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3 years ago

Consider the rigid pressure vessel containing air in the diagram at the

Engineering

1 answer:

vredina [299]3 years ago

8 0

Answer:

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Explanation:

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A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo

Natalija [7]

Answer:

For $20^{\circ}$ - 5.556 lb/s

For $15^{\circ}$ - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = $20^{\circ}$

Temperature rise, T' = $15^{\circ}$

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, $C_{p} = 0.24 btu/lb ^{\circ}F$

Now, for a rise of $20^{\circ}$ in temeprature:

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s$

Now, for $15^{\circ}$ :

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s$

4 0

2 years ago

Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with

Scilla [17]

Answer:

DIAMETER = 9.797 m

POWER = $\dot W = 28.6 kW$

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

$v =\frac{RT}{P}$

where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

$v = \frac{0.287\times 293}{100}$

v = 0.8409 m^3/ kg

from continuity equation

$A_1 V_1 = A_2 V_2$

$\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2$

$D_2 = D_1 \sqrt{\frac{V_1}{V_2}}$

$D_2 = 8 \times \sqrt{\frac{12}{8}}$

$D_2 = 9.797 m$

mass flow rate is given as

$\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}$

$\dot m = 717.309 kg/s$

the power produced $\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]$

$\dot W = 28.6 kW$

8 0

3 years ago

A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o

grigory [225]

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

7 0

3 years ago

A long corridor has a single light bulb and two doors with light switch at each door.

Tpy6a [65]

Answer:

Light = A xor B

Explanation:

If switches A and B produce True or False, then Light will be True for ...

Light = A xor B

8 0

2 years ago

A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r

allochka39001 [22]

Answer: 37.4K

Explanation:

See attachment

7 0

2 years ago