Page Level Ads

Question From Readers # 4: How Long To Empty Tanks


How long will it take to empty it if the 3 pumps are working together


 If a pump can empty a tank in 16 hrs, another pump can empty the same tank in 8 hrs, and another caempty this tank in 12 hrs. How long will it take to empty it if the 3 pumps are working together on this tank.

What is asked?

 How long will it take to empty it if the 3 pumps are working together on this tank.

 Analysis:

 In this problem, it is not necessary to get or to know  the volume of the tank. So you can put any figure of volume of the tank - be it 1,000 or 10,000, the answer is the same.  
Solution:

Let us say the volume of the tank is 1000 cubic meter. 

We have to get the rate of each tank. 

Pump A        1000 ­m³ ­÷ 16 hr = 62.5 m³/hr
Pump B        1000 m³ ÷   8 hr = 125 m³/hr
PumpC        1000 m³ ÷ 12 hr = 83.33 m³/hr
                                                 270.83 m³/hr is the combined rate of 3 pumps

Now we have to divide the volume by the combined rate of 3 pumps.

                    1000 m³ ÷ 270.83 m³/hr = 3.69 hrs is the time for the 3 pumps to empty the tank.  

But if you want to solve this as it should be, I mean by letting X as the volume since it is unknown, so this is it.

Let x be the volume.

Pump A     x ÷ 16 =  x 
                               16
Pump B     x ÷ 8 =  x 
                              8
Pump C     x ÷ 12 =  x  
                               12

 x  + x  + x  = 3x + 6x + 4x  = 13x
16   8     12           48               48

Now, x ÷ 13x 
                48

     

 x  ÷  13x  
1       48   
x   x   48   = 48x  = 3.69
1       13x     13x 

If you find this helpful, please click the share button below to share this post to your friends. Sharing is caring...
1 comment:

Question From Readers # 3: Reduction In GM Due To Free Surface

What is the reduction in GM due to free surface

A 7000 ton displacement tank ship carries two slack tanks of alcohol with a SG of 0.8 each tank i 50 ft. long 30 ft. wide. What is the reduction in GM due to free surface with the vessel floating in sea water, SG is 1.026?

Solution:

I will try to answer your question.

 This is the formula we are going to use FSC =           r L B³          
                                                                       420 x Displacement  

1. Divide SG of liquid inside the tank by the density of water where the vessel is floating to get the value of   "r".
                            r =   0.8   = 0.78
                                 1.026

2. Proceeding now to the formula to get the reduction in GM.

                     FSC =           r L B³          
                                420 x Displacement
                           
                             = 0.78 x 50 x 30³
                                 420 x 7000
                             = 0.78 x 50 x 27000
                                 420 x 7000
                             = 1053000 
                                 2940000
                             = 0.36 ft      * as is if there is only one tank. But since there are two slack tanks,
                                  x 2             multiply it by 2.
                                0.72 ft    is our final answer.

I hope this answer to your question. If ever I am wrong in some way, feel free to correct me. And if you may, share this blog to your friends.


You Might Also Like
2 comments:
Labels:

Question From Readers # 2: Parallel Sailing: Calculate Latitude

Parallel Sailing Formula

  A ship "A" is on the equator steering 090°T at 16 knots ; while a ship"B" is on the parallel of north latitude, steering 270°T at 12 knots. When "A" makes Dlo of 1', "B" makes a Dlo of 48'. Calculate the latitude of "B".

Given:

Ship A
               Course = 090°
                Speed = 16 knots

Ship B

               Course = 270°
                Speed = 12 knots

What is asked?

Latitude of ship B

Analysis:

The relationship between ships is that when "A" makes Dlo of 1', "B" makes a Dlo of 48'.

This is not quite an easy problem compared to other parallel sailing questions. However, if we could solve for a distance traveled by ship A when she makes a Dlo of 1 nautical mile at the equator, then that will be our key to solving for the latitude of ship B.

This is how...

Solution:

1. Solving for a distance traveled by ship A at the equator when she makes a dlo of 1'.

            Distance = Dlo x Cosine Latitude
                          = 1 nm x Cos 0°               (equator is zero latitude)
                          = 1 nm x 1
            Distance = 1 nm

2. Second, we have to get how much time when ship A travel for 1 nm with the speed of 16 kts.

   Steaming Time = 1 nm               (1 knot is equal to 1 nm/hr)
                             16 nm/hr
                         = 1 nm        
                            16 nm/hr   
                         = 0.625 hr       (convert it to minutes)
                         = 0.625 hr x 60 min = 3.75 mins = 3.75 mins
                                              1 hr       1
                         = 3.75 mins

   So, for ship A with the speed of 16 knots, she could travel 1 nautical mile at the equator for 3.75 mins and makes a  Dlo of 1'. How about ship B?

3. Now for the distance traveled for ship B, we have to use the same steaming time 3.75 mins because of this relationship that when "A" makes Dlo of 1', "B" makes a Dlo of 48'. In parallel sailing, distance is equal to the departure.

           Distance = Speed x Time
                         = 12 x 3.75
                         = 45            is the departure of ship B. Dep = Dist
     
                                               Distance = Dlo x Cos Lat
                                               Depature = Dlo x Cos Lat

4. And now for solving for the latitude of ship B, we have to divide the departure by Dlo, and then inverse cosine the answer.

                Dep = Dlo x Cos Lat

 Dlo x Cos Lat = Dep

 Dlo x Cos Lat = Dep 
        Dlo             Dlo

 Dlo x Cos Lat = Dep 
        Dlo             Dlo

         Cos Lat = Dep    
                        Dlo

         Cos Lat = 45
                        48
                     = 0.9375
                Lat = 0.9375 inv Cos
                Lat = 20° 22' N   is the latitude of ship B

P.S. Like what you read? Please click the share button below to share this post to your friends. Sharing is caring...
No comments:
Labels:

Gyro Error Calculation

Find the gyro error if the setting sun was observed to bear 290° PGC at Lat 37° 10' S. Declination 13° 37' N.

Given:

     Latitude = 37° 10' S
Declination = 13° 37' N

What is asked?

Gyro error

Solution:

                Sin amplitude = Sin Declination
                                         Cos Latitude
                                     = Sin 13° 37' 
                                        Cos 37° 10'
                                     = 0.23542
                                        0.79689
                Sin Amplitude = 0.29542
                                     = 0.29542 inv Sin 
                      Amplitude = W 17.18° N          (West because the sun is setting)
                                          +    270°             (plus 270° because it is in the 4th quadrant)
                  True azimuth = 287.18°
                        per Gyro = 290°               (subtract to get gyro error)
                      Gyro Error = 2.8° W         (Compass best, error west; compass least, error east)

P.S. Like what you read? Please clink the share button below to share this post to your friends. Sharing is caring...

You Might Also Like
No comments:

Find The True Azimuth At Sunrise

At Lat 55° 53' N, Long 130° 18 W, sun's declination is 12° 40' S. Find the true azimuth (Zn) at sunset.

Given:

     Latitude = 55° 53'
Declination = 12° 40' S

What is asked?

True azimuth at sunset

Solution:

 1. We have to get first the amplitude. This is the formula for amplitude:

                Sin amplitude = Sin Declination
                                         Cos Latitude
                                     = Sin 12° 40' 
                                        Cos 55° 53'
                                     = 0.21928
                                        0.56088
               Sin Amplitude = 0.39096
                                     = 0.39396 inv Sin 
                     Amplitude = W 23 S             (West because the sun is setting)
                                                                   (3rd quadrant)
2. Solving for true azimuth. Since the sun is setting, we have to subtract 270° by our amplitude.

                            270°
                          -   23°       (minus because it is in the 3rd quadrant)
                            247° T is the true azimuth of the sun at sunset

If you find this helpful, please click the share button below to share this post to your friends. Sharing is caring...

You Might Also Like

               

No comments:
Labels:

Finding Sun's Amplitude

On Nov. 21, the sun bore 101° psc to an observer in DR Position Lat 41° 29' N, Long 120° 30' E, declination is 14° 32.2' S. Find the amplitude.

Given:

    Latitude = 41° 29' N
declination = 14° 32.2' S

What is asked?

The amplitude

Solution:

Sin Amplitude = Sin declination
                        Cos latitude

                     = Sin 14° 32.2'
                        Cos 41° 29'
                     = 0.25099
                        0.74914
Sin Amplitude = 0.33504
      Amplitude = 0.33504 inv Sin
                     = 19.57 or simply
     Amplitude  = E 20 S 

* Why E 20 S? EAST because the sun bears 101°, which means it is rising. And SOUTH because that is the sign of the declination. So, just copy the sign of the declination.

If you find this helpful, please click the share button below to share this post to your friends. Sharing is caring...

You Might Also Like
2 comments:
Labels:

Finding Latitude Of The Observer

On Sept. 12, at GMT 10h 35m 00s LZT in longitude 057° 58' W, the Ho of star Polaris was 35° 50'. The correction values from the Polaris Tables: Ao = 1° 22.1', A1 = 0.4', A2 = 0.9'. Find the latitude of the observer.

Given:
 
  Observed altitude (Ho) Polaris = 35° 50'
                                         Ao = 1° 22.1'
                                         A1 = 0.4'
                                         A2 = 0.9'

What is asked?

The latitude of the observer

Solution:

Add all the values from Polaris Table to the observed altitude of Polaris. And then minus 1 to get the latitude of the observer.
                  
                                 
  Observed altitude (Ho) Polaris = 35° 50'
                                         Ao = 1°   22.1'
                                         A1 =         0.4'
                                         A2 =         0.9' 
                                                  37° 13.4'
                                               -    1°          
                                                  36° 13.4' N is the latitude of the observer      

You might wonder why the sign of the latitude is N. It must be north because you can not observe Polaris in the southern hemisphere. That is, if you can see or observed polaris, know that you are at northern hemisphere.

If you find this helpful, please click the share button below to share this post to your friends. Sharing is caring...

You Might Also Like

1 comment:
Labels:

Ship's Squat Formula

Find the approximate calculated squat of your vessel is proceeding to a channel not enclosed with a width of 90 meters deep and dredge surrounding depths of 20 meters. Your vessel's draft is 11 meters and beam of 27 meters, speed 7 knots and block coefficient is 0.8.

Given:

               Speed = 7 knots
Block Coefficient = 0.8

What is asked?

Ship's squat

Solution:

This is the ship's squat formula on open waters and the unit of the answer will be meter.
      
                       Squat = Block Coefficient x Speed²
                                                   100
                                 = 0.8 x 7²
                                        100
                                 = 0.8 x 49
                                        100
                                 = 39.2
                                    100
                       Squat = 0.392 m      

Again, the example above is that we get  the value of the ship's squat in open waters. You might asked then, if there's a formula for squat on open waters, what will be the formula for squat on enclosed water?

To get the value of squat on enclosed water is to multiply block coefficient by the square of speed, and then times 2, then divide the product by 100.

Here is the problem for squat on closed water. Try to solve it yourself.

A container vessel of 12,000 tons displacement is approaching her berth at speed of 4 knots. Its block coefficient is 0.78. Calculate the value of squat.

Answer: 0.25 meters

If you find this helpful, please click the share button below to share this post to your friends. Sharing is caring...

You Might Also Like
4 comments:
Labels:

Rolling Period Formula In Still Water

Find the still water period of roll for a ship when the radius of gyration is 6 meters and the metacentric height is 0.5 meters.

Given: 

                           Radius = 6 m
Metacentric height (GM) = 0.5 m

What is asked?    

Rolling Period in still water              

Solution:

Still water period of roll = 2Ï€ x Radius
                                       √g x GM

                                     = 2 x 3.1416 x 6 m   
                                        √9.81 m/sec² x 0.5

                                     = 37.6992 m       
                                         √4.905 m/sec²

                                     = 17.02 seconds

If you find this helpful, please click the share button below to share this post to your friends. Sharing is caring...

You Might Also Like
                                       
2 comments:
Labels:

At What Speed Must You Proceed To Reach Your Next Port

Your vessel arrive in port with sufficient fuel to steam 1175 miles at 19 knots. If you are unable to take on bunkers, at what speed must you proceed to reach your next port, 1341 miles distance?

 Given:

   Old distance = 1175 miles
New distance = 1341 miles
     Old speed = 19 knots

  What is asked?

 New speed

Solution:
 
  New Speed = √ Speed2 x Old Distant
                              New Distance 

                          = √192 kts x 1175 mi 
                               1341 mi 

                    = √361 kts x 1175 mi
                             1341 mi

                    = √424175
                          1341

                    = √316.31

 New Speed = 17.8 knots
If you find this blog helpful, please click the share button below to share this post to your friends. Sharing is caring...
2 comments:
Labels:

Calculating Ship's Maximum Speed To A Desired Fuel Consumption Per Day

While steaming at 15 knots, your vessel consumes 326 barrels of fuel oil per day. In order to reduce consumption to 178 barrels of fuel oil per day, what is the maximum speed a vessel can turn for?

 Given:

   Old Consumption = 326 bbls
  New Consumption = 178 bbls
             Old Speed = 15 kts

What is asked?

New speed

Solution:  


                     New Consumption = New Speed3
                       Old Consumption   Old Speed3

New Speed3 x Old Consumption = New Consumption x Old Speed3
New Speed3 x Old Consumption = New Consumption x Old Speed3
                       Old Consumption         Old Consumption

New Speed3 x Old Consumption = New Consumption x Old Speed3
                       Old Consumption         Old Consumption

                              New Speed3 = New Consumption x Old Speed3
                                                             Old Consumption
                                                 
                                                 = 178 x 153
                                                                               326
                                                 = 178 x 3375
                                                           326
                                                 = 600750   
                                                         326
                              New Speed3 = 1843      Apply cube root to both sides

                           ³√New Speed3 = ³√1843 
                              

                                New Speed = 12.2 knots

If you find this blog helpful, please click the share button below to share this post to your friends. Sharing is caring...
No comments:
Subscribe to: Posts (Atom)