STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

MWD and LWD 969
(4-203)
(4-204)
Solution
1. Dissipated energy: 852,000 J/min
Massic mud flow: 2721 kg/min
Calories required to raise temperature by IOC: 2,095,170 cal/min
Calories available: 203,828 cal/min
Temperature rise: AT = 0.1"C = 0.17"F
2. Heat given up by the inflowing water equals heat received by the mud.
AT = 2.5"F
3. Bottomhole pressure: 6240 psi
Gas flow at downhole conditions: 286,000 ft3/d = 8,099 mg/d = 5.62 m3/min
Gas pressure decrease: 760 psi = 5,239,440 Pa
a. Energy absorbed by the gas if E = VdP = 29,468,211 J/min = 7,049,811
cal/min
Temperature decrease of the mud: 3.36"C = 6°F
b. Energy absorbed by the gas in isentropic process = 600 Btu/lb mole
(See [113], p. 96)
When converted and for massic flowrate of 443.6 lbm/min = 2,501,393
cal/min
Temperature decrease of the mud: 1.19"C = 2.15"F
(second calculation is probably more correct)
Example 14: Drilling Parameters-Drill Collar Pressure Drop
The following data characterize a well during drilling:
depth: 10,000 ft
43-in. drillpipes (ID = 3.64 in.)
mud specific weight: 12 lb/gal
flowrate: 500 gal/min
three-bit nozzles: %-in. diameter
mud viscosity: 12 cp
nozzle factor: C = 1.0
hole diameter: 8.5 in.
1. Assuming no cutting in the annulus, compute the pressure recorded inside
the drill collars downhole and the pressure in the standpipe at surface
using the formula given hereafter.
2. A leak develops in the pipe string. The standpipe pressure reading drops
to 1,896 psi with the same mud flowrate and the downhole drill collar
inside pressure drops to 6,700 psi. What is the flowrate of the leak? What
is the area of the leaking hole assuming it is located at 3,000 or 5,000 or
7,000 ft? (Assume that AP annulus does not change.)