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

a)
`P=73225\ W`

b)
`v=25.4647\ cm.s^(-1)`

Step-by-step 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* 4186* (55-20)`

`\dot{Q}=4395300\ J.min^(-1)`

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

`P=(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 (d^2)/(4)`

`a=\pi* (5^2)/(4)`

`a=19.635\  cm^2`

∴Flow velocity

`v=\dot{V}/ a`

`v=30000/19.635`

`v=1527.884\ cm.min^(-1)`

`v=25.4647\ cm.s^(-1)`