Answer:
474.59 mg/L
Explanation:
Given that
BOD = 30 mg/L
Original BOD = 30 mg/L × dilution factor
Original BOD = 30 mg/L × 10 = 300 mg/L
here is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day
Consider a normal shock wave in air. The upstream conditions are given by M1=3, p1 = 1 atm, and r1 = 1.23 kg/m3. Calculate the d
ownstream values of p2, T2, r2, M2, u2, p02 and T02. (Anderson 3.4)
1 answer:
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Answer:
coupling is in tension
Force = -244.81 N
Explanation:
Diameter of Hose ( D1 ) = 35 mm
Diameter of nozzle ( D2 ) = 25 mm
water gage pressure in hose = 510 kPa
stream leaving the nozzle is uniform
exit speed and pressure = 32 m/s and atmospheric
<u>Determine the force transmitted by the coupling between the nozzle and hose </u>
attached below is the remaining part of the detailed solution
Inlet velocity ( V1 ) = V2 ( D2/D1 )^2
= 32 ( 25 / 35 )^2
= 16.33 m/s
Answer:
778.4°C
Explanation:
I = 700
R = 6x10⁻⁴
we first calculate the rate of heat that is being transferred by the current
q = I²R
q = 700²(6x10⁻⁴)
= 490000x0.0006
= 294 W/M
we calculate the surface temperature
Ts = T∞ +
Ts =
The surface temperature is therefore 778.4°C if the cable is bare