Pipa dengan jari jari $40\,\text{mm}$ dan panjang $4\,\text{m}$ ketebalan $4\,\text{cm}$ dengan koefisien pindah panas $k = 0.744\,\text{W/m K}$ memiliki suhu $20\,\text{C}$ dan $30\,\text{C}$ . tentukan pindah panas yang terjadi!

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Q. Pipa dengan jari jari

40\,\text{mm}

dan panjang

4\,\text{m}

ketebalan

4\,\text{cm}

dengan koefisien pindah panas

k = 0.744\,\text{W/m K}

memiliki suhu

20\,\text{C}

dan

30\,\text{C}

. tentukan pindah panas yang terjadi!

Calculate Surface Area: First, we need to find the surface area of the pipe. The formula for the surface area of a cylinder is $A = 2 \pi r (r + h)$ , where $r$ is the radius and $h$ is the height (length of the pipe). $\newline$ $A = 2 \pi \times 0.04 \, \text{m} \times (0.04 \, \text{m} + 4 \, \text{m})$
Calculate Temperature Difference: Now, calculate the surface area. $\newline$ $A = 2 \times \pi \times 0.04 \times (0.04 + 4)$ $\newline$ $A = 2 \times \pi \times 0.04 \times 4.04$ $\newline$ $A = 0.3216 \times \pi$
Calculate Heat Transfer: Next, we need to calculate the temperature difference ( $\Delta T$ ) between the inside and outside of the pipe. $\newline$ $\Delta T = T_2 - T_1$ $\newline$ $\Delta T = 30 \, \text{C} - 20 \, \text{C}$ $\newline$ $\Delta T = 10 \, \text{C}$
Calculate Heat Transfer: Next, we need to calculate the temperature difference ( $\Delta T$ ) between the inside and outside of the pipe. $\newline$ $\Delta T = T_2 - T_1$ $\newline$ $\Delta T = 30 \, \text{C} - 20 \, \text{C}$ $\newline$ $\Delta T = 10 \, \text{C}$ Now, we use the formula for heat transfer through a cylindrical wall, which is $Q = \frac{k \cdot A \cdot \Delta T}{\ln(\frac{r_2}{r_1})}$ , where $k$ is the thermal conductivity, $A$ is the surface area, $\Delta T$ is the temperature difference, $r_2$ is the outer radius, and $r_1$ is the inner radius. Since the thickness is $\Delta T = T_2 - T_1$ $0$ , the outer radius $r_2$ is $\Delta T = T_2 - T_1$ $2$ . $\newline$ $\Delta T = T_2 - T_1$ $3$
Calculate Heat Transfer: Next, we need to calculate the temperature difference ( $\Delta T$ ) between the inside and outside of the pipe. $\newline$ $\Delta T = T_2 - T_1$ $\newline$ $\Delta T = 30 \, \text{C} - 20 \, \text{C}$ $\newline$ $\Delta T = 10 \, \text{C}$ Now, we use the formula for heat transfer through a cylindrical wall, which is $Q = \frac{k \cdot A \cdot \Delta T}{\ln(\frac{r_2}{r_1})}$ , where $k$ is the thermal conductivity, $A$ is the surface area, $\Delta T$ is the temperature difference, $r_2$ is the outer radius, and $r_1$ is the inner radius. Since the thickness is $\Delta T = T_2 - T_1$ $0$ , the outer radius $r_2$ is $\Delta T = T_2 - T_1$ $2$ . $\newline$ $\Delta T = T_2 - T_1$ $3$ Calculate the heat transfer. $\newline$ $\Delta T = T_2 - T_1$ $4$ $\newline$ $\Delta T = T_2 - T_1$ $5$ $\newline$ $\Delta T = T_2 - T_1$ $6$ $\newline$ $\Delta T = T_2 - T_1$ $7$