Question

It is proposed to have a water heater that consists of an insulated pipe of 5-cm diameter and an electric resistor inside. Cold water at 20°C enters the heating section steadily at a rate of 30 L/min. If water is to be heated to 55°C, determine (a) the power rating of the resistance heater (b) the average velocity of the water in the pipe.

Answer 1

Answer:

a) $P=73225\ W$

b) $v=25.4647\ cm.s^{-1}$

Explanation:

Given:

• initial temperature of water, $T_i=20^{\circ}C$

• volume flow rate of water in to the heating section, $\dot{V}=30\ L.min^{-1}=0.03\ m^3.min^{-1}$

∵1L of water = 1 kg by mass

∴$\dot{m}= 30\ kg.min^{-1}$

• final temperature of water, $T_f=55^{\circ}C$

• diameter of insulated pipe, $d=0.05\ m$

(a)

we have,

Specific heat capacity of water, $c=4186\ J.kg^{-1}. ^{\circ}C^{-1}$

The amount of heat to be supplied per min to the water:

$\dot{Q}=\dot{m}.c.\Delta T$

$\dot{Q}=30\times 4186\times (55-20)$

$\dot{Q}=4395300\ J.min^{-1}$

$\therefore P=\frac{\dot{Q}}{60} \ watts$

$P=\frac{4395300}{60}$

$P=73225\ W$

∴Power required is 73.225 kW.

(b)

∵Volume flow rate, $\dot{V}=0.03\ m^3.min^{-1}=30000\ cm.min^{-1}$

Now area of pipe:

$a=\pi \frac{d^2}{4}$

$a=\pi\times \frac{5^2}{4}$

$a=19.635\  cm^2$

∴Flow velocity

$v=\dot{V}\div a$

$v=30000\div19.635$

$v=1527.884\ cm.min^{-1}$

$v=25.4647\ cm.s^{-1}$