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I figured this is probably an okay place to post this question.

From Rocket Propulsion Elements: Chapter 2, Problems

I managed to get (a), (c) and (d) (quite simply), but I have absolutely no idea why I can't work out (b).

I used $$ I_{sp}=\frac{I_t}{g_0\dot{m}\Delta t} $$

I assumed thrust was constant (is this wrong?). I did also check my units.

Thanks. :)

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You got caught by those silly English units.

Your expression, $I_t = I_{sp}g_0 \dot m \Delta t$, works just fine when you use metric units. The mass flow rate, in metric units, is 127 kg/s. The total impulse, in metric units, is $$I_t = (217.5\,\text{s})(9.80665\,\text{m}/\text{s}^2)(127\, \text{kg}/\text{s})(65\,\text{s}) = 1.76\times 10^7\,\text{N}\,\text{s}$$ Converting this to customary units yields $3.96\times 10^6\,\text{lbf}\,\text{s}$.

Doing the exact same calculation, this time using customary units, yields $$I_t = (217.5\,\text{s})(32.17405\,\text{ft}/\text{s}^2)(280\, \text{lbm}/\text{s})(65\,\text{s}) = 1.2736\times 10^8\,\text{but in what units?}$$ The dimensional analysis says the units of this expression are lbm·ft/s. Those aren't the desired units; you want lbf·s. The metric F=ma becomes F=kma in customary units. Dividing by the numerical value of g0 converts lbm·ft/s2 to lbf. That same division here converts lbm·ft/s to lbf·s. And indeed, 1.2736×108/32.17405 is 3.96×106.

This suggests an alternate expression, $I_t = I_{sp} \dot m \Delta t$, where $I_{sp}$ is in seconds.This yields a numerical value of 217.5·280·65=3.96×106. That's the right value (at least numerically). Note well: This alternate expression does not work in metric units. In metric, this calculation yields a value of 1.80×106 kg·s. The correct value is 1.76×107 N·s.

Strictly speaking, the above does not have the correct units; it has units of mass*time. Numerically, it yields the right value in customary units because dividing by the numerical value of g0 cancels the explicit use of g0 in the numerator.

This calculation is not correct in metric units. To get the right metric units from this calculation one needs to multiply the numerical value of that result (1.80×106 kg·s) by the numerical value of g0. Indeed, 1.80×106 * 9.80665 = 1.76×107.

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    $\begingroup$ Argh! I hate English units! Thanks so much :) $\endgroup$
    – dunxen
    Commented Jan 6, 2015 at 17:43
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    $\begingroup$ Just wanted to chime in as an Englishman - we don't use imperial units anymore. We went formally went metric back in 1965! :p $\endgroup$
    – ThePlanMan
    Commented Jan 6, 2015 at 18:19
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    $\begingroup$ @DuncanDean - Yes. Mars Climate Orbiter, en.wikipedia.org/wiki/Mars_Climate_Orbiter . $\endgroup$ Commented Jan 6, 2015 at 18:30
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    $\begingroup$ Wow, that is interesting! However, I'm probably not going to stray too far at all from SI units doing a degree in physics. I don't know why I didn't convert everything in the question anyway. $\endgroup$
    – dunxen
    Commented Jan 6, 2015 at 19:11
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    $\begingroup$ That's my preferred technique. Convert all those nasty non-SI values to SI, do everything in metric where F=ma etc., and convert back if needed as output. $\endgroup$ Commented Jan 6, 2015 at 19:22
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You take the effective exhaust velocity and multiply it by the mass flow. That gives you the thrust. Multiply that by the operating time and you get the total impulse.

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  • $\begingroup$ This works great in metric. It doesn't with customary units: numerically, 7000*280*65=1.2736×10$^8$, but in what units? It's in the wrong units. Stupid English system strikes again! $\endgroup$ Commented Jan 6, 2015 at 17:41
  • $\begingroup$ The reason this doesn't work in English units is because exhaust velocity (in feet/second) times flow rate (in pounds/second) doesn't give thrust (in pounds force). You need to divide by the numerical value of $g_0$. Thrust in pounds is (7000 ft/s)*(280 lbm/s)*(1/32.17405 (lbf/(lbm·m/s^2))), or 60919 lbf. Multiplying this by 65 seconds yields the correct answer of 3.96 million lbf·s. $\endgroup$ Commented Jan 6, 2015 at 17:56
  • $\begingroup$ @DavidHammen Hammen: Well, sure you have to do your unit math. I don't think the answer deserves downvotes, though. $\endgroup$ Commented Jan 7, 2015 at 18:15

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