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kiruha [24]
3 years ago
11

The air contained in a room loses heat to the surroundings at a rate of 50 kJ/min while work is supplied to the room by computer

, TV, and lights at a rate of 1.2 kW. What is the net amount of energy change of the air in the room during a 30-min period?

Engineering
2 answers:
Mkey [24]3 years ago
8 0

Answer:

660KJ

Explanation:

Given

Let Q = Heat Loss from room = 50kj/min

Let W = Work Supplied to room = 1.2KW

1 kilowatt = 1 kilojoules per second

So, W = 1.2KJ/s

In heat and work (Sign Convention)

We know that

1. Heat takes positive sign when it is added to the system

2. Heat takes negative sign when it is removed from the system.

3. Work done is considered positive when work is done by the system

4. Work done is considered negative when work is done on the system.

From the above illustration, heat loss (Q) = -50KJ/Min

In 30 minutes time, Q = -50Kj/Min * 30 Min

Work done in 30 minutes = -1500 KJ

Also, work supplied = -1.2Kj/s

Work supplied to the system in 30 minutes = -1.2Kj/s * 30 minutes

W = -1.2 KJ/s * 30 * 60 seconds

W = -2160KJ

In thermodynamics (First Law)

Q = W + ΔU

-1500 = -2160 + ΔU

∆U = 2160 - 1500

∆U = 660KJ

Artemon [7]3 years ago
4 0

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

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Answer:

y  ≈ 2.5

Explanation:

Given data:

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discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

v =1/n R^{2/3} s^{1/2}

where R is hydraulic radius

S = bed slope

Q = Av =A 1/n R^{2/3} s^{1/2}

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P is perimeter =  (B+2\sqrt{5} y)

R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}

Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}

solving for y100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}

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A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r
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Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.
Rufina [12.5K]

Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0

Energy Balance

-q_{loss} - w_{out} + h_{in} - h_{out} = 0

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\nu_{in} = 0.055665\,\frac{m^{3}}{kg}

h_{in} = 3650.6\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mix)

\nu_{out} = 5.89328\,\frac{m^{3}}{kg}

h_{out} = 2500.2\,\frac{kJ}{kg}

a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

v_{out} = 680.590\,\frac{m}{s}

c) The power output of the steam turbine is:

\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

\dot W_{out} = 18276.307\,kW

6 0
3 years ago
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