Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 14

Q. 14.3

Water at 40 °C at a mass flow rate of 0.5 kg/s enters a 2.5 cm ID tube whose wall is maintained at a uniform temperature of 90 °C. Calculate the tube length required for heating the water to 60 °C. Also determine the pressure drop across the length of pipe.

Step-by-Step

Verified Solution

Given data is written down

\dot{m}=0.5 kg / s

D = 25 mm

T_{B, 0}=40^{\circ} C T_{B, L}=60^{\circ} C T_{w}=90^{\circ} C

Water properties are evaluated at T_{m}=\frac{T_{B, 0}+T_{B, L}}{2}=\frac{40+60}{2}=50^{\circ} C. From table of properties of water, we have

\nu=0.568 \times 10^{-6} m ^{2} / s, k=0.064 W / m ^{\circ} C , P r=3.68 and \rho=990 kg / m ^{3}

Hence, the dynamic viscosity of water is \mu=\rho \nu=990 \times 0.568 \times 10^{-6}=5.623 \times 10^{-4}kg/ms

The velocity of water in the tube may be calculated as

U=\frac{4 \dot{m}}{\pi D^{2} \rho}=\frac{4 \times 0.5}{\pi \times 0.025 \times 990}=1.029 m / s

The Reynolds number is then given by

R e_{D}=\frac{U D}{\nu}=\frac{1.029 \times 0.025}{0.568 \times 10^{-6}}=45285

The flow is turbulent and hence we may use the Sieder and Tate correlation, taking into account the variation of properties, to calculate the Nusselt number. From tables, we read off the kinematic viscosity and density at wall temperature of 90 °C as

\nu_{w}=0.329 \times 10^{-6} m ^{2} / s ^{\prime} \rho_{w}=967.4 kg / m ^{3}

Hence, the dynamic viscosity of water at the wall temperature is

\mu_{w}=\rho_{w} \nu_{w}=967.4 \times 0.329 \times 10^{-6}=3.183 \times 10^{-4} kg / m s

Hence, we have

\frac{\mu}{\mu_{w}}=\frac{5.623 \times 10^{-4}}{3.183 \times 10^{-4}}=1.767

Clearly the constant property assumption would not be valid. The Sieder and Tate correlation is appropriate. Hence, using Eq. 14.28, we have

N u_{D}=0.027 \operatorname{Re}_{D}^{0.8} \operatorname{Pr}^{\frac{1}{3}}\left(\frac{\mu}{\mu_{w}}\right)^{0.14}  (14.28)

N u_{D}=0.027 \times 45285^{0.8} \times 3.68^{\frac{1}{3}} \times 1.767^{0.14}=239.5

The mean value of the heat transfer coefficient is then given by

\bar{h}=\frac{N u_{D} k}{D}=\frac{239.5 \times 0.640}{0.025}=6131.2 W / m ^{2}{ }^{\circ} C

With the various temperatures that are specified, we may calculate the temperature differences ratio as

\frac{T_{w}-T_{B, L}}{T_{w}-T_{B, 0}}=\frac{90-60}{90-40}=\frac{30}{50}=0.6

The Stanton number may be calculated as

S t=\frac{N u_{D}}{\operatorname{Re}_{D} P r}=\frac{239.5}{45285 \times 3.68}=0.001437

Hence, using Eq. 14.32,

\frac{T_{w}-T_{B, L}}{T_{w}-T_{B, 0}}= e ^{-\left(4 \times S t \times \frac{L}{D}\right)}  (14.32)

e^{-4 \cdot S t \cdot \frac{L}{D}}=0.6

or solving for pipe length

L=-\frac{0.025 \times \ln (0.6)}{4 \times 0.001437}=2.22 m

Pressure drop is calculated assuming that the tube is smooth. Equation 14.24(b) is appropriate and hence

f=\frac{0.184}{R e_{D}^{0.2}} \text { for } R e_{D}>2 \times 10^{4}  (14.24)(b)

f=\frac{0.184}{45285^{0.2}}=0.0216

The pressure drop over the length of the tube is then given by

-\Delta p=f \cdot \frac{L}{D} \cdot \frac{\rho U^{2}}{2}=0.0216 \times \frac{2.2}{0.025} \times \frac{990 \times 1.029^{2}}{2}=1005.3 Pa