152 TRIANGULAR ORIFICE IN THE SIDE OF A VESSEL. 



Q= r*- 



*/n h 



dx 



(1) 



If the mean velocity is v, we have 

 Q = &hv, 

 which in (1) gives v = 



(2) 



(2) When the base of the triangle is in the surface 

 of the liquid. 



Let KEH be the triangular orifice, KE = b, and KH 

 = h. Then the quantity discharged through KEH will 

 equal the discharge through the rectangle KHFE, minus 

 that through the triangle EHF ; therefore subtracting (1) 

 of this Art. from (2) of Art. 85, we have 



Q = 



and 



COR. 1. If the orifice be a trapezoid ABCD, 

 whose upper base AB = b 1 lies in the sur- 

 face of the liquid, whose lower base CD = b z , 

 and whose altitude is DF = //, the discharge 

 may be found by combining the discharge 

 through the rectangle ECDF with those 

 through the two triangles ADF and BCE. 

 Hence, combining (3) with (2) of Art. 85, we have 



Q = 



(3) 

 (4) 

 A F K E B 



(5) 



