Question

The density of mercury is 13.546 g/cm3 . Calculate the pressure exerted by a column of mercury 76 cm high. Give your answer in Pascals and lbf/in2 . 2. The density of water is 62.43 lbm/ft3 . Calculate the pressure exerted by a column of water 25 ft high. Give your answer in Pascals and lbf/in2 3. What is the power required to pump 10 kg/s water from a height of 5 meters to a height of 30 meters? Report the power in Watts and hp. 4. Water is being pumped by the application of pressure at point 1, see below, up to a height of 50 m at a rate of 1 kg/s. At the top, point 2 the pressure is nearly one atmosphere (105 N/m2 )

Answer 1

1. p = 14.63 lb/in² or 100890.608 Pa

2. p = 74676 Pa or 10.83 lb/in²

3. P = 2450 W or 3.28 hp

4.
`p_(1)` = 490105 N/m²

Step-by-step explanation:

1. Let's begin by listing out the given parameters:

density of mercury = 13.546 g/cm³ = 13546 kg/m³,

height of column = 76 cm = 0.76 m, acceleration due to gravity = 9.8m/s²

Using Pressure = density * acceleration due to gravity * height of column

p = ρ g h = 13546 * 9.8 * 0.76

p = 100890.608 Pa

To get the answer in lb/in², divide by 6895

p = 100890.608 ÷ 6895 = 14.632

p = 14.63 lb/in²

2. Let's list out the parameters given:

density of water = 62.43 lbm/ft³ = 62.43 * 16.018 = 1000kg/m³,

height of column = 25 ft = 25 ÷ 3.281 = 7.62 m,

acceleration due to gravity = 9.8m/s²

Using Pressure = density * acceleration due to gravity * height of column

p = ρ g h = 1000 * 9.8 * 7.62

p = 74676 Pa

To convert from Pa to lb/in², divide by 6895

p = 74676 ÷ 6895

p = 10.83 lb/in²

3. Let's list out the parameters given:

mass flow rate (ṁ) = 10 kg/s,
`h_(1)` = 5 m,
`h_(2)` = 30 m, Δh = 30 - 5 = 25 m, g = 9.8 m/s²

Using Power = Energy (Potential Energy) ÷ Time

Energy (Potential Energy) = m g h

Power = mgΔh ÷ t; m÷ t = ṁ

Substitute ṁ into the equation

P = ṁ g h = 10 * 9.8 * 25

P = 2450 W

To convert from W to hp, divide by 746

P = 2450 ÷ 746 = 3.284

P = 3.28 hp

4. Let's list out the parameters given:

height (Δh) = 50 m, ṁ = 1 kg/s, g = 9.8 m/s²,

p2 = 105N/m², ρ = 1000 kg/m³

Using Bernoulli's Equation,

p1 + ½ρ(
`v_(1)`)² + ρgh1 = p2 + ½ρ(
`v_(2)`)² + ρgh2

Assuming steady state flow;
`v_(2)` =
`v_(1)` ⇒
`v_(2)` -
`v_(1)` = 0

`p_(1)` -
`p_(2)` = ½ρ(
`v_(2)` -
`v_(1)`)² + ρg(
`h_(2)` -
`h_(1)`)

`p_(1)` -
`p_(2)` = ρgΔh

`p_(1)` - 105 = 1000 * 9.8 * 50

`p_(1)` = 490000 + 105 = 490105

`p_(1)` = 490105 N/m²

Answer 2

The required solution is 100890 Pa and 14.3lb/in²

Step-by-step explanation:

See attached image