2020-06-19_REVISION - M1973021

• Permit No: M1973021
• IBM Index Class Name: REVISION
• Doc Date: 6/19/2020
• Doc Name: Request for Technical Revision
• From: Aggregate Industries
• To: DRMS
• Type & Sequence: TR7
• Email Name: ECS
• Email Name: MAC
• Email Name: AWA
• Media Type: D
• Archive: No

' Rock_Chute.xls Page 2 of 3
Rock Chute Design Calculations
' (Version W I-July-2010, Based on Design of Rock Chutes by Robinson, Rice, Kadavy, ASAE, 1998)
Project: Morrison Quarry- Basin B Outlet County: Jefferson
Designer: TPY Checked by:
' Date: 5/18/2020 Date:
I. Calculate the normal depth in the inlet channel
' High Flow Low Flow
yn= 0.33 ft, yn= 0.21 ft. (Normal depth)
Area= 0.2 ft2 Area = 0.1 ft2 (Flow area in channel)
' Qhigh= 0.6 cfs QicW = 0.2 cfs (Capacity in channel)
Scupstreamchannel= 0.111 ft/ft
' II. Calculate the critical depth in the chute
High Flow Low Flow
yc= 0.35 ft, yc= 0.22 ft. (Critical depth in chute)
Area= 0.3 ft2 Area= 0.1 ft2 (Flow area in channel)
Qhigh= 0.6 cfs QIOW= 0.2 cfs (Capacity in channel)
Hce= 0.44 ft. HCe= 0.27 ft. (Total minimum specific energy head)
' hcv= 0.09 ft. hcv= 0.05 ft. (Velocity head corresponding to yc)
10yc= 3.54 ft. (Required inlet apron length)
0.715yc= 0.25 ft. 0.715yc= 0.16 ft. (Depth of flow over the weir crest or brink)
' III. Calculate the tailwater depth in the outlet channel
High Flow Low Flow
' Tw= 0.49 ft. Tw= 0.31 ft. (Tailwater depth)
Area= 0.5 ft2 Area= 0.2 ft2 (Flow area in channel)
Qhigh = 0.6 cfs Clow= 0.2 cfs (Capacity in channel)
H2= 0.00 ft. H2= 0.00 ft. (Downstream head above weir crest,
H2=0, if Hz<0.715*yc)
IV. Calculate the head for a trapezoidal shaped broadcrested weir
Cd = 1.00 (Coefficient of discharge for broadcrested weirs)
High Flow
HP= 0.69 ft. 0.69 ft. (Weir head)
Area= 0.9 ft2 1.0 ft2 (Flow area in channel)
' Vo= 0.00 fps 0.63 fps (Approach velocity)
hPV= 0.00 ft. 0.01 ft. (Velocity head corresponding to Hp)
Qhigh= 0.6 cfs 0.6 cfs (Capacity in channel)
' Trial and error procedure solving simultaneously for velocity and head
Low Flow
HP= 0.43 ft. 0.43 ft. (Weir head)
Area= 0.4 ft2 0.4 ft2 (Flow area in channel)
' Vo = 0.00 fps 0.49 fps (Approach velocity)
hpv= 0.00 ft. 0.00 ft. (Velocity head corresponding to HP)
Q10W= 0.2 cfs 0.2 cfs (Capacity in channel)
' Trial and error procedure solving simultaneously for velocity and head