Answer:

Explanation:
Given data:
P_1 power = 20 dBm = 0.1 watt
coupling factor is 20dB
Directivity = 35 dB
We know that
coupling factor 
solving for final power




Directivity 


output Power 


The total amount of daily heat transfer is 1382.38 M w.
The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.
<u>Explanation:</u>
Given data,
= 10° C
= 250 w/
k
Pipe length = 20 m
Inner diameter
= 6 cm,
= 3 cm
Outer diameter
= 8 cm,
= 4 cm
The thickness of insulation is 4 cm.
=
+ 4
= 4+4
= 8 cm
is the heat transfer coefficient of convection inside,
is the heat transfer coefficient of convection outside.
The heat transfer rate between ambient and steam is
watt
=
watt
=
watt
q = 15999.86 watt
The total amount of daily heat transfer = 15999.86 × 86400
= 1382.387904 watt
= 1382.38 M w
The total amount of daily heat transfer is 1382.38 M w.
b) The temperature on the outside surface of the gypsum plaster insulation.
q = 
15999.86 
- 10 = 7.96
= 17.96 ° C.
Answer:
Explanation:
A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq
˙
q
q
˙
, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT
∞
=20
∘
C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx
2
where a=82.0∘C,b=−210∘C/m,c=−2×104C/m2a=82.0
∘
C,b=−210
∘
C/m,c=−2×10
4
C/m
2
, and x is in meters. The origin of the x-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation q in the wall? (c) Determine the surface heat fluxes, q
′′
x
(−L)q
x
′′
(−L) and q
′′
x
(+L)q
x
′′
(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q
′′
x
(x)q
x
′′
(x). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated (q=0), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with q=0? How much energy must be removed by the fluid per unit area of the wall (J/m2)(J/m
2
) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m
3
and 800J/kg⋅K800J/kg⋅K, respectively.
Answer:
1.505
Explanation:
cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.
stress is force per unit area
stress=P/A
A = πd^2/4.
uncertainty of axial force P= +/-.11
s=+/-.20, strength
d=+/-.04 diameter
fail load/max allowed
minimum design=fail load/max allowed
minimum design =s/(P/A)
sA/P
A=(
.96d^2)/4, so Amin=
(because the diameter at minimum is (1-0.04=0.96)
minimum design=Pmax/(sminxAmin)
1.11/(.80*.96^2)=
1.505