The Idea |

Safe Harbor Statement: This was/is done for fun. I’m not by any means an engineer – in fact I was a CDAT in the Army (Computerized Dumb A$$ Tanker). I figure if I can violate reality so that I have a 30-in narrow gauge railroad still in operation in the later 30’s/early 40’s .. then I can violate other things .. like .. physics. 🙂
This was originally a thread on Railroad Line forums titled – The Caddo Creek Dam
While at the NNGC I was talking with someone about my plans for the coke oven bank. I explained that in the design of the ovens at the Wharton Coke Works in PA they used 70-lb rail as cross-rails to carry the larry rail. I was going on about going to have to use plastic I-Beams for the cross-rail to keep from shorting the larry-rail. I said that I would prob use a gas-mechanical of some sort to move the larry .. that in the ‘real world’ they used electric w/trolley poles but that I didn’t think my little bank of 16 ovens had a reason to get the electric out to my ovens. Well then .. it was suggested. You have a river .. why not build a small power station and .. DIY? Well .. OH! Says I .. and the idea of a small dam to supply an equally small power station appealed to me. So .. rather then just toss up a hunk of stone and go .. “that’s a dam” I decided to do a little research .. after all .. how much power was needed? How much power could my little stream provide? Let’s see .. |

Where to put it? |

If I put the dam JUST (and I mean JUST) upstream of the pumphouse that would make for some cool modeling .. said I to me. In the photo .. where the blue is.
That’s 6-in wide there. I thought that a 3-in high dam would be 12-ft high (in REAL life) and not overwhelm the pumphouse. The next couple of questions of course are inter-twinned .. each feeds the other. What size and kind of dam?, how much water is flowing, how much power can be produced, how much power is needed .. and so on. Luckily, a book published in 1920 and titled “Power development of small streams” answers most of these questions. |

How much power? |

Confession: I enjoy researching these projects as much I think as actually building them. Finding how it was done originally – well .. that’s being allowed to look over some one’s shoulder ‘back when’. The math bit .. well .. again that’s just something I find mildly amusing. Someone can walk up and say “You know .. that little dam is nice but it would NEVER provide the power that your coke ovens require..” .. to which .. I have an answer – backed by the ‘amusing math’.
Understand that all a dam does is .. well .. dam up water. The dam doesn’t change the amount of water flowing in the stream .. it just provides a head .. the higher water falls the more pressure it exerts at the turbine. More on that later but first the question was .. how much water DOES my little stream flow? My stream is 6-in wide on my O scale module. That’s set by the rock walls of the ravine just upstream of the pump-house. Later it widens where it passes under the bridge but .. the amount of water that flows remains constant. Since my stream is .. well .. imaginary .. it could be any depth and the water could flow at any speed (ok. REASONABLE depth and speed). Time for some numbers: Assumption: Measuring the stream immediately up-stream of the pump-house it has a width 24-ft (6-in). Now, let’s ‘suppose’ that it is 3-ft deep at the center of the stream there and the banks slope at 45°. That makes it simple to calculate the sq.ft. of that section .. (24 x 3)-3^2. That would give me 63 sq.ft. in the section. Measure back up-river on foot and we have a slice of 63 cu.ft. Toss a stick into the river at that point and how fast does it move? Let’s ‘assume’ (like that since the numbers can be adjusted to make stuff fit later!) .. that the stick moves down the stream at THIS point .. at 2-ft a second which seems reasonable. That would mean that the 63 cu.ft. slice of water is also moving at 2-ft a second. We get rate of flow by multiplying the speed, 2-ft a second, by the quantity of water, 63 cu.ft. and find that the stream flows 126 cu.ft. of water a second. Multiply by 60 and we find it flows 7,560 cu.ft. of water a minute. So. How much power can we get from this? Each cu.ft. of water weighs 62.5-lbs .. so .. 7,560 x 62.5 gives us 472,500 pounds per minute. One horse power equals 33,000 pounds dropping one foot in one minute. Dividing 472,500 by 33,000 gives us 14.32-hp! That’s without any |

[1] Archie – see comment below |

Hi

Sorry to correct you but your 14.32HP is NOT for no head. It is for unit head ie 1ft. If the head were 6″ then power would be halved, if for 2ft then 28.64hp.

Second. This assumes 100% efficiency. A small turbine like this might expect 80-85% efficiency so reduce figures accordingly.

Archie

Archie. Thanks for your reply. You’re correct. That was the ‘non-engineer’ in me talking. I DID mention taking a 1-ft thick ‘slice’ of the water in my calculations to figure the water available to power the turbine. Unfortunately logic is often left to fight with a nice cold beer when I am going this kind of stuff .. the results you see. 🙂

I’ll re-write that bit then.

I’m not sure if that 80-85% efficiency was addressed at some point in the bit of the thread I haven’t posted here yet on the turbines but I will make a point of addressing that also. I looked back and I did say that “

Of course there will be loss from friction and you have to take the efficiency of the turbine into account”Thanks for the input.