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3 years ago

10

Make a sketch of a simple mechanically expanded brake and indicate the forces acting on the leading shoe when the brake is a

pplied.

1 answer:

Pavel [41]3 years ago

4 0

Answer: sorry I just need points

Explanation

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A mass of 12 kg saturated refrigerant-134a vapor is contained in a piston-cylinder device at 240 kPa. Now 300 kJ of heat is tran

Ket [755]

Answer:

I = 12.706 Amps

Explanation:

Given:

- The mass of saturated R-134a m = 12 kg

- The initial Conditions

P_1 = 240 KPa

Saturated Vapor

- The final Conditions

P_2 = 240 KPa

T_2 = 70° C

- The amount of Heat transferred Q_in = 300 KJ

- The voltage of current source V = 110 V

- The current supplied by the source = I

- The time duration Δt = 6 min

Find:

Determine the current supplied I.

Solution:

- Look-up enthalpies h_1 and h_2 at both states using Tables A-11 and A-13.

P_1 = 240 KPa

Saturated Vapor ----------> h_1 = h_g = 247.32 KJ/kg

P_2 = 240 KPa

T_2 = 70° C ----------> h_2 = 314.53 KJ/kg

- Using First Thermodynamic Law, set up an energy balance:

E_in - E_out = ΔE_system

Q_in + W_electric,in - W_out = Δ U

Q_in + V*I*Δt - W_out = Δ U

Q_in + V*I*Δt = Δ H

Q_in + V*I*Δt = m*( h_2 - h_1 )

- Make the current I the subject of the expression above:

V*I*Δt = m*( h_2 - h_1 ) - Q_in

I = [ m*( h_2 - h_1 ) - Q_in ] / V*Δt

- Plug in the values in the expression derived above and evaluate current of source I:

I = [ 12*( 314.53 - 247.32 ) - 300 ]*1000 / 110*6*60

I = 12.706 Amps

- The required source of current is I = 12.706 Amps.

4 0

3 years ago

lorasvet [3.4K]

Answer: observational error

Explanation:

6 0

3 years ago

You’re engineering an energy-efficient house that will require an average of 6.85 kW to heat on cold winter days. You’ve designe

Oxana [17]

Answer:

2.95 approximately 3

Explanation:

For a heat pump,

COP = Q/W

Where Q = power needed for heating process

W = power input into heat pump.

Power for heating Q = 6.85 kW

Proposed power input to heat pump W = 2.32 kW

Minimum COP = 6.85/2.32 = 2.95

Approximately 3

8 0

4 years ago

A compressor operates at steady state with Refrigerant 134a as the working fluid. The refrigerant enters at 0.24 MPa, 0°C, with

marissa [1.9K]

Answer:

POWER INPUT = 82.989 KW

Explanation:

For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg

For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg

Heat loss Q = 0.05w

Inlet diameter = 3 cm

exit diamter = 1.5 cm

volume of tank will be v = area * velocity

velocity at inlet $= \frac{0.64\60 m/s}{ \frac{\pi}{4} (3\times 10^{-2})^2} = 15.09 m/s$

velocity at outlet $= \frac{0.64\60 m/s}{ \frac{\pi}{4} (1.5\times 10^{-2})^2} = 60.36 m/s$

steady flow energy equation

$E_{IN} = E_{OUT}$

$h_1 + \frac{v_1^2}{2g} +wc = h_2 + \frac{v_2^2}{2g} + 0.05wc$

$248.89 + \frac{15.09^2}{2} + wc = 280.18 + \frac{60.36^2}{2} + 0.05 wc$

solving wc = 1830.64 kJ/kg

wc in KWH

we know that $wc = \dot m wc$

$\dot m = 4.25 kg/m3 \times (0.64/60) m^3/s$

$\dot m = 0.04533 kg/s$

$wc = 0.04533 \times 1830.64 = 82.989 kW$

5 0

4 years ago

What are employers required to do to keep employees safe from caught-in and -between hazards from hand-held power tools?

nika2105 [10]

Answer:

Employees who use hand and power tools and who are exposed to the hazards of falling, flying, abrasive and splashing objects, or exposed to harmful dusts, fumes, mists, vapors, or gases must be provided with the appropriate equipment needed, including Personal Protective Equipment, to protect them from the hazard.

Explanation:

8 0

3 years ago