Step by step solution to "A pump is used to transport water to..."

High School
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Stacie Worsley

Answered question

2021-12-18

A pump is used to transport water to a higher reservoir. If the water temperature is

20^{\circ} C

, determine the lowest pressure that can exist in the pump without cavitation.

Answer & Explanation

Suhadolahbb

Beginner2021-12-19Added 32 answers

In this issue, a pump is working to transport water while its temperature is (20°C), The smallest possible pressure inside the pump is what we can anticipate without cavitation as following.
Cavitation is the phenomenon, which is a frequent reason for performance decline and even impeller blade erosion, when the pressure during liquid flow system drop below the vapor pressure at some places, and it resulted in unpredictable vaporization for the liquid, Cavitation bubbles collapse because the are move from the low-pressure regions, producing a high destructive pressure waves which damage the propeller.
Therefore, the flow pressure must be greater than the vapor pressure at the condition temperature in order to avoid cavitation.
At the given condition ( 20°C ) From TABLE (A-4):
$P_{v} = P_{s a t} = 2.3392 k P a$
Thus, the minimum pressure that can exist in the pump to avoid cavitation occurrence at any position on the pump surface should not be allowed to drop below $P_{v} = P_{s a t} = 2.3392 k P a$

Neil Dismukes

Beginner2021-12-20Added 37 answers

Step 1
Cavitation is the phenomenon which causes drop in performance or erosion of impeller blades, when the pressure during liquid flow system drop below the vapor pressure at some locations.
Step 2
To prevent cavitation, the pressure of the flow should be above the vapor pressure at given temperature.
Hence, at 20°C, the saturation pressure is,

P_{s a t} = 2.3392 k P a

Answer
The minimum pressure that can exist in the pump to avoid cavitation to occur, should not be dropped below 2.3392 kPa, in the given case.

Don Sumner

Skilled2023-05-27Added 184 answers

The vapor pressure of water can be approximated by the Antoine equation:

\log_{10} (P) = A - \frac{B}{T + C}

where

P

is the vapor pressure in mmHg,

T

is the temperature in degrees Celsius, and

A

B

, and

C

are constants specific to water.
At 20 degrees Celsius, the vapor pressure of water is approximately 17.5 mmHg. Therefore, the lowest pressure that can exist in the pump without cavitation is 17.5 mmHg.

Vasquez

Expert2023-05-27Added 669 answers

Answer:

2.3392

kPa
Explanation:
First, we need to find the saturation pressure of water at 20 degrees Celsius. This can be calculated using the Antoine equation:

P_{sat} = 10^{A - \frac{B}{T + C}}

where

P_{sat}

is the saturation pressure in kPa,

T

is the temperature in degrees Celsius, and

A

B

, and

C

are constants for water.
For water, the Antoine equation constants are:

A = 8.07131

B = 1730.63

C = 233.426

.
Substituting the values into the equation, we have:

P_{sat} = 10^{8.07131 - \frac{1730.63}{20 + 233.426}}

Calculating the value, we get:

P_{sat} = 2.3392

kPa.
Therefore, the lowest pressure that can exist in the pump without cavitation is

2.3392

kPa.

RizerMix

Expert2023-05-27Added 656 answers

To determine the lowest pressure (

P_{min}

) that can exist in the pump without cavitation, we need to consider the relationship between pressure and temperature known as the vapor pressure. Vapor pressure is the pressure at which a liquid turns into vapor at a given temperature.
The relationship between vapor pressure (

P_{vap}

) and temperature (

T

) can be described by the Clausius-Clapeyron equation:

P_{vap} = P_{ref} \cdot e^{(\frac{Δ H_{vap}}{R}) \cdot (\frac{1}{T_{ref}} - \frac{1}{T})}

Where:
-

P_{ref}

is the reference pressure (usually atmospheric pressure).
-

Δ H_{vap}

is the enthalpy of vaporization.
-

R

is the gas constant.
-

T_{ref}

is the reference temperature.
To prevent cavitation, the pressure in the pump must be greater than the vapor pressure of the water at the given temperature. Therefore, we can rearrange the Clausius-Clapeyron equation to solve for

P_{min}

P_{min} = P_{vap} = P_{ref} \cdot e^{(\frac{Δ H_{vap}}{R}) \cdot (\frac{1}{T_{ref}} - \frac{1}{T})}

Substituting the values for water at 20 degrees Celsius, we have:
-

P_{ref}

= atmospheric pressure (e.g., 101325 Pa)
-

Δ H_{vap}

= enthalpy of vaporization of water (e.g., 2257 kJ/kg)
-

R

= gas constant (8.314 J/(mol·K))
-

T_{ref}

= reference temperature (e.g., 298 K)
-

T

= temperature of water (20 degrees Celsius = 293.15 K)
Now, plugging in the values, we can calculate

P_{min}

P_{min} = 101325 \cdot e^{(\frac{2257 \cdot 10^{3}}{8.314}) \cdot (\frac{1}{298} - \frac{1}{293.15})}

Simplifying the equation and evaluating it will give us the numerical value for

P_{min}

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