Respuesta :
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.
Explanation:
Given data,
[tex]T_{\infty}[/tex] = 10° C
[tex]h_{0}[/tex] = 250 w/ [tex]m^{2}[/tex] k
Pipe length = 20 m
Inner diameter [tex]d_{1}[/tex] = 6 cm, [tex]r_{1}[/tex] = 3 cm
Outer diameter [tex]d_{2}[/tex] = 8 cm, [tex]r_{2}[/tex] = 4 cm
The thickness of insulation is 4 cm.
[tex]r_{3}[/tex] = [tex]r_{2}[/tex] + 4
= 4+4
[tex]r_{3}[/tex] = 8 cm
[tex]h_{0}[/tex] is the heat transfer coefficient of convection inside, [tex]h_{i}[/tex] is the heat transfer coefficient of convection outside.
The heat transfer rate between ambient and steam is
[tex]q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}[/tex] watt
= [tex]\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}[/tex] watt
= [tex]\frac{190}{0.0003317+0.0000458+0.0110+0.0004976}[/tex] 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 = [tex]\frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}[/tex]
15999.86 [tex]=\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}[/tex]
[tex]T_{3}[/tex] - 10 = 7.96
[tex]T_{3}[/tex] = 17.96 ° C.
In this exercise we have to use the knowledge of temperature in this way we can calculate the heat transferred per day, in this way we find that:
A)The total heat transfer is 1382.38 M w.
B) The temperature is 17.96 ° C.
So from the information given in the statement we find that:
- [tex]T[/tex] = 10° C
- [tex]h_0[/tex]= 250 w/ k
- Pipe length = 20 m
- Inner diameter [tex]d_1[/tex]= 6 cm, [tex]r_1[/tex]= 3 cm
- Outer diameter [tex]d_2[/tex]= 8 cm, [tex]r_2[/tex] = 4 cm
- The thickness of insulation is 4 cm.
- [tex]r_3=8 cm[/tex]
A) The heat transfer coefficient of convection inside, is the heat transfer coefficient of convection outside. The heat transfer can be calculeted as:
[tex]q= \frac{1}{800(2\pi x* 3*20}+\frac{ln(4/3)}{2\pi*50*20}+\frac{ln(8/4)}{2\pi*0.5*20}+\frac{1}{200(2*\pi x*08*20)} \\ q=15999.86 Watts\\ h_0=1382.38Mw[/tex]
b) The coldness of some degree outside surface of the thick plaster the act of insulating:
[tex]T_3[/tex]- 10 = 7.96
[tex]T_3[/tex]= 17.96 ° C.
See more about temperature at brainly.com/question/7510619
