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 mgd? and a effluent flow of 900 gpm?

Explanation:

Step1: Calculate the volume of the tank

The tank is cylindrical, so the volume \( V \) of a cylinder is given by the formula \( V=\pi r^{2}h \), where \( r \) is the radius and \( h \) is the height (depth). The diameter is 30 feet, so the radius \( r = \frac{30}{2}=15 \) feet, and the depth \( h = 20 \) feet.
\[ V=\pi\times(15)^{2}\times20=\pi\times225\times20 = 4500\pi\approx4500\times3.1416 = 14137.2\text{ cubic feet} \]
We know that 1 cubic foot is approximately 7.48052 gallons. So convert the volume to gallons:
\[ V_{gallons}=14137.2\times7.48052\approx105737.4\text{ gallons} \]

Step2: Determine the net inflow rate

The inflow rate is 1 MGD (million gallons per day). First, convert 1 MGD to gallons per minute (gpm). There are 1440 minutes in a day.
\[ \text{Inflow rate (gpm)}=\frac{1000000}{1440}\approx694.44\text{ gpm} \]
The effluent flow is 900 gpm. Wait, if the effluent flow is higher than the inflow, there might be a mistake? Wait, maybe the inflow is 1 MGD (1,000,000 gpd) and effluent is 900 gpm. Wait, let's re - check the net flow. Net inflow \( Q_{net}=\) Inflow - Effluent. But if inflow is 1 MGD = 1000000 gpd=\(\frac{1000000}{1440}\approx694.44\) gpm and effluent is 900 gpm, then the net inflow is negative, which doesn't make sense for detention time (detention time is for when there is a storage, maybe the question has a typo, or maybe the inflow is 1 MGD and effluent is 900 gpd? Wait, no, the user wrote "a flow of 1 MGD? and a effluent flow of 900 gpm". Let's assume that maybe it's a typo and the inflow is 1000 gpm (close to 1 MGD is 1000000 gpd≈694 gpm, but maybe the inflow is 1000 gpm). Wait, or maybe the question is about detention time which is volume divided by net flow. Wait, maybe the inflow is 1 MGD (inflow) and effluent is 900 gpm, but let's proceed with the given numbers. Wait, 1 MGD is 1,000,000 gallons per day. Let's convert 1 MGD to gpm: \( \frac{1000000}{1440}\approx694.44\) gpm. The effluent is 900 gpm. So the net flow \( Q = 694.44-900=- 205.56\) gpm. Negative flow means the tank is emptying, which is not for detention time. So maybe the inflow is 1000 gpm (approx 1.44 MGD) and effluent is 900 gpm. Let's assume that the inflow is 1000 gpm (maybe a mis - write of 1 MGD as 1000 gpm? No, 1 MGD is 694 gpm). Alternatively, maybe the question is that the inflow is 1 MGD and the effluent is 900 gpd? No, the user wrote 900 gpm. Wait, maybe the diameter is 30 feet, depth 20 feet, inflow 1 MGD, effluent 900 gpm. Let's calculate the volume again. Radius \( r = 15\) ft, height \( h = 20\) ft. Volume \( V=\pi r^{2}h=\pi\times15^{2}\times20 = 4500\pi\approx14137.2\) cubic feet. Convert to gallons: \( 14137.2\times7.48052\approx105737\) gallons. Now, if the inflow is 1 MGD = 1000000 gpd=\(\frac{1000000}{1440}\approx694.44\) gpm and effluent is 900 gpm, the net flow is \( 694.44 - 900=-205.56\) gpm. Detention time \( t=\frac{V}{|Q|}=\frac{105737}{205.56}\approx514.4\) minutes. But this is when the tank is emptying. Alternatively, if the inflow is 1000 gpm (and effluent 900 gpm), then net flow \( Q = 1000 - 900 = 100\) gpm. Then detention time \( t=\frac{105737}{100}\approx1057.37\) minutes. But since the original inflow is 1 MGD, let's check the correct conversion. 1 MGD = 1,000,000 gallons per day. 1 day = 1440 minutes. So inflow rate \( Q_{in}=\frac{1000000}{1440}\approx694.44\) gpm. Effluent rate \( Q_{out}=900\) gpm. Net inflow \( Q = Q_{in}-Q_{out}=694.44 - 900=-205.56\) gpm. The negative sign indicates that the outflow is greater than inflow. Detention time is calculated as \( t=\frac{V}{|Q|}\), where \( V \) is the volume of the tank. So \( t=\frac{105737}{205.56}\approx514\) minutes. But this is a bit confusing. Alternatively, maybe the question has a typo and the effluent is 900 gpd instead of gpm. Let's try that. 900 gpd=\(\frac{900}{1440}=0.625\) gpm. Then net flow \( Q = 694.44-0.625 = 693.815\) gpm. Then detention time \( t=\frac{105737}{693.815}\approx152.4\) minutes. But since the user wrote 900 gpm, we have to go with that. Wait, maybe the diameter is 30 feet, depth 20 feet, inflow 1 MGD, effluent 900 gpm. Let's recalculate the volume: \( V=\pi r^{2}h=\pi\times15^{2}\times20 = 4500\pi\approx14137.2\) cubic feet. Convert to gallons: \( 14137.2\times7.48052\approx105737\) gallons. Inflow rate: 1 MGD = 1000000 gpd=\(\frac{1000000}{1440}\approx694.44\) gpm. Effluent rate: 900 gpm. Net flow \( Q = 694.44 - 900=-205.56\) gpm. Detention time \( t=\frac{V}{|Q|}=\frac{105737}{205.56}\approx514\) minutes.

Answer:

Approximately \(\boldsymbol{514}\) minutes (Note: There might be a typo in the flow rates as the effluent flow is higher than the inflow, but based on the given numbers, this is the result)