Question

what would the detention time be, in minutes for a tank with a diameter of 30 feet and a depth of 20 feet with a flow of 1,000,000 gallons per day and a effluent flow of 900 gpm?

Explanation:

Step1: Calculate tank volume (cubic feet)

Tank is cylindrical, volume formula \( V = \pi r^2 h \). Diameter \( d = 30 \) ft, so radius \( r=\frac{30}{2}=15 \) ft, depth \( h = 30 \) ft.
\( V=\pi\times15^2\times30 = \pi\times225\times30 = 6750\pi \approx 6750\times3.1416 \approx 21205.75 \) cubic feet.

Step2: Convert volume to gallons

1 cubic foot ≈ 7.48052 gallons.
Tank volume in gallons: \( 21205.75\times7.48052 \approx 21205.75\times7.48052 \approx 158630 \) gallons (approximate, more precise: \( 6750\pi\times7.48052 \approx 6750\times3.14159\times7.48052 \approx 6750\times23.4582 \approx 158343 \) gallons).

Step3: Determine inflow rate in gpm

Inflow is 1,000,000 gallons per day. There are \( 24\times60 = 1440 \) minutes in a day.
Inflow rate \( Q_{in}=\frac{1000000}{1440}\approx 694.44 \) gpm.

Step4: Determine net inflow rate

Effluent flow \( Q_{out}=900 \) gpm. Since \( Q_{in}<Q_{out} \), net inflow is negative (tank is draining), but detention time is based on volume and net flow? Wait, maybe misinterpretation: maybe inflow is 1,000,000 gpd (gallons per day) and effluent is 900 gpm. Wait, 900 gpm is \( 900\times1440 = 1,296,000 \) gpd, which is more than inflow. So tank is losing water. But detention time is usually \( T=\frac{V}{Q_{net}} \), where \( Q_{net}=Q_{in}-Q_{out} \). If \( Q_{in}=1,000,000 \) gpd = \( \frac{1,000,000}{1440}\approx694.44 \) gpm, \( Q_{out}=900 \) gpm, so \( Q_{net}=694.44 - 900=-205.56 \) gpm (outflow). But detention time for a tank with inflow and outflow: if we consider the time to fill or empty? Wait, maybe the problem has a typo, or maybe inflow is 1,000,000 gpd and effluent is 900 gpm, but let's check units again. Wait, 1,000,000 gpd is 694.44 gpm, effluent 900 gpm: so net outflow is 205.56 gpm. But detention time is volume divided by net flow (absolute value if we consider time to empty? No, detention time is usually \( T = \frac{V}{Q_{in}} \) if no outflow, but with outflow, it's \( T=\frac{V}{Q_{in}-Q_{out}} \), but if \( Q_{in}<Q_{out} \), it's negative, which doesn't make sense. Alternatively, maybe the inflow is 1,000,000 gpm? No, that's too high. Wait, maybe the problem meant inflow is 1,000,000 gpd and effluent is 900 gpd? No, 900 gpd is too low. Wait, maybe I misread: "flow of 1,000,000 gallons per day and a effluent flow of 900 gpm". Let's recast:

Wait, maybe the tank is being filled at 1,000,000 gpd and drained at 900 gpm. Let's compute the time to fill the tank (if inflow > outflow) or empty (if outflow > inflow).

First, confirm tank volume:

Diameter 30 ft, depth 30 ft: cylindrical tank, volume \( V = \pi r^2 h = \pi (15)^2 (30) = 6750\pi \approx 21205.75 \) cubic feet.

Convert to gallons: 1 cubic foot = 7.48052 gallons, so \( V = 21205.75 \times 7.48052 \approx 158630 \) gallons (as before).

Inflow rate: 1,000,000 gpd = \( \frac{1,000,000}{24 \times 60} \approx 694.44 \) gpm.

Effluent rate: 900 gpm.

Net flow rate: \( 694.44 - 900 = -205.56 \) gpm (negative means outflow, so the tank is emptying at 205.56 gpm).

Detention time (time to empty the tank) would be \( T = \frac{V}{|Q_{net}|} = \frac{158630}{205.56} \approx 771.7 \) minutes? Wait, no, that can't be. Wait, maybe the inflow is 1,000,000 gpm? No, that's impossible. Wait, maybe the effluent is 900 gpd? No, 900 gpd is 0.625 gpm, which is too low.

Wait, maybe the problem has a mistake, and the inflow is 1,000,000 gpm (but that's 1.44e9 gpd, too big) or effluent is 900 gpd. Alternatively, maybe I misread the numbers: maybe inflow is 1,000,000 gpd and effluent is 900 gpd? No, 900 gpd is 0.625 gpm.

Wait, another approach: maybe the question is about detention time as \( T = \frac{V}{Q} \), where \( Q \) is the inflow rate (ignoring effluent, but that's not standard). If we take inflow as 1,000,000 gpd = 694.44 gpm, then \( T = \frac{158630}{694.44} \approx 228.4 \) minutes. But if effluent is 900 gpm, which is higher than inflow, the tank can't fill, it's emptying.

Wait, maybe the numbers are reversed: inflow 900 gpm (which is 900*1440=1,296,000 gpd) and effluent 1,000,000 gpd (694.44 gpm). Then net inflow is 900 - 694.44 = 205.56 gpm. Then detention time \( T = \frac{158630}{205.56} \approx 771.7 \) minutes.

But the problem says "inflow of 1,000,000 gallons per day and effluent flow of 900 gpm". Let's check units again:

1,000,000 gpd = 1,000,000 / 1440 ≈ 694.44 gpm (inflow).

Effluent: 900 gpm (outflow).

Net flow: 694.44 - 900 = -205.56 gpm (outflow, so tank is emptying).

Volume of tank: ~158,630 gallons.

Time to empty: 158630 / 205.56 ≈ 771.7 minutes, ~12.86 hours.

But maybe the problem meant inflow is 1,000,000 gpm (but that's impossible) or effluent is 900 gpd. Alternatively, maybe the depth is 30 feet, diameter 30 feet, so volume is \( \pi*(15)^2*30 = 6750\pi \) cubic feet ≈ 21205.75 cubic feet ≈ 21205.75*7.48052 ≈ 158,630 gallons (correct).

Inflow: 1,000,000 gpd = 694.44 gpm.

Effluent: 900 gpm.

Net flow: -205.56 gpm (outflow).

Detention time (time to empty) = 158630 / 205.56 ≈ 772 minutes (approx).

But this seems odd. Maybe the problem has a typo, and the effluent is 900 gpd, not gpm. Let's check that: 900 gpd = 0.625 gpm. Then net flow = 694.44 - 0.625 ≈ 693.815 gpm. Detention time = 158630 / 693.815 ≈ 228.6 minutes, ~3.8 hours.

Alternatively, maybe the inflow is 1,000,000 gpm (but that's 1.44e9 gpd, impossible).

Wait, maybe the question is about retention time, which is \( T = \frac{V}{Q} \), where \( Q \) is the average flow. If we take the inflow as 1,000,000 gpd = 694.44 gpm, and ignore effluent (but that's not right), then \( T = 158630 / 694.44 ≈ 228 \) minutes.

But given the numbers, the most plausible (assuming a typo where effluent is 900 gpd instead of gpm, or inflow is 1,000,000 gpm which is wrong), but based on given numbers:

Wait, maybe I misread the effluent flow: is it 900 gpm or 900 gpd? The handwriting says "900gpm". So with that, net flow is negative, so detention time (time to empty) is ~772 minutes.

But let's recalculate:

Tank volume: \( V = \pi r^2 h = \pi (15)^2 (30) = 6750\pi \approx 21205.75 \) ft³.

Gallons: \( 21205.75 \times 7.48052 \approx 158630 \) gal.

Inflow rate: \( Q_{in} = \frac{1,000,000}{24 \times 60} \approx 694.44 \) gpm.

Effluent rate: \( Q_{out} = 900 \) gpm.

Net flow: \( Q_{net} = Q_{in} - Q_{out} = 694.44 - 900 = -205.56 \) gpm (outflow, so the tank is emptying at 205.56 gpm).

Detention time (time to empty the tank) = \( \frac{V}{|Q_{net}|} = \frac{158630}{205.56} \approx 771.7 \) minutes, approximately 772 minutes.

Answer:

Approximately \(\boxed{772}\) minutes (note: depends on interpretation, may have typo in problem).