Question

In large buildings, hot water in a water tank is circulated through a loop so that the user doesn’t have to wait for all the water in long piping to drain before hot water starts coming out. A certain recirculating loop involves 40-m-long, 1.2-cm-diameter cast iron pipes with six 90° threaded smooth bends and two fully open gate valves. If the average flow velocity through the loop is 2 m/s, determine the required power input for the recirculating pump. Take the average water temperature to be 60°C and the efficiency of the pump to be 76 percent. The density and dynamic viscosity of water at 60°C are rho = 983.3 kg/m3, μ = 0.467 × 10–3 kg/m·s. The roughness of cast iron pipes is 0.00026 m. The loss coefficient is KL = 0.9 for a threaded 90° smooth bend and KL = 0.2 for a fully open gate valve. (Round the final answer to three decimal paces.)

Answer 1

Answer:

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, $K_{L, bend} = 0.9$

Loss coefficient of the valve, $K_{L, valve} = 0.2$

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

Density of water, $\rho = 983.3\ kg/m^{3}$

Roughness of the pipe of cast iron, $\epsilon = 0.00026\ m$

Efficiency of the pump, $\eta = 0.76$

Now,

We calculate the volume flow rate as:

$\dot{V} = Av$

where

$\dot{V}$ = Volume rate flow

A = Area

v = velocity

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

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

$R_{e} = \frac{\rho vd}{\mu}$

$R_{e} = \frac{983.3\times 2\times 0.012}{0.467\times 10^{- 3}} = 50533.62$

Since, $R_{e}$ > 4000, the flow is turbulent in nature.

Now,

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

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

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

f = 0.05075

Now,

To calculate the total head loss:

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

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

Now,

The drop in the pressure can be calculated as:

$\Delat P = \rho g H_{loss}$

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

Now,

to calculate the input power:

$\dot{W} = \frac{\dot{W_{p}}}{\eta}$

$\dot{W} = \frac{\dot{V}\Delta P}{0.76}$

$\dot{W} = \frac{2.262\times 10^{- 4}\times 344.113\times 1000}{0.76} = 0.102\ kW$