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( 
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
i want coins sorry use a calculator or sum
Explanation:
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Answer:
The young lady was his daughter.The shoemaker was frightened when he saw that she wants to sit near him and took his knife to frighten her and leave him alone to do his work
Explanation:
could uh name them since if i know any i would surely tryin help
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=
=500×
×
=2×
=2 pF
