38 - Dr. Gundry's Diet Evolution by Steven R. Gundry, M.D.

1. "Indeed, whether you realize it or not, your fate is being controlled by a hidden system that runs most of your cellular, hormonal, nervous system and aging processes without your conscious input"
2. " We are what we eat"
3. "We consider any male patient who weighs in at 220 pounds a member of "big boy club"
4. Myth Busting:

Myth: Drink three glasses of milk or the equivalent each day

Reality : Most people are not programmed to drink milk

Myth: Orange Juice is a great source of vitamins

Reality: Drinking any fruit juice is like drinking liquid sugar

Diet in three messages three phases:

1. You don't have to store fat for winter, Phase 1 (6 weeks) detaching phase
2. You are not a threat to future generations, Phase 2 (6 weeks) , restoration phase 
3. Your staying alive ensures your genes future, Phase 3, Permanent way of eating

You can accomplish the first two phases in as little as 90 days and move to a permanent way of eating in phase 3, unlike a diet you leave behind after getting results you were looking for or not . This is possible because new cell replaces 90 % of your existing cells every three months.

37 - Boiler operation Q&A - T # 1 - Boiler Auxiliaries - Posted by Kambiz Ehsani on 5/17/2018

Q: What kind of pump can be used as Boiler feed water pump?
A: Centrifugal pump

Q: Why?
A:

1. A centrifugal pump is capable of delivering steady flow of BFW
2. Supplying the largest quantity of BFW under a given head
3.  Accepting load variation most easily
4. Trouble free and smooth operation 
5. Less floor space is required
6. Less maintenance is required

Q:Suppose both steam and electricity are available. Which centrifugal pump would you prefer to switch on?

A: Turbo driven

Q: Why?

Steam is cheaper than electricity

Q: The size of ID fan is always large than the FD fan. Why?

A:ID fan is to handle a fluid (flue gas) with higher specific volume than the air handled by the FD fan. Greater specific volume of the gas handled. 

36 - T # 8 - What is Specific Humidity ? - Posted by Kambiz Ehsani - 6/16/2018

The amount of moisture present in the air is expressed in grains of moisture per pound of dry air and referred to as specific humidity. In a pound of water there are 7,000 grains of moisture. Pounds of moisture per pound of dry air could be used to indicate how much moisture is present, but generally , grains of moisture are used because the numbers are larger and easier to work with.

35 - T # 7 - What is relative Humidity ? - Posted by Kambiz Ehsani - 5/16/2018

This is the ratio of the amount of moisture present in the air to the amount it can hold at saturation.

34 - T # 6 - Total heat of Air formula - Posted by Kambiz Ehsani - 5/16/2018

When dealing with changes in both sensible and latent heat such as in heating and humidification or in cooling, use the total heat formula for calculating the heat change in the air which states that:


Total Heat in Btu/hr = Specific Density x 60 min/hr x CFM x Delta H

Total Heat in Btu/hr = 0.075 x 60 x CFM x Delta H

Total Heat in Btu/hr = 4.5 x CFM x Delta H

Delta H = Change in the enthalpy of entering and leaving

33 - T # 5 - Moisture in the air - Post by Kambiz Ehsani - 5/16/2018

Moisture is always present in the air and has a heat content of its own called Latent Heat.
The latent heat of water vapor added to the sensible heat of dry air gives us the total heat of the quantity of air.
We use a Wet Bulb thermometer to measure enthalpy.

32 - T # 4 - Air Sensible Heat Formula - Posted by Kambiz Ehsani - 5/16/2018

Sensible Heat of Air = Btu/hr = Sp. Heat x Sp. Gravity  x 60 Min/Hr x CFM x Delta T

Sensible Heat of Air =  0.24 x 0.075 x 60 x CFM x Delta T

Sensible Heat of Air =  1.08 x CFM x Delta T

Temperature readings are Dry Bulb

31 - T # 3 - Air Properties - Posted by Kambiz Ehsani - 5/16/2018

What is specific heat of air?

Looking at the ability of air to get hot. It is measured in the amount of heat in Btu's required to raise one (1) pound of a substance one (1) degree F. Standard air has a specific heat of 0.24 Btu's per pound per degree F.

30 - Air Properties - T # 2 - Posted by Kambiz Ehsani - 5/16/2018

As Air is heated, It expands. the specific volume would increase and the Specific density would decrease. At higher temperatures, air weighs less per cubic foot of volumes. This is why warm air rises. Example is the balloons that go up with passengers.

29 - Fundamentals - Posted by Kambiz Ehsani - 5/15/2018

How much pressure is a column of water 1 foot high exerting at the base of this column?

Solution:
The answer is 0.433 psi. So 119 psi pressure is equal to 119 / 0.433 = 275 feet

28 - Air Properties T #1 - Posted by Kambiz Ehsani - 5/15/2018

1. Air is a mixture of many gasses and is classified as either Dry or Moist air
2. Dry air contains approx. 78% nitrogen , 20.9% Oxygen, 1% Argon and other gasses (1% other gasses
3. Moist air is a mixture of dry air and water vapor. The amount of moisture present in the mixture may vary from zero to all that air can hold, called its saturation point.
4. If dry air and water vapor were mixed together in a container, the pressure of this mixture is the summation of the pressure exerted by each of the two gasses. This is known as Dalton's Law.
5. Dry air is a gas and follow the gas laws of Charles and Boyle. This gas law states that if air were heated and maintained at a constant pressure, the air would expand and weigh less per cubic foot of volume.This property of air is known as specific density.
6. At sea level, the pressure exerted on the surface of the mercury of 70 degree dry air is 14.696 psi absolute. This will maintain a level of mercury in the tube at 29.92 inches.
7. Specific Density equals one over the specific volume. Or, Specific volume equals one over the specific density. For example, the specific volume of standard air is 13.33 cubic feet per pound, so its density will be one over 13.33 or 0.075 lb per cubic foot.  

27 - Health recommendation pick for the week - Benefits of antioxidants - Posted by Kambiz Ehsani - 5/13/2018

Benefits of antioxidants:

https://www.webmd.com/food-recipes/antioxidants-your-immune-system-super-foods-optimal-health

26 - Question 20 - Boiler Deaerator energy balance - Post by Kambiz Ehsani - 5/12/2018

A boiler generates 50,000 lb/hr of saturated steam at 300 psia, out of which 10,000 lb/hr is taken for process and is returned to the deaerator as condensate at 180 degree F, the remainder is consumed. Makeup water enters the DA tank at 70 degree F and steam is available at 300 psia for deaeration. The DA tank operates at 25 psia. The blow down has TDS of 1,500 ppm by weight and make up has a TDS of 100 ppm. Evaluate the blow down and deaeration steam quantities.

Solution:

If boiler concentration is 0.5 ppm solid and water hardness concentration is 2,500 ppm: then the percent moisture in steam = Steam Purity, ppm / Boiler water concentration, ppm or

percent moisture in steam = (0.5 / 2,500) * 100 = 0.02 so the steam quality = 100 – 0.02 = 99.98 %

DA Energy balance:

The mass balance around two control volumes gives us:

 If M is the make up water to DA, D is the steam for deaeration to DA, F the BFW and B is the Blowdown

For the boiler 50,000 + B = F and for DA: 10,000 + M + D = F                         Eq (1)

by equating this two we will have: 50,000 + B = 10,000 + M + D  

 Energy Balance around the two control volumes gives:

10,000 * 148 + 1,202.8 * D + M * 38 = 209 * F = 209 * (50,000 + B)           Eq (2)

Balance of the solids in the makeup and blowdown gives 100 * M = 1,500 B   Eq (3)

25 - Question 19 - General Physics - gym lifting energy burned - Post by Kambiz Ehsani - 5/9/2018 - Solved

You are at gym and working out. You are lifting weight. What is the work for lifting 30 pounds weight for 1 foot? How much energy or calories will you burn if you do this exercise for 10 minutes and you do it 500 times? Although it looks impossible :-)

Solution:

Work for lifting a weight @ gym :
30 lbf lifted 1 ft = 30 * 1 = 30 lbf-ft = 30 * 1.3558 = 41 J

Total work including 500 times lifting = 41 * 500 = 20,500 J

Power = 34 J/sec = Watts  or 5 kcal

Assuming the human body efficiency in converting the Fat to exercise of 0.25

Power converted to to burned fat = 5 / 0.75 = approximately 7 Kcal

24 - Question 18 - Entropy Calculation - posted by Kambiz Ehsani - 5/8/2018

A peice of metal with the weight of 100 kg and temperature of 500 K falls into a big lake and gets cooled down with the lake temperature of 285 K. Find the entropy change of the lake water.

23 - Question 17 - Carnot engine - Posted by Kambiz Ehsani - 5/5/2018 - Solved

Consider three carnot machines, where first one works between Th and Tl, second one works between Ti and Tc the third one between Th and Tc. Find the relationship between the efficiencies of the three carnot engines (Eff 1, Eff 2 and Eff 3):

Solution:
Efficiency of the Carnot machine is a function of  Tl/Th ;

Efficiency of the First machine = Eff 1 = Tl / Th
Efficiency of the First machine = Eff 2 = Tc / Tl
Efficiency of the First machine = Eff 3 = Tc / Th

Therefore :

Eff 3 = Eff 1 *  Eff 2

22 - Question 16 - Braking Standard Fuel Consumption (BSFC) - Posted by Kambiz Ehsani - 5/5/2018 - Solved

If the thermal efficiency of an engine equals 25% and the heating value of the fuel 36,000 kJ/kg, calculate the Braking Standard Fuel Consumption (BSFC) in kgkW-hr.

Solution:

W = 0.25 * 36,000 = 9,000 kJ/Kg fuel
1 kW - hr = 1 kJ /sec  * 3,600 Sec = 3,600 kJ


Therefore the BSFC = 0.4  kg/kW-hr

21 - Question 15 - The triple point of the water - Posted by Kambiz Ehsani - 5/5/2018 - Solved

The temperature of the objects on a winter night reaches zero degree C. The moisture in the air is relatively high. Explain which one of the options below is the right process for the moisture on the objects phase change?

1) From vapor to solid
2) From vapor to solid to liquid
3) From vapor to liquid
4) From vapor to liquid to solid


Solution:
Because the temperature of the objects on the surface are below the temperature of the triple point of the water the vapor will become solid directly.

Photo source :

https://www.quora.com/What-is-the-meaning-of-triple-point-of-water

20 - Question 14 - The Second Law and Carnot Engine - Posted by Kambiz Ehsani - 5/5/2018 - Solved

Is this possible to build a heat engine that works between 800 and 300 degree K and the heat transfer from the hot source Qh = 500 kJ , Ql = 187.5 kJ and produce 200 kJ? Explain your answer.

Solution:

NO beacuse: the First Law is not met here.

Qh - Ql  must be the W => this becomes 500 - 187.5 which is not equal the 200 given in the assumptions of the problem and based on the First Law this is not possible because the delta of the Q must be equal to the work created on the shaft and in this example it is not the case.

19 - Question 13 - Thermodynamics - Post by Kambiz Ehsani - 5/5/2018 - Solved

Consider a balloon with the initial internal pressure of P1 when the outside pressure is P0. If air is pumped so that the internal balloon pressure becomes P2, Find the work of system on the surrounding.

Solution:
Assuming the air is the ideal gas and neglect the heat transfer. Therefore the Pv = m RT, since the P is not changing, the work will be applied to increase the volume from V1 to V2 therefore the work will be :
1W2 = P0 (V2 -V1)