Unit conversion:lbm/lbf*h to (N/s)/N (2024)

  • Forums
  • Homework Help
  • Introductory Physics Homework Help
  • Thread starteryecko
  • Start date
  • Tags
    Unit

In summary: Lift/drag and the log term are both dimensionless .Range = a dimensionless constant x Velocity/TSFCDimensionally : L = L T-1 / Dimensions of TSFC(Edit) or dimensions of TSFC = T-1Incidentally fuel 'weight' is sometimes expressed in Kg rather than N because it makes the unit independent of g . This is important in some flight dynamics calculations because g force as experienced by the fuel in a flight vehicle can vary considerably...In summary, the conversation discusses the conversion of thrust-specific fuel consumption from imperial units to SI units. It is mentioned that the conversion should be done by multiplying by gravitational acceleration, but it is unclear why. The conversation also clarifies that the lift

  • #1

yecko

Gold Member

279
15

Homework Statement

Given Thrust-specific fuel consumption cT=0.5279lb/lbf*h... as other units in the question are in SI unit, I have to convert it to (N/s)/N first...

Homework Equations

1 lbf = 4.448 N
1 lbm = 0.4536 kg
Thrust-specific fuel consumption = weight of fuel burned per unit time (N/s) per unit thrust (N)

The Attempt at a Solution

cT=0.5279lb/lbf*h= (0.5279lb/lbf*h)* (4.448 N/lbf) / (0.4536 kg/lbm) / (3600s/h) = 1.49e-5 kgN-1s-1
which the unit of my answer stands for mass per force per time (kg/Ns) instead of force per time per force(N/Ns)...
where in the unit conversion have i done wrong? thank you very much!

  • #2

mfb

Mentor

Insights Author

37,232
14,076

Oh the joy of exotic unit systems. Not only the SI part doesn’t match, the imperial units don’t cancel each other either: As an example, you have lbf in the denominator twice and nothing to cancel it with. You’ll have to reconsider what to multiply and what to divide.

  • #3

yecko

Gold Member

279
15

I have also tried to put 1lbf=1lbm*g=32.2 lbm^2 ft^-2,
then, the whole unit become lbm ft^2 h^-1
which also do nothing in manipulating the units at the end...

to be honest, this is my first course which i have to deal with imperial units, so i am not sure if my work is correct...

  • #4

jbriggs444

Science Advisor

Homework Helper

12,795
7,428

yecko said:

The Attempt at a Solution

cT=0.5279lb/lbf*h= (0.5279lb/lbf*h)* (4.448 N/lbf) / (0.4536 kg/lbm) / (3600s/h) = 1.49e-5 kgN-1s-1

It might be easier to read and understand if you multiplied by every conversion factor instead of multiplying by some and dividing by others.

Typesetting that can be a pain. But maybe you could use [CODE] and [/CODE] tags to improve the presentation like this.

[CODE]

Code:

0.5279 lb 4.448 N 1 lbm--------- * ------- * --------- * ... 1 lbf h 1 lbf 0.4536 kg

[/CODE]

Or you could get fancy and use LaTex. Key in

$\$\frac{0.529\ lb}{1\ lbf\ h} \cdot \frac{4.448\ N}{1\ lbf} \cdot \frac{1\ lbm}{0.4536\ kg} \cdot\ ...$\$​

and get
$$\frac{0.529\ lb}{1\ lbf\ h} \cdot \frac{4.448\ N}{1\ lbf} \cdot \frac{1\ lbm}{0.4536\ kg} \cdot\ ...$$

Last edited:

  • #5

Ray Vickson

Science Advisor

Homework Helper

Dearly Missed

10,704
1,723

jbriggs444 said:

It might be easier to read and understand if you multiplied by every conversion factor instead of multiplying by some and dividing by others.

Typesetting that can be a pain. But maybe you could use [CODE] and [/CODE] tags to improve the presentation like this.

[CODE]

Code:

0.5279 lb 4.448 N 1 lbm--------- * ------- * --------- * ... 1 lbf h 1 lbf 0.4536 kg

[/CODE]

Or you could get fancy and use LaTex. Key in

$\$\frac{0.529\ lb}{1\ lbf\ h} \cdot \frac{4.448\ N}{1\ lbf} \cdot \frac{1\ lbm}{0.4536\ kg} \cdot\ ...$\$​

and get
$$\frac{0.529\ lb}{1\ lbf\ h} \cdot \frac{4.448\ N}{1\ lbf} \cdot \frac{1\ lbm}{0.4536\ kg} \cdot\ ...$$

The "CODE" version actually looks better than the LaTeX version if we don't take care to use the "\text" command to convert math font to text font. When we do that we get
$$\frac{0.529\,\text{lb}}{1\, \text{lbf h}} \cdot \frac{4.448\, \text{N}}{1\, \text{lbf}} \cdot \frac{1\, \text{lbm}}{0.4536\, \text{kg}} \cdots $$

  • #6

jbriggs444

Science Advisor

Homework Helper

12,795
7,428

Ray Vickson said:

The "CODE" version actually looks better than the LaTeX version if we don't take care to use the "\text" command

Thank you for that tip!

  • #7

yecko

Gold Member

279
15

regardless the way I typed the equation, is there ways so that I can obtain the unit as (N/S)/N? (I am not able to corrected the original post...)
Thank you!

  • #9

mfb

Mentor

Insights Author

37,232
14,076

You still have lbf in the denominator twice. But it looks like the original value has lbf in the nominator as well, so this conversion is much easier than you think.

  • #10

yecko

Gold Member

279
15

or is lb=lbf?
$$\ cT=\frac{0.5279lb}{lbf\cdot h}=\left(\frac{0.5279lbf}{lbf\cdot h}\right)\cdot(\frac{4.448N}{lbf})\cdot(\frac{0.4536kg}{/lbm})^{-1}\cdot\left(\frac{3600s}{h}\right)^{-1}=1.49e-5\ kgN^{-1}s^{-1}$$

  • #11

yecko

Gold Member

279
15

or like this??
$$\ cT=\frac{0.5279lb}{lbf\cdot h}=\left(\frac{0.5279lbf}{lbf\cdot h}\right)\cdot\left(\frac{3600s}{h}\right)^{-1}=1.467e-4\ s^{-1}$$

  • #12

mfb

Mentor

Insights Author

37,232
14,076

Looks like it. N/N=1 so that can be included easily.

  • #13

yecko

Gold Member

279
15

yecko said:

0.5279lb

but when I searched online,the numerator lb is lbm instead of lbf... so it can't be cancel out that easily
or else lb=lbf or kg doesn't really matter in the unit, and there is only unit of time^-1 there

  • #15

Nidum

Science Advisor

Gold Member

2,993
853

Specific fuel consump:(jet a/c) lb/hr. lb thrust = 0.283 x 10-4 kg/Ns

That's the aircraft industry standard conversion . Note the use of Kg and not N in top line on RHS .

  • #16

Nidum

Science Advisor

Gold Member

2,993
853

In any conversion with mixed or odd units always check for consistency of dimensions on both sides of the conversion .

  • #17

yecko

Gold Member

279
15

Nidum said:

Specific fuel consump:(jet a/c) lb/hr. lb thrust = 0.283 x 10-4 kg/Ns

yecko said:

cT=0.5279lb/lbf*h= (0.5279lb/lbf*h)* (4.448 N/lbf) / (0.4536 kg/lbm) / (3600s/h) = 1.49e-5 kgN-1s-1

That should means my numeral calculation is correct :-)

Nidum said:

That's the aircraft industry standard conversion . Note the use of Kg and not N in top line on RHS .

Unit conversion:lbm/lbf*h to (N/s)/N (1)

This is the relevant lecture slide of my course... Do you think what mentioned by that is wrong? or they are somehow equivalent but by interpreting in different way? Thank you very much!

  • #18

mfb

Mentor

Insights Author

37,232
14,076

It is the same quantity but multiplied by g, the gravitational acceleration. Why? Who knows. For rockets this is very common (Isp in seconds).

  • #19

yecko

Gold Member

279
15

mfb said:

It is the same quantity but multiplied by g, the gravitational acceleration

multiply by g? then it has difference of 9.81!
when i use Breguet Range Equation (see the picture below), should i use cT with unit of N/Ns or kg/Ns?

Unit conversion:lbm/lbf*h to (N/s)/N (2)

thanks.

  • #20

Nidum

Science Advisor

Gold Member

2,993
853

Lift/drag and the log term are both dimensionless .

Range = a dimensionless constant x Velocity/TSFC

Dimensionally : L = L T-1 / Dimensions of TSFC

(Edit) or dimensions of TSFC = T-1

Incidentally fuel 'weight' is sometimes expressed in Kg rather than N because it makes the unit independent of g . This is important in some flight dynamics calculations because g force as experienced by the fuel in a flight vehicle can vary considerably .

Last edited:

  • #21

mfb

Mentor

Insights Author

37,232
14,076

yecko said:

multiply by g? then it has difference of 9.81!

No, the difference is 9.81 m/s2.

when i use Breguet Range Equation (see the picture below), should i use cT with unit of N/Ns or kg/Ns?
View attachment 213073
thanks.

That needs N/(Ns).

  • #22

yecko

Gold Member

279
15

Nidum said:

Incidentally fuel 'weight' is sometimes expressed in Kg rather than N because it makes the unit independent of g . This is important in some flight dynamics calculations because g force as experienced by the fuel in a flight vehicle can vary considerably .

So I have to multiply g in order to calculate range, right? Let me type my unit conversion here again:
$$\ cT=\frac{0.5279lb}{lbf\cdot h}=\frac{(0.5279lb}{lbf\cdot h})\cdot(\frac{4.448N}{lbf})\cdot(\frac{0.4536kg}{lbm})^{-1}\cdot\left(\frac{3600s}{h}\right)^{-1}$$
$$\ =1.49e-5\ kgN^{-1}s^{-1}\ \cdot\frac{9.81m}{s^2}=1.46e-4\ \frac{N}{Ns}$$
Is it correct and be ready to substitute it into Breguet Range Equation? Thank you so much!

  • #23

jbriggs444

Science Advisor

Homework Helper

12,795
7,428

yecko said:

So I have to multiply g in order to calculate range, right? Let me type my unit conversion here again:
$$\frac{0.5279 \text{ lb}} {\text{lbf} \cdot \text{h}} \cdot \frac{4.448 \text{ N}}{\text{lbf}}$$

[TeX formula cleaned up and trimmed down]

As has been pointed out repeatedly, this is the wrong way to proceed. You have lbf in the denominator and are multiplying it by a term with lbf in the denominator.

Also, your habit of taking the conversion terms and applying a (-1) exponent seems silly. Just turn the fraction upside down instead, putting the numerator on the bottom and the denominator on the top. That makes it much easier to read and keep track of everything.

  • #24

Chestermiller

Staff Emeritus

Science Advisor

Homework Helper

Insights Author

2023 Award

23,500
5,776

As a guy who has been working with these unit all my life, I can tell you that 1 lbm/hr is the same as 0.4536 kg/hr. The weight of this is (0.4546)(9.8)=4.448 N/hr. And we know that 1 lbf=4.448 N. So, to convert from imperial to metric in this case, all you need to do is divide by 3600.

  • #25

yecko

Gold Member

279
15

$$\ cT=\frac{0.5279lb}{lbf\cdot h}=\frac{(0.5279lb}{lbf\cdot h})\cdot\left(\frac{3600s}{h}\right)^{-1}=1.46e−4 N/Ns $$​

  • #26

yecko

Gold Member

279
15

Chestermiller said:

As a guy who has been working with these unit all my life, I can tell you that 1 lbm/hr is the same as 0.4536 kg/hr. The weight of this is (0.4546)(9.8)=4.448 N/hr. And we know that 1 lbf=4.448 N. So, to convert from imperial to metric in this case, all you need to do is divide by 3600.

yecko said:

cT=0.5279lblbf⋅h=(0.5279lblbf⋅h)⋅(4.448Nlbf)⋅(0.4536kglbm)−1⋅(3600sh)−1 cT=0.5279lblbf⋅h=(0.5279lblbf⋅h)⋅(4.448Nlbf)⋅(0.4536kglbm)−1⋅(3600sh)−1​

\ cT=\frac{0.5279lb}{lbf\cdot h}=\frac{(0.5279lb}{lbf\cdot h})\cdot(\frac{4.448N}{lbf})\cdot(\frac{0.4536kg}{lbm})^{-1}\cdot\left(\frac{3600s}{h}\right)^{-1}

=1.49e−5 kgN−1s−1 ⋅9.81ms2=1.46e−4 NNs​

Thanks... I finally found that the two ways finally give the same answer!

  • #27

jbriggs444

Science Advisor

Homework Helper

12,795
7,428

yecko said:

$$\ cT=\frac{0.5279lb}{lbf\cdot h}=\frac{(0.5279lb}{lbf\cdot h})\cdot\left(\frac{3600s}{h}\right)^{-1}=1.46e−4 N/Ns $$​

You are equating ##\frac {\text{lb}_{\text{force}}} {\text{lb}_{\text{mass}}}## with ##\frac {\text{N}_{\text{mass}}} {\text{N}_{\text{force}}}##.

That appears to be OK, since the intended output units do appear to be implicitly using Newton-mass. [Ordinarily, I'd run screaming if someone tried to use the Newton as a unit of mass instead of the kilogram.

  • #28

Chestermiller

Staff Emeritus

Science Advisor

Homework Helper

Insights Author

2023 Award

23,500
5,776

jbriggs444 said:

You are equating ##\frac {\text{lb}_{\text{force}}} {\text{lb}_{\text{mass}}}## with ##\frac {\text{N}_{\text{mass}}} {\text{N}_{\text{force}}}##.

That appears to be OK, since the intended output units do appear to be implicitly using Newton-mass. [Ordinarily, I'd run screaming if someone tried to use the Newton as a unit of mass instead of the kilogram.

Actually, the weight of 1 lbm is 1 lbf. So the original units were really ##\frac{lb_f}{lb_f-h}=1/h##

  • #29

yecko

Gold Member

279
15

Chestermiller said:

the weight of 1 lbm is 1 lbf

oh i see... i thought 1lbf is only equal to 1 slug and equal to 1 lgm times g (gravity)...
sorry for my confusion about imperial units...

  • #30

mfb

Mentor

Insights Author

37,232
14,076

That’s one of the problems of imperial units, they often lead to confusion.

FAQ: Unit conversion:lbm/lbf*h to (N/s)/N

1. What is the formula for converting lbm/lbf*h to (N/s)/N?

The formula for converting lbm/lbf*h to (N/s)/N is (lbm * 0.45359237 * 9.80665) / (lbf * 4.4482216).

2. Why is it necessary to convert units from lbm/lbf*h to (N/s)/N?

It is necessary to convert units from lbm/lbf*h to (N/s)/N in order to accurately measure and compare values in the metric system. The lbm/lbf*h unit is commonly used in the Imperial system, while (N/s)/N is the equivalent unit in the metric system.

3. How do I convert a specific value from lbm/lbf*h to (N/s)/N?

To convert a specific value from lbm/lbf*h to (N/s)/N, simply multiply the value in lbm/lbf*h by the conversion factor of (lbm * 0.45359237 * 9.80665) / (lbf * 4.4482216).

4. Can I use an online converter to convert units from lbm/lbf*h to (N/s)/N?

Yes, there are many online converters available that can easily convert units from lbm/lbf*h to (N/s)/N. However, it is always recommended to double check the results and calculations for accuracy.

5. Are there any other common units that can be converted to (N/s)/N?

Yes, there are several other common units that can be converted to (N/s)/N, such as kg/N*s, g/N*s, and mg/N*s. These units are also commonly used in the metric system and can be easily converted using the appropriate conversion factors.

Similar threads

Determine if a car can pop a wheelie?

    Replies
    15
    Views
    2K

    What Do Lbf, Lb, and Lbm Really Mean in American Units?

      Replies
      15
      Views
      92K

      Calculating average horsepower of an elevator help

        Replies
        4
        Views
        13K

        Force and acceleration of a rocket

          Replies
          3
          Views
          1K

          Stimulated emission per second per atom?

            Replies
            2
            Views
            1K

            Basic physics units problems involving the Ideal Gas Law

              Replies
              5
              Views
              1K

              Momentum and impulse of a football

                Replies
                2
                Views
                2K

                Calculating Aircraft Range and Endurance Using De Breguet Formula

                  Replies
                  1
                  Views
                  2K

                  What is the exhaust pipe diameter required for the given exhaust speed?

                    Replies
                    5
                    Views
                    2K

                    What Exhaust Pipe Diameter is Needed for a Given Exhaust Speed?

                      Replies
                      1
                      Views
                      1K
                      • Forums
                      • Homework Help
                      • Introductory Physics Homework Help
                      Unit conversion:lbm/lbf*h to (N/s)/N (2024)
                      Top Articles
                      Latest Posts
                      Recommended Articles
                      Article information

                      Author: Nicola Considine CPA

                      Last Updated:

                      Views: 5243

                      Rating: 4.9 / 5 (49 voted)

                      Reviews: 88% of readers found this page helpful

                      Author information

                      Name: Nicola Considine CPA

                      Birthday: 1993-02-26

                      Address: 3809 Clinton Inlet, East Aleisha, UT 46318-2392

                      Phone: +2681424145499

                      Job: Government Technician

                      Hobby: Calligraphy, Lego building, Worldbuilding, Shooting, Bird watching, Shopping, Cooking

                      Introduction: My name is Nicola Considine CPA, I am a determined, witty, powerful, brainy, open, smiling, proud person who loves writing and wants to share my knowledge and understanding with you.