## Name and some simplification finally gives us, ?*Cp*

Name – Akash Sonowal

Roll No. – 160107003

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Assignment 1   Section 1

1. 3D Heat Conduction Equation for Cylindrical Coordinates

Consider a volume element in cylindrical coordinates as shown in the figure.

The energy balance on this element during a small time interval dt can be expressed as

?stored=?in-?out+?gen                                  …………?

?gen= ?*Volume of the element

= ?*dr*dz*rd?

(?=energy generated per unit volume, W/m3)

?stored=?*(dr*rd?*dz)*Cp*

Now, plugging in the values in the energy balance equation i.e., ?

?*Cp**(dr*rd?*dz) = qr + q? + qz  – qr+dr  – q?+d? – qz+dz + ?*dr*rd?*dz

………?

By Taylor’s Series Expansion

qr+dr = qr+*dr

q?+d? = q?+*d?

qz+dz = qz+*dz                                                                                                x

Now, ?=>    ?*Cp**(dr*rd?*dz) = – *dr – *d? – *dz + ?*dr*rd?* dz                ……….?

By Fourier’s law of Heat Conduction,

qr =  – k *   * rd? *dz

q?= –  *  * dr * dz

qz = – k *  * dr * rd?

And now finally plugging in all the values for qr, q?, qz in ? and some simplification finally gives us,

?*Cp* =  *   +    +  + ?

3D Heat Conduction Equation for Spherical Coordinates

Consider a volume element in spherical coordinates as shown in the figure.

The energy balance on this element during a small time interval dt can be expressed as

?stored= ?in – ?out + ?gen                                  …………?

?gen= ?*Volume of the element

= ?*dr*rd?*rsin?d?

(?=energy generated per unit volume, W/m3)

?stored=?*( dr*rd?*rsin?d?)*Cp*

Now, plugging in the values in the energy balance equation i.e., ?

?*Cp**(dr* rd?*rsin?d?) = qr + q? + q?  – qr+dr  – q?+d? – q?+d? + ? * dr * rd? * rsin?d?       ………?

By Taylor’s Series Expansion

qr+dr = qr+*dr

q?+d? = q?+*d?

q?+d? = q?+* d?

Now, ?=>

?*Cp**(dr* rd?*rsin?d?) = – *dr – *d? – *d? + ?*dr* rd? * rsin?d?                                                                                       ……….?             x

By Fourier’s law of Heat Conduction,

qr =  – k *   * rd? * rsin?d?

q?= –  *  * rd? * dr

q? = – k *  * dr * rsin?d?

And now finally plugging in all the values for qr, q?, q? in ? and some simplification finally gives us,

?*Cp* =  *   + +  + ?

2. Given, Thickness of the flat plate = L

Temperature difference between the plates =

Thermal conductivity of material, k =

From Fourier’s Law of Heat Conduction,

q = – KA

Now, we can write

q?x = – KA ?x

Integrating both the sides,

= –

q =

3. Assuming steady heat flow condition

Given, Temperature difference across the fiberglass layer = 85°C

Thickness = 15 cm.

Thermal Conductivity (k) = 0.035 W/m °C

From Fourier’s Law of Heat Conduction,

q = – KA

= 0.035 *

= 19.8334

= 71400.24

4. Given, Thickness of the plane wall = 14 cm

Temperature difference = (376 – 86) °C

= 290 °C

K = 0.79

? = 2800 kg /

From Fourier’s Law of Heat Conduction,

q = – KA

q = – KA

q = 0.79 *

q = 1636.4286

5. Given, k = 12

?T = 400°C

q = 400

From Fourier’s Law of Heat Conduction,

q = – KA

q = – KA

?x =

=

= 12 m

6. Given, water flow rate = 0.8 kg/s

Length of the tube = 3 m

Diameter of the tube = 3 cm

Temperature difference between tube wall and water = 40°C

Here the mode of heat transfer is convection

From Newton’s Law of Cooling,

q = h*A*(T – )

q = 3501.6719 *  *40                             ()

q = 39582.9 W

Assuming all heat is being used in raising the temperature of water

q = m  dT

39582.9 = 0.8*4180*                                 (

= 11.84°C

7. Given,

P = 1 atm

= h

h =                          (

=

= 272.73

8. Given, Thickness = 40 cm

Thermal conductivity = 2 W/m°C

Surrounding Temperature = 10°C

h = 10

Wall Temperature on one side () = 10°C

Wall Temperature on the other side ( =

Fourier’s Law of Heat Conduction,

= – K A

Newton’s Law of Cooling,

= h A ()

Since,  =

The heat transfer processes are in series

K A  = h A ()

The heat transfer rate through the wall,

= – K

=  2 *

= 300

9. Given, Surface Temperature ( = 333.15 k

Surrounding Temperature ( = 294.15 K

Surface emissivity = 0.81

The mode of heat transfer will be radiation

Applying Stefan Boltzmann Law,

q = ? ? A (

q =

= 28.523

The total heat loss by both sides of the plate = 2 *q

= 2 * 28.523

= 57.046

10. Warm water due to it’s low density than cold water evaporates rapidly and this process is endothermic, thus causing it to freeze faster than cold water.

11. Here the mode of heat transfer will be the combination of both convection and radiation

Given, h = 35

Emissivity, ? = 0.6

A = 0.02

By Newton’s Law of Cooling,  = 35 * 0.02 *

By Stefan’s Boltzmann’s law,  = 0.6 * 0.02 * 5.669 *

Now,

800 = 35 * 0.02 *  + 0.6 * 0.02 * 5.669 *

= 421.32 K

12. Here, heat transfer takes place by both convection and radiation

Given, A = 75

? = 0.174

= 2.5 * 75 * (100 – 70) + 0.1714

= 5625 + 10.846

= 5635.846 Btu/hr