260 Professor Gordon on the Discharge of Water through Pipes, 



that under a constant head or pressure of ten feet of water it may 

 discharge neither more nor less than 100 imperial gallons per minute? 

 In the case proposed by Mr. Wilson to test the formula here arrived 

 at, the velocity through the pipe must evidently be very considerable, 

 and hence it was tliought best to choose the constant k = -006. 



Now, if Q = the quantity of water = -— — - c, by substituting this 



144 Q I /144 0\ ^^^ 



in the last we have c = „ and hence h = -006 - + ( — -r-^ ) 



. •. A = Yg Q^ ~ "T^Ts Q ^^^ ^^^ ^^^^ ^^^s latter we have r ^ 



= — r— Q^ a formula very convenient for calculation by logarithms, 



as appears by the following actual working of the question proposed to 

 test the formula : — 



Q = 100 gallons '2^^ cubic feQi per second. 

 I = 8 feet 

 A= 10 feet. 



Log. -2666 = 1'4259677 



7 



I power = 4 | 5.9817739 



Log. 98967 = —2-9954434 



Log. 4-84 X 8 =.38-72 = 1-5879353 



0-5833787 

 Log. 10 = 1-0000000 



1-5833787 

 4 



j% root =19 1 2-3335148 

 Log. -81713 = 1-9122902 

 Then -81713 + -01 = -827 = r, therefore 



-827 X 2 = 1-654 is the diameter of the pipe 

 according to the formula. 



According to the subjoined report of the engineer who made the 

 experiments for Mr. Wilson, the pipe found experimentally to fulfil 

 the required conditions was one inch and |^ = 1-6875 

 Formula gives, . . . 1'6544 



DiflPerence, . . . 00331 



In a second case, where all other data remain the same, save that 

 the discharge is limited to 80 gallons per minute, the pipe fitted ex- 

 perimentally was . . l'5Q25 inches diameter. 

 By formula it should be . . 1-5315 



Difference, 0.0310 



