Simply throw the gateway into a river.

Let's say there's 2000 cubic meters of water per second flowing out the river. After 12 seconds, there's 4000 cubic meters of flow, after 24 seconds there's 6000 and so on.

As the duplicates are all created at once, there won't be any backpressure reducing the flow.

After a day, you'll have imported 622 cubic kilometers of water. After a year, 82 million cubic kilometers of water. After 8 years and 5.3 billion cubic kilometers of water, Mount Everest would be under water.

Mission accomplished.

Edit: Some people in the comments don't understand how I've arrived at these numbers, so here's how it breaks down:

• The question, at the time I've answered it, doesn't specify a time or length limit for objects passing through the portal, or that the first duplicate has to finish coming out before the second duplicate starts coming out.
• However, it does specify the same thing will be produced over and over, at a rate of about 5 per minute (i.e. 1 every 12 seconds)
• Therefore, if you divert a river with flow $n$ cubic meters per second into the portal, at the output, the flow rate after $t$ seconds will be $n*((t/12)+1)$ i.e. a flow of $n$ at zero seconds, $2n$ after 12 seconds when the first clone starts emerging from the portal, $3n$ after 24 seconds when the second clone also starts emerging, and so on.
• And the cumulative volume after $u$ seconds will be the integral of the flow rate - $$\int_0^u n*((t/12)+1) dt = n*u^2/24 + n*u$$
• Hence, if the flow rate $n=2000$ cubic meters per second, after 8 years $u=252288000$ giving a cumulative volume in cubic meters of $$2000*252288000^2/24 + 2000*252288000 = 5304103416576000000$$
• Divide by $10^9$ to get cubic kilometers, and $10^9$ again to get billions of cubic kilometers, and you arrive at 5.3 billion cubic kilometers.
• Granted, 2000 cubic meters of water per second is on the high side - but the Hoover dam's spillway has a capacity of $11,000 m^3/s$ - and thanks to the exponential, even if you had a more modest 150 cubic meters per second you could still have Everest under water in less than 90 years and massive flooding, food shortages and population displacement well before that.

Simply throw the gateway into a river.

Let's say there's 2000 cubic meters of water per second flowing out the river. After 12 seconds, there's 4000 cubic meters of flow, after 24 seconds there's 6000 and so on.

As the duplicates are all created at once, there won't be any backpressure reducing the flow.

After a day, you'll have imported 622 cubic kilometers of water. After a year, 82 million cubic kilometers of water. After 8 years and 5.3 billion cubic kilometers of water, Mount Everest would be under water.

Mission accomplished.

Edit: Some people in the comments don't understand how I've arrived at these numbers, so here's how it breaks down:

• The question, at the time I've answered it, doesn't specify a time or length limit for objects passing through the portal, or that the first duplicate has to finish coming out before the second duplicate starts coming out.
• However, it does specify the same thing will be produced over and over, at a rate of about 5 per minute (i.e. 1 every 12 seconds)
• Therefore, if you divert a river with flow $n$ cubic meters per second into the portal, at the output, the flow rate after $t$ seconds will be $n*((t/12)+1)$ i.e. a flow of $n$ at zero seconds, $2n$ after 12 seconds when the first clone starts emerging from the portal, $3n$ after 24 seconds when the second clone also starts emerging, and so on.
• And the cumulative volume after $u$ seconds will be the integral of the flow rate - $$\int_0^u n*((t/12)+1) dt = n*u^2/24 + n*u$$
• Hence, if the flow rate $n=2000$ cubic meters per second, after 8 years $u=252288000$ giving a cumulative volume in cubic meters of $$2000*252288000^2/24 + 2000*252288000 = 5304103416576000000$$
• Divide by $10^9$ to get cubic kilometers, and $10^9$ again to get billions of cubic kilometers, and you arrive at 5.3 billion cubic kilometers.
• Granted, 2000 cubic meters of water per second is on the high side - but the Hoover dam's spillway has a capacity of $11,000 m^3/s$ - and thanks to the exponential, even if you had a more modest 150 cubic meters per second you could still have a meter of sea level rise in a year (resulting in massive food shortages and and population displacement) and Everest under water in less than 90 years

Simply throw the gateway into a river.

Let's say there's 2000 cubic meters of water per second flowing out the river. After 12 seconds, there's 4000 cubic meters of flow, after 24 seconds there's 6000 and so on.

As the duplicates are all created at once, there won't be any backpressure reducing the flow.

After a day, you'll have imported 622 cubic kilometers of water. After a year, 82 million cubic kilometers of water. After 8 years and 5.3 billion cubic kilometers of water, Mount Everest would be under water.

Mission accomplished.

Simply throw the gateway into a river.

Let's say there's 2000 cubic meters of water per second flowing out the river. After 12 seconds, there's 4000 cubic meters of flow, after 24 seconds there's 6000 and so on.

As the duplicates are all created at once, there won't be any backpressure reducing the flow.

After a day, you'll have imported 622 cubic kilometers of water. After a year, 82 million cubic kilometers of water. After 8 years and 5.3 billion cubic kilometers of water, Mount Everest would be under water.

Mission accomplished.

Edit: Some people in the comments don't understand how I've arrived at these numbers, so here's how it breaks down:

• The question, at the time I've answered it, doesn't specify a time or length limit for objects passing through the portal, or that the first duplicate has to finish coming out before the second duplicate starts coming out.
• However, it does specify the same thing will be produced over and over, at a rate of about 5 per minute (i.e. 1 every 12 seconds)
• Therefore, if you divert a river with flow $n$ cubic meters per second into the portal, at the output, the flow rate after $t$ seconds will be $n*((t/12)+1)$ i.e. a flow of $n$ at zero seconds, $2n$ after 12 seconds when the first clone starts emerging from the portal, $3n$ after 24 seconds when the second clone also starts emerging, and so on.
• And the cumulative volume after $u$ seconds will be the integral of the flow rate - $$\int_0^u n*((t/12)+1) dt = n*u^2/24 + n*u$$
• Hence, if the flow rate $n=2000$ cubic meters per second, after 8 years $u=252288000$ giving a cumulative volume in cubic meters of $$2000*252288000^2/24 + 2000*252288000 = 5304103416576000000$$
• Divide by $10^9$ to get cubic kilometers, and $10^9$ again to get billions of cubic kilometers, and you arrive at 5.3 billion cubic kilometers.
• Granted, 2000 cubic meters of water per second is on the high side - but the Hoover dam's spillway has a capacity of $11,000 m^3/s$ - and thanks to the exponential, even if you had a more modest 150 cubic meters per second you could still have Everest under water in less than 90 years and massive flooding, food shortages and population displacement well before that.