# In large buildings, hot water in a water tank is circulated through a loop so that the user doesn’t

###### Question:

## Answers

The power input is 0.102 kW

Solution:

As per the question:

Length of the loop, L = 40 m

Diameter of the loop, d = 1.2 cm

Velocity, v = 2 m/s

Loss coefficient of the threaded bends, [tex]K_{L, bend} = 0.9[/tex]

Loss coefficient of the valve, [tex]K_{L, valve} = 0.2[/tex]

Dynamic viscosity of water, [tex]\mu = 0.467\times 10^{- 3}\ kg/m.s[/tex]

Density of water, [tex]\rho = 983.3\ kg/m^{3}[/tex]

Roughness of the pipe of cast iron, [tex]\epsilon = 0.00026\ m[/tex]

Efficiency of the pump, [tex]\eta = 0.76[/tex]

Now,

We calculate the volume flow rate as:

[tex]\dot{V} = Av[/tex]

where

[tex]\dot{V}[/tex] = Volume rate flow

A = Area

v = velocity

[tex]\dot{V} = \frac{\pi}{4}d^{2}\times 2 = 2.262\times 10^{- 4}\ m^{3}/s[/tex]

For this, Reynold's N. is given by:

[tex]R_{e} = \frac{\rho vd}{\mu}[/tex]

[tex]R_{e} = \frac{983.3\times 2\times 0.012}{0.467\times 10^{- 3}} = 50533.62[/tex]

Since, [tex]R_{e}[/tex] > 4000, the flow is turbulent in nature.

Now,

With the help of the Colebrook eqn, we calculate the friction factor as:

[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{\epsilon}{d}}{3.7} + \frac{2.51}{R_{e}\sqrt{f}}][/tex]

[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{0.00026}{0.012}}{3.7} + \frac{2.51}{50533.62\sqrt{f}}][/tex]

f = 0.05075

Now,

To calculate the total head loss:

[tex]H_{loss} = (\frac{fL}{d} + 6K_{L, bend} + 2K_{L, valve})\farc{v^{2}}{2g}[/tex]

[tex]H_{loss} = (\frac{0.05075\times 40}{0.012} + 6\times 0.9 + 2\times 0.2)\farc{2^{2}}{2\times 9.8} = 35.71\ m[/tex]

Now,

The drop in the pressure can be calculated as:

[tex]\Delat P = \rho g H_{loss}[/tex]

[tex]\Delat P = 983.3\times 9.8\times 35.71 = 344.113\ kN/m^{2}[/tex]

Now,

to calculate the input power:

[tex]\dot{W} = \frac{\dot{W_{p}}}{\eta}[/tex]

[tex]\dot{W} = \frac{\dot{V}\Delta P}{0.76}[/tex]

[tex]\dot{W} = \frac{2.262\times 10^{- 4}\times 344.113\times 1000}{0.76} = 0.102\ kW[/tex]

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

Amethod of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.