Quiz 3 – 2013.11.29 SURVIVAL MODE

Question

Liquid water at 1 atm is being pumped at a rate of 0.189 m3/s from a large storage tank by a pump with a rating of 2 kW. The water is pumped through a heat exchanger, where it gives up 758 kW of heat and is then delivered to a storage tank at 18.76 m above the first tank. What is the total change in temperature for the water? For water: = 1 g/cm3, cP = 4.184 J/g K.∙

TIME IS UP!!!

Solution

Assumptions:1. Steady state process2. No heat loss in the tanks and along the pipes3. Uniform pipe diameter4. Constant fluid properties

Q ∆z

WS

Solution

2

2cv

Sc c

d mU gm H z Q Wdt g g

0 0

3

3kg m kg1000 0.189 189m s s

m V

18.76 mc

gzg

2

m9.81s

2mkgs1.00

N

1 kJ1000 N m

kJ0.184kg

Solution

kJ kJ kJ758 2 756s s sSQ W

S

c

gm H z Q Wg

S

c

Q W gH zm g

kJ756 kJ kJs 0.184 4.184kg kg kg189s

H

kJ kJ4.184 4.184kg kg K

H T 1 KT

Outline

1.Mass Balance

2.Energy Balance

3.Momentum Balance

Momentum Balance

2 21 1 1 1 2 2 2 2 1 1 1 2 2 2

s f tot

d P v Au v Au P Au P Audt

F m g

Rate of increase of momentum Rate of change

in momentum

Change in pressure forces

Force of solid surface on fluid

Force of gravity on fluid

Momentum Balance

We can rewrite the following terms as:

Substituting into the balance equaton:

mAv

2 21 21 1 2 2 1 1 1 2 2 2

1 2

s f tot

m md P v u v u P Au P Audt v v

F m g

Momentum Balance

For turbulent flows:

2v vv

v

0.95 0.99 for turbulent0.75 for laminar

Momentum velocity correction factor

Momentum Balance

Rewriting:

1 1 1 2 1 2 1 1 1 2 2 2

s f tot

d P mv u mvu P Au P Audt

F m g

s f totd P mv PA u F m gdt

Momentum Balance

Important Notes:1. All terms are considered vectors, so the direction

must be specified (x, y, or z).2. The force due to gravity only acts along the y-

direction.3. This equation assumes that the flow is turbulent,

and the velocity profile is flat.

s f totd P mv PA u F m gdt

Exercise

A diagram of a liquid-liquid ejector is shown in the figure below. It is desired to analyze the steady-state mixing of two streams, both of the same fluid, by means of overall balances. At plane 1 the two fluids merge. Stream 1a has a velocity v0 and a cross-sectional area (1/3)A1, and Stream 1b has a velocity (1/2)v0 and a cross-sectional area (2/3)A1. Plane 2 is chosen far enough downstream so that the two streams have mixed and the velocity is almost uniform at v2. The flow is turbulent and the velocity profiles at planes 1 and 2 are assumed to be flat. Neglect Fs→f and gravity effects.

Liquid-Liquid Ejector

We can rewrite the entire system like this:

v0

(1/2)v0 Plane 1 Plane 2

Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fs→f

Overall Mass Balance

v0

(1/2)v0 Plane 1 Plane 2

Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional

area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fs→f

1 1 2

1 1 1 1 1 1 2 2 2

1 1 1 1 2 2since incompressible flow:

a b

a a a b b b

a a b b

m m mv A v A v A

v A v A v A

00 1 1 2 2

1 23 2 3

vv A A v A

1 2A A2 0

23v v

Overall Momentum Balance

Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional

area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fs→f

2 21 21 1 2 2 1 1 1 2 2 2

1 2s f tot

m md P v u v u P Au P Au F m gdt v v

Since the flow is turbulent and unidirectional:

0 mv PA

1 1 1 1 2 2 1 1 2 20 a a b bm v m v mv P A P A

2 2 1 1 1 1 1 1 1 1 2 2 2a a a b b bP A P A v A v v A v v A v

1 1 2 1 2 22 2 1 1 0 1 0 0 1 0 0 2 03 2 3 2 3 3P A P A v A v v A v v A v

2 2 21 1 42 2 1 1 0 1 0 1 0 23 6 9P A P A v A v A v A

Overall Momentum Balance

Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fsf

2 2 21 1 42 2 1 1 0 1 0 1 0 23 6 9P A P A v A v A v A

1 2since :A A

2 2 21 1 42 1 0 0 03 6 9P P v v v

212 1 018P P v

Questions:1. What conclusion can be made from the above result?2. If we are to carry out an MEB on the system (ΣF is significant),

what result should we expect? What is ΣF is negligible?

Exercise

Fluid is flowing at steady state through a reducing pipe bend, as shown in the figure below. Turbulent flow will be assumed with frictional forces negligible. The volumetric flow rate of the liquid and the pressure p2 at point 2 are known, as are the pipe diameters at both ends. Derive the equations to calculate the forces on the bend. Assume that the fluid density is constant.

Exercise

Required Quantity:Force of the fluid on the surface

f s s fF F

Assumptions:1. Steady-state flow2. Incompressible fluid3. Only the x- and the y-

direction are involved4. Significant gravity effects5. No friction loss

Overall Mass Balance

1 2

1 1 2 2 (incompressible flow)m mv A v A

From the mass balance, given the volumetric flowrate and areas of the bend, we can obtain the velocities at the two points.

Mechanical Energy Balance

2

2 Sv Pg z F W

Assumptions:1. Steady-state flow2. Incompressible fluid3. Only the x- and the y-

direction are involved4. Significant gravity effects5. No friction loss

2 22 1 2 1

2 1 02v v P Pg z z

2 1since is given, we need to solve for :P P

2 211 2 1 2 1 22P g z z v v P

From the energy balance, we now have pressure values.

Overall Momentum Balance

s f totd P mv PA u F m gdt

Assumptions:1. Steady-state flow2. Incompressible fluid3. Only the x- and the y-

direction are involved4. Significant gravity effects5. No friction loss

f s totF mv PA u m g

Resolving the forces into its x- and y-components:

, 1 1 1 2 2 2 1 1 1 2 2 2

,y 1 1 1 2 2 2 1 1 1 2 2 2

f s x x x x x

f s y y y y tot

F mv u mv u P Au P AuF mv u mv u P Au P Au m g

Overall Momentum Balance

Based on the figure, the unit vectors are:

1 1

2 2

1 0cos sin

x y

x y

u uu u

Plugging in the unit vectors:

, 1 1 2 2 1 1 2 2

,y 2 2 2 2

cos cossin sin

f s x

f s tot

F mv mv P A P AF mv P A m g

Overall Momentum Balance

, 1 1 2 2

1 1 2 2

,y 2 2

2 2

coscos

sinsin

f s x

f s

tot

F mv mvP A P A

F mvP A m g

The force exerted by the fluid on the bend have components:

Questions:1. What would be the magnitude and direction of this force?2. What will be the force exerted by the bend on the fluid?

Exercise

Water at 95°C is flowing at a rate of 2.0 ft3/s through a 60° bend, in which there is a contraction from 4 to 3 inches internal diameter. Compute the force exerted on the bend if the pressure at the downstream end is 1.1 atm. The density and viscosity of water at the conditions of the system are 0.962 g/cm3 and 0.299 cp, respectively.