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Goshia
1 month ago
6

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

.0 m). The pipe is insulated with a 4-cm thick layer of gypsum plaster (k=0.5 W/mK). The insulated pipe is placed horizontally inside a warehouse where the average air temperature is 10 C. The steam and the air heat transfer coefficients are estimated to be 800 and 200 W/m^2 K, respectively. Calculate (a) the daily rate of heat transfer from the superheated steam, and (b) the temperature on the outside surface of the gypsum plaster insulation.
Engineering
1 answer:
iogann1982 [368]1 month ago
4 0

The daily heat transfer totals 1382.38 M w.

The temperature on the outer layer of the gypsum plaster insulation registers at 17.96 ° C.

Explanation:

Given data,

= 10° CT_{\infty}

= 250 w/

kh_{0}m^{2}Pipe length = 20 m

Inner diameter

= 6 cm,

= 3 cmd_{1}Outer diameter r_{1} = 8 cm,

= 4 cm

The insulation thickness measures 4 cm.d_{2}r_{2}

=

+ 4

= 4+4r_{3}r_{2} = 8 cm

refers to the heat transfer coefficient of  convection internally, and

refers to the heat transfer coefficient of convection externally.

r_{3}The rate of heat transfer between the ambient environment and steam is

h_{0} watth_{i} =

watt

=

watt

q = 15999.86 wattq=\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)}}

The total daily heat transfer = 15999.86 × 86400\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}

= 1382.387904 watt

\frac{190}{0.0003317+0.0000458+0.0110+0.0004976} = 1382.38 M w

The final temperature on the outer surface of the gypsum plaster insulation is found by

q = \frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}

15999.86   =\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi. 08 \times 20)}}

T_{3} - 10 = 7.96

T_{3} = 17.96 ° C.

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Answer:

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#include <iostream>

using namespace std;

int main()

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