A coil with an average diameter of 5 inch will have an area of ""blank"" square meters
1 answer:
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Answer:
The thickness of the material is 6.23 cm
Explanation:
Given;
quantity of heat, Q = 6706.8 *10⁶ kcal
duration of the heat transfer, t = 5 months
thermal conductivity of copper, k = 385 W/mk
outside temperature of the heater, T₁ = 30° C
inside temperature of the heater, T₂ = 50° C
dimension of the rectangular heater = 450 cm by 384 cm
1 kcal = 1.163000 Watt-hour
6706.8 *10⁶ kcal = 7800008400 watt-hour
I month = 730 hours
5 months = 3650 hours
Rate of heat transfer, P =
Rate of heat transfer,
where;
P is the rate of heat transfer (W)
k si the thermal conductivity (W/mk)
ΔT is change in temperature (K)
A is area of the heater (m²)
L is thickness of the heater (m)
L = 6.23 cm
Therefore, the thickness of the material is 6.23 cm
Answer:
i) SF:
ii) BM :
Explanation:
Let's take,
Making y the subject of formula, we have :
For shear force (SF), we have:
This is the area of the diagram.
The shear force equation =
For bending moment (BM):
The bending moment equation =