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Chapter 1: Problem 25

An open storage tank is placed at the top of a building. The tank contains water up to a depth of \(1.5 \mathrm{~m}\). Calculate the pressure at the bottom of the tank. It is given that atmospheric pressure is \(101.3 \mathrm{kPa}\) and density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The pressure at the bottom of the tank is 116.015 kPa.

Step by step solution

01

Identify the given values

The depth of the water in the tank is given as 1.5 meters. The atmospheric pressure is given as 101.3 kPa, and the density of water is 1000 kg/m^3.
02

Use the formula for pressure due to a liquid column

The pressure at the bottom of a liquid column can be calculated using the formula: \[ P_{\text{liquid}} = \rho g h \] where \(\rho\) is the density of the liquid, \(g\) is the acceleration due to gravity (9.81 m/s^2), and \(h\) is the height (depth) of the liquid column.
03

Substitute the given values

Using the given values, the pressure due to the water column is: \[ P_{\text{liquid}} = (1000 \, \mathrm{kg}/\mathrm{m}^3) \cdot (9.81 \mathrm{m}/\mathrm{s}^2) \cdot (1.5 \, \mathrm{m}) = 14715 \, \mathrm{Pa} = 14.715 \, \mathrm{kPa} \]
04

Calculate the total pressure at the bottom of the tank

The total pressure at the bottom of the tank is the sum of the atmospheric pressure and the pressure due to the water column: \[ P_{\text{total}} = P_{\text{atm}} + P_{\text{liquid}} \] Given that \( P_{\text{atm}} = 101.3 \mathrm{kPa} \), the total pressure is: \[ P_{\text{total}} = 101.3 \, \mathrm{kPa} + 14.715 \, \mathrm{kPa} = 116.015 \, \mathrm{kPa} \]

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Key Concepts

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

fluid mechanics
Fluid mechanics is the study of how fluids behave when they are in motion and at rest. Fluids are substances that can flow, like liquids and gases. Using principles of fluid mechanics, engineers and scientists can predict and control how fluids will flow in various situations, such as through pipes, around aircraft wings, or in water tanks.
In our exercise, we dealt with a water tank, which is a common application of fluid mechanics. When analyzing the pressure at the bottom of the tank, one of the key factors is the depth of the water. As the depth increases, so does the pressure, due to the weight of the water above. This concept is critical in fluid mechanics for designing safe and efficient systems involving fluids.
pressure calculation
Calculating pressure in a fluid involves understanding the forces exerted by the fluid. Pressure is defined as the force per unit area exerted by a fluid against a surface. It can be calculated using the formula:
The pressure at the bottom of the tank was calculated using the formula for pressure due to a liquid column: where:

    • Once the pressure due to the water column is found, we add the atmospheric pressure to get the total pressure at the bottom of the tank. This step accounts for the fact that the water is exposed to the atmosphere at the surface.
density and pressure relationship
Density is a measure of mass per unit volume. In our case, the density of water is given as 1000 kg/m鲁. Density plays a crucial role in determining pressure in fluid mechanics. The pressure at a certain depth in a fluid is directly proportional to its density.
For example, if the density of the fluid were to increase, the pressure at the same depth would also increase proportionally. The relationship between density and pressure is shown in the formula used in the solution:
Here, higher density () would result in higher pressure () at the same depth ( ). This principle is used in many practical applications, like designing dams, submarines, and even understanding the behavior of natural bodies of water.
thermodynamics applications
Thermodynamics is the study of energy, heat, and work, and how they interact. Understanding the principles of thermodynamics can help in analyzing systems involving fluids. For instance, in the given problem, knowing how temperature affects water density could refine our pressure calculations.
In thermodynamics, every fluid鈥檚 behavior can change with temperature variations. For water in a tank, changes in temperature could cause expansion or contraction, altering its depth and pressure at the bottom. Applications of thermodynamics include engines, refrigerators, and even atmospheric sciences.
  • By integrating thermodynamics into fluid mechanics, we get a comprehensive understanding of how fluids respond to various conditions, whether in industrial processes or natural phenomena.

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Most popular questions from this chapter

A gas initially at \(p_{1}=1\) bar and occupying a volume of 1 liter is compressed within a piston-cylinder assembly to a final pressure \(p_{2}=4\) bar. (a) If the relationship between pressure and volume during the compression is \(p V=\) constant, determine the volume, in liters, at a pressure of 3 bar. Also plot the overall process on a graph of pressure versus volume. (b) Repeat for a linear pressure-volume relationship between the same end states.

Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at \(8 \mathrm{MPa}\), \(440^{\circ} \mathrm{C}\), and the condenser pressure is \(8 \mathrm{kPa}\). Steam expands through the first- stage turbine to \(0.5 \mathrm{MPa}\) and then is reheated to \(440^{\circ} \mathrm{C}\). Determine for the cycle (a) the rate of heat addition, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of steam entering the first-stage turbine. (b) the thermal efficiency. (c) the rate of heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per \(\mathrm{kg}\) of steam entering the first-stage turbine.

Convert the following temperatures from \({ }^{\circ} \mathrm{C}\) to \(\mathrm{K}:\) (a) \(21^{\circ} \mathrm{C}\). (b) \(-40^{\circ} \mathrm{C}\), (c) \(500^{\circ} \mathrm{C}\), (d) \(0^{\circ} \mathrm{C}\), (e) \(100^{\circ} \mathrm{C}\), (f) \(-273.15^{\circ} \mathrm{C}\).

Water is the working fluid in an ideal regenerative Rankine cycle with one open feedwater heater. Superheated vapor enters the first-stage turbine at \(14 \mathrm{MPa}, 520^{\circ} \mathrm{C}\), and the condenser pressure is \(6 \mathrm{kPa}\). The mass flow rate of steam entering the first-stage turbine is \(90 \mathrm{~kg} / \mathrm{s}\). Steam expands through the first-stage turbine to \(0.9 \mathrm{MPa}\), where some of the steam is extracted and diverted to an open feedwater heater operating at \(0.9 \mathrm{MPa}\). The remainder expands through the second-stage turbine to the condenser pressure of \(6 \mathrm{kPa}\). Saturated liquid exits the feedwater heater at \(0.9 \mathrm{MPa}\). Determine the rate of exergy input to the working fluid passing through the steam generator, in MW. Let \(T_{0}=298 \mathrm{~K}\) and \(p_{0}=1\) bar. Also, determine the rate of exergy destruction in the open feedwater heater, in \(\mathrm{MW}\).

Water is the working fluid in a Rankine cycle modified to include one closed feedwater heater and one open feedwater heater. Superheated vapor enters the turbine at \(16 \mathrm{MPa}, 560^{\circ} \mathrm{C}\), and the condenser pressure is \(8 \mathrm{kPa}\). The mass flow rate of steam entering the first- stage turbine is \(120 \mathrm{~kg} / \mathrm{s}\). The closed feedwater heater uses extracted steam at \(4 \mathrm{MPa}\), and the open feedwater heater uses extracted steam at \(0.3\) MPa. Saturated liquid condensate drains from the closed feed water heater at \(4 \mathrm{MPa}\) and is trapped into the open feedwater heater. The feedwater leaves the closed heater at \(16 \mathrm{MPa}\) and a temperature equal to the saturation temperature at \(4 \mathrm{MPa}\). Saturated liquid leaves the open heater at \(0.3\) MPa. Assume all turbine stages and pumps operate isentropically. Determine (a) the net power developed, in kW. (b) the rate of heat transfer to the steam passing through the steam generator, in \(\mathrm{kW}\). (c) the thermal efficiency. (d) the mass flow rate of condenser cooling water, in \(\mathrm{kg} / \mathrm{s}\), if the cooling water undergoes a temperature increase of \(18^{\circ} \mathrm{C}\) with negligible pressure change in passing through the condenser.
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