Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into the tank at rate 4 liter per minute. The tank is kept perfectly mixed and also drains at a rate of 4 liter per minute, so the volume stays constant. State the initial value problem modeling Q(t), the amount of grams of salt dissolved in the solution in the tank at time t minutes.

Sagot :

oladayoademosu

Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.