Question

Water at 158C (r 5 999.1 kg/m3 and m 5 1.138 3 1023 kg/m·s) is flowing steadily in a 30-m-long and 5-cm-diameter horizontal pipe made of stainless steel at a rate of 9 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

(a) the pressure drop is
`\Delta P_L = 100.185\ kPa`

(b) the head loss is the
`h_L =10.22\ m`

(c) the pumping power requirement to overcome this pressure drop
`W_p = 901.665\ W`

Step-by-step explanation:

So the question we are told that the water is at a temperature of 15°C

And also we are told that the density of the water is
`\rho=999.1\ kg/m^3`

We are also told that the dynamic viscosity of the water is
`\mu = 1.138 *10^(-3) kg/m \cdot s`

From the diagram the length of the pipe is
`L=` 30 m

The diameter is given as
`D=` 5 cm
`(5)/(100) m = 0.05m`

The volumetric flow rate is given as
`Q=` 9 L/s
`= (9)/(1000) m^3/ s = 0.009\ m^3/s`

Now the objective of this solution is to obtain

i the pressure drop, ii the head loss iii the pumping power requirement to overcome this pressure drop

to obtain this we need to get the cross-sectional area of the pipe which will help when looking at the flow analysis

So mathematically the cross-sectional area is

`A_s =(\pi)/(4) D^2`

`= (\pi)/(4) (5*10^(-2))`

`=1.963*10^(-3) m^2`

Next thing to do is to obtain average flow velocity which is mathematically represented as

`v = (Q)/(A_s)`

`v =(9*10^(-3))/(1.963*10^(-3))`

`=4.585 \ m/s`

Now to determine the type of flow we have i.e to know whether it a laminar flow , a turbulent flow or an intermediary flow

We use the Reynolds number

if it is below
`4000` then it is a laminar flow but if it is higher then it is a turbulent flow ,now when it is exactly the value then it is an intermediary flow

This Reynolds number is mathematically represented as

`Re = (\rho vD)/(\mu)`

`=((999.1)(4.585)(0.05))/(1.138*10^(-3))`

`=2.0127*10^5`

Now since this number is greater than 4000 the flow is turbulent

So we are going to be analyse the flow using the Colebrook's equation which is mathematically represented as

`(1)/(√(f) ) =-2.0\ log [(\epsilon/D_h)/(3.7) +(2.51)/(Re√(f) ) ]`

Where f is the friction factor ,
`\epsilon` is the surface roughness ,

Now generally the surface roughness for stainless steel is

`\epsilon = 0.002 mm = 2*10^(-6) m`

Now substituting the values into the equation we have

`(1)/(√(f) ) =-2.0\ log [(2*10^(-6)/5*10^(-2))/(3.7) +(2.51)/((2.0127*10^(5))√(f) ) ]`

So solving to obtain f we have

`(1)/(√(f) ) =-2.0\ log [(4*10^(-5))/(3.7) +(2.51)/((2.0127*10^(5))√(f) ) ]`

`= -2.0log[1.0811 *10^(-5) +(1.2421*10^(-5))/(√(f) ) ]`

`f = 0.0159`

Generally the pressure drop is mathematically represented as

`\Delta P_L =f\ (L)/(D) \ (\rho v^2)/(2)`

Now substituting values into the equation

`= 0.0159 [(30)/(5*10^(-5)) ][((999.1)(4.585))/(2) ]`

`=100.185 *10^3 Pa`

`=100.185 kPa`

Generally the head loss in the pipe is mathematically represented as

`h_L =(\Delta P_L)/(\rho g)`

`= (100.185 *10^3)/((999.1)(9.81))`

`=10.22m`

Generally the power input required to overcome this pressure drop is mathematically represented as

`W_p = Q \Delta P_L`

`=(9*10^(-3)(100.185*10^3))`

`= 901.665\ W`