Chapter 9, Problem 1P

To determine

Find the time required to increase the oil temperature from $20°\text{C}$ to $100°\text{C}$.

Answer to Problem 1P

The time required to increase the oil temperature from $20°\text{C}$ to $100°\text{C}$ is $\underset{¯}{14.7\text{sec}}$.

Explanation of Solution

Given information:

The width of the parabolic trough $\left(l\right)$ is $1.0\text{m}$.

The insider diameter of the pipe $\left(d\right)$ is $1\text{cm}$.

Initial temperature $\left(T_1\right)$ is $20°\text{C}$  and the final temperature $\left(T_2\right)$ is $100°\text{C}$.

Calculation:

Take the mid day irradiance of incident radiation as $\text{674}{\text{W/m}}^{\text{2}}$ and incident power of the 1 m long parabolic trough $\left(P\right)$ as 674 W.

The inside diameter of the pipe is 1 cm.

Find the volume of oil in a 1 m pipe section $\left(V\right)$.

$\begin{array}{c}V=\frac{\pi }{4}{d}^2\times l\\ =\frac{\pi }{4}{\left(1\text{cm}\right)}^2\times \left(1\text{m}\right)\\ =\frac{\pi }{4}{\left(1\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\times \left(1\text{m}\right)\\ =7.85\times {10}^{-5}{\text{m}}^{\text{3}}\end{array}$

Take the density of oil $\left(\rho \right)$ as $\text{900}{\text{kg/m}}^{\text{3}}$ and the value of specific heat $\left(C\right)$ as $1,758\text{J/}\left(\text{kg}\cdot \text{°C}\right)$.

Express the formula for energy $\left(Q\right)$ as follows:

$Q=mC\Delta T$                                                                                                                   …... (1)

Here, the mass is $m$ and the change in temperature is $\Delta T$.

Write the expression for power $\left(P\right)$ as follows:

$\begin{array}{l}P=\frac{Q}{t}\\ Q=Pt\end{array}$

Here, $t$ is the time required to raise the oil temperature.

The mass is the product of density and the volume. Therefore, $m=\rho V$.

Substitute $Pt$ for $Q$ and $\rho V$ for $m$ in equation (1).

$\begin{array}{l}Pt=\rho VC\Delta T\\ t=\frac{\rho VC\left(T_2-T_1\right)}{P}\\ t=\frac{\text{900}{\text{kg/m}}^{\text{3}}\times 7.85\times {10}^{-5}{\text{m}}^{\text{3}}\times 1,758\text{J/}\left(\text{kg}\cdot \text{°C}\right)\times \left(100°\text{C}-20°\text{C}\right)}{674\text{W}}\end{array}$

$t=14.7\text{sec}$

Therefore, the time required to raise the oil temperature is $\underset{¯}{14.7\text{sec}}$.