The ultimate BOD of a river just below a sewage outfall is 50.0 mg/L, and the oxygen deficit at the outfall D0 is 2.0 mg/L. The
saturation value of dissolved oxygen is 11.0 mg/L. The deoxygenation rate coefficient kd is 0.30/day, and the reaeration rate coefficient kr is 0.90/day. The river is flowing at the speed of 45.0 miles per day. The only source of BOD on this river is this single outfall.
A. Find the critical distance downstream at which DO is a minimum.
B. Find the minimum O.
A. Find the critical distance downstream at which DO is a minimum.
B. Find the minimum O.
1 answer:
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Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:
It is necessary to get the Reynold's number:
Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:
The overall heat transfer coefficient:
Here
Substituting values:
Answer:
save space = 1.9626 Gibibits
Explanation:
given data
recording song time = 3 minute = 180 seconds
song per sample = 24 bits per sample
frequency = 96 kHz
surround sound system = 5.1
solution
first we get here required space for 24 bits at 96 kHz that is
required space = 24 × 96 × 10³ × 6 × 180
required space = 2.48832 × bits
as 1 Gib bit = bits
so required space = 2.48832 × bits ÷
bits
required space = 2.3174 Gibibits ...............1
and
space required to save record 16 bit at 44.1 kHz
space required = 16 × 44.1 × 10³ × 3 × 180
space required = 0.381024 × bits
space required = 0.381024 × bits ÷
bits
space required = 0.3548 Gibibits ...........2
so
we get here that save space in 16 bit at 44.1 kHz
save space = 2.3174 Gibibit - 0.3548 Gibibits
save space = 1.9626 Gibibits