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Chapter 4: Problem 7

A cylindrical tank contains \(1500 \mathrm{~kg}\) of liquid water. It has one inlet pipe through which water is entering at a mass flow rate of \(1.2 \mathrm{~kg} / \mathrm{s}\). The tank is fitted with two outlet pipes and the water is flowing through these exit pipes at the mass flow rates of \(0.5 \mathrm{~kg} / \mathrm{s}\) and \(0.8 \mathrm{~kg} / \mathrm{s}\). Determine the amount of water that will be left in the tank after thirty minutes.

Short Answer

Expert verified
1320 kg water

Step by step solution

01

Calculate the Total Inlet Flow Rate

The inlet flow rate is provided as 1.2 kg/s. This step does not require further calculations.
02

Calculate the Total Outlet Flow Rate

The outlet flow rates are given as 0.5 kg/s and 0.8 kg/s. Add these together to get the total outlet flow rate:ening:\[ 0.5 \, \text{kg/s} + 0.8 \, \text{kg/s} = 1.3 \, \text{kg/s} \].
03

Determine the Net Flow Rate

Subtract the total outlet flow rate from the inlet flow rate to find the net flow rate:\[ 1.2 \, \text{kg/s} - 1.3 \, \text{kg/s} = -0.1 \, \text{kg/s} \]The negative sign indicates that the tank is losing water at a rate of 0.1 kg/s.
04

Calculate the Water Loss in Thirty Minutes

Convert the time from minutes to seconds:\[ 30 \, \text{minutes} \times 60 \, \text{seconds/minute} = 1800 \, \text{seconds} \]Multiply the net flow rate by the total number of seconds to determine the amount of water lost:\[ 0.1 \, \text{kg/s} \times 1800 \, \text{seconds} = 180 \, \text{kg} \]Thus, the tank loses 180 kg of water in 30 minutes.
05

Calculate the Final Amount of Water in the Tank

Subtract the water loss from the initial amount of water in the tank:\[ 1500 \, \text{kg} - 180 \, \text{kg} = 1320 \, \text{kg} \]There will be 1320 kg of water left in the tank after 30 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mass balance
The mass balance principle is fundamental in understanding many thermodynamic problems. At its core, it states that mass cannot be created or destroyed in a closed system. This means that the total mass entering a system minus the total mass leaving must equal the rate of accumulation of mass within the system.
Consider the given problem where water flows into and out of a tank. By applying the mass balance principle, we analyze the system by calculating the rates at which mass enters and exits the tank.
In this problem:
  • Inlet flow rate = 1.2 kg/s
  • Total outlet flow rate = 1.3 kg/s
These values show the mass balance in action. Applying this concept correctly determines whether the tank gains or loses mass over time.
thermodynamic systems
A thermodynamic system in this context is our cylindrical tank of water. It's also considered a control volume where mass can cross its boundaries.
In the analysis of such systems, we always look at:
  • Inlet 鈥� where the mass enters
  • Outlet 鈥� where the mass exits
For our problem, the tank is a simple, open system, where water flows in and out. The water's mass is impacted by these flows. By understanding and defining our system's boundaries (the tank edges and pipes), we can apply thermodynamic principles to solve for the system's behavior over time, like how much water remains after 30 minutes.
rate of change
The rate of change is a critical part of solving mass flow rate problems.
Here, the net flow rate reflects how fast the mass inside the tank changes. This is calculated by subtracting the total outlet flow rate from the inlet flow rate:
  • Net flow rate = Inlet flow rate - Total outlet flow rate
  • Net flow rate = 1.2 kg/s - 1.3 kg/s = -0.1 kg/s
The negative value indicates that the tank is losing water.
Then, we use this rate to find out how much water is lost over a specified time (30 minutes). We work in consistent units, converting 30 minutes to seconds and multiplying the rate by this time:
inlet and outlet flows
Understanding inlet and outlet flows is essential in mass flow rate problems. For the given exercise:
  • The inlet flow rate is the rate at which water enters the tank, given as 1.2 kg/s.
  • The outlet flow rates are the rates at which water exits the tank, given as 0.5 kg/s and 0.8 kg/s respectively.
Adding the outlet flows gives us the total flow rate out of the tank:
  • 0.5 kg/s + 0.8 kg/s = 1.3 kg/s
To find out whether the tank is gaining or losing water, we compare these values. In this scenario, the tank is losing water at a rate of 0.1 kg/s, as the outlet flows are greater than the inlet flow.
This understanding is vital for systematically and correctly solving such problems.

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Most popular questions from this chapter

Helium gas flows through a well-insulated nozzle at steady state. The temperature and velocity at the inlet are \(333 \mathrm{~K}\) and \(53 \mathrm{~m} / \mathrm{s}\), respectively. At the exit, the temperature is \(256 \mathrm{~K}\) and the pressure is \(345 \mathrm{kPa}\). The mass flow rate is \(0.5 \mathrm{~kg} / \mathrm{s}\). Using the ideal gas model, and neglecting potential energy effects, determine the exit area, in \(\mathrm{m}^{2}\).

Air enters a water-jacketed air compressor operating at steady state with a volumetric flow rate of \(37 \mathrm{~m}^{3} / \mathrm{min}\) at \(136 \mathrm{kPa}, 305 \mathrm{~K}\) and exits with a pressure of \(680 \mathrm{kPa}\) and a temperature of \(400 \mathrm{~K}\). The power input to the compressor is \(155 \mathrm{~kW}\). Energy transfer by heat from the compressed air to the cooling water circulating in the water jacket results in an increase in the temperature of the cooling water from inlet to exit with no change in pressure. Heat transfer from the outside of the jacket as well as all kinetic and potential energy effects can be neglected. (a) Determine the temperature increase of the cooling water, in \(\mathrm{K}\), if the cooling water mass flow rate is \(82 \mathrm{~kg} / \mathrm{min}\). (b) Plot the temperature increase of the cooling water, in \(\mathrm{K}\), versus the cooling water mass flow rate ranging from 75 to \(90 \mathrm{~kg} / \mathrm{min}\).

Why is it that when air at \(1 \mathrm{~atm}\) is throttled to a pressure of \(0.5 \mathrm{~atm}\), its temperature at the valve exit is close to the temperature at the valve inlet, yet when air at \(1 \mathrm{~atm}\) leaks into an insulated, rigid, initially evacuated tank until the tank pressure is \(0.5 \mathrm{~atm}\), the temperature of the air in the tank is greater than the air temperature outside the tank?

Why might a computer cooled by a constant-speed fan operate satisfactorily at sea level but overheat at high altitude?

In a steam power plant, steam flows steadily from the boiler to the turbine through a \(0.25 \mathrm{~m}\) diameter pipe. Steam conditions at the boiler end are found to be \(P=4.5 \mathrm{Mpa}\), \(T=400^{\circ} \mathrm{C}, h=3204.7 \mathrm{~kJ} / \mathrm{kg}\) and \(v=0.06475 \mathrm{~m}^{3} / \mathrm{kg}\). At the turbine end, steam conditions are found to be \(P=4 \mathrm{MPa}\), \(T=393^{\circ} \mathrm{C}, h=3196.4 \mathrm{~kJ} / \mathrm{kg}\) and \(v=0.073 \mathrm{~m}^{3} / \mathrm{kg}\). There is a heat loss of \(6 \mathrm{~kJ} / \mathrm{kg}\) from the pipeline. Compute the steam flow rate.
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