What tool should be used to loosen or tighten brake or fuel lines?

Can high throttle torque tend to open the throttle plate

koban [17]

Answer:

no

Explanation:

7 0

3 years ago

Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces

RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

mh1 + W = mh2 + Q where W = Q.

Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0

3 years ago

Given the following phasors and the information related to the frequency of that phasor, provide the corresponding time-domain r

Evgesh-ka [11]

Answer:

Explanation:

In Engineering and Physics a Phasor That is a portmanteau of phase vector, is a complex number that represents a sinusoidal function whose Amplitude (A), Angular Frequency (ω), and Initial Phase (θ) are Time-invariant.

For the step by step solution to the question you asked, go through the attached documents.

4 0

3 years ago

Air enters a turbine operating at steady state at 440 K, 20 bar, with a mass flow rate of 6 kg/s, and exits at 290 K, 5 bar. The

marysya [2.9K]

Answer:

The rate of heat transfer is - 935.392 kW

Explanation:

Given;

inlet pressure, P₁ = 20 bar

inlet temperature, T₁ = 440 k

outlet pressure, P₂ = 5 bar

outlet temperature, T₂ = 290 k

inlet velocity, v₁ = 18 m/s

outlet velocity, v₂ = 30 m/s

mass flow rate, Q = 6kg/s

developed power, W = 815 kW

From steam table;

at P₁ and T₁, interpolating between T = 400 and T = 450, h₁ = 3335.52 kJ/kg

at P₂ and T₂, interpolating between T= 250 and T= 300, h₂ = 3043.5 kJ/kg

Applying steady state energy balance equation;

$\frac{Q}{m} -\frac{W}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2})\\\\\frac{Q}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2}) +\frac{W}{m}$

substitute values above into this equation and evaluate the rate of heat transfer

$\frac{Q}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2}) +\frac{W}{m}\\\\\frac{Q}{m} = 3043.5 - 3335.52 +\frac{1}{2000} (30{^2} -18{^2}) + \frac{815}{m} \\\\\frac{Q}{m} = -291.732 + \frac{815}{m}\\\\\frac{Q}{m} = \frac{815}{m} -291.732\\\\Q = 815 - m(291.732)\\Q = 815 - 6*291.732\\\\Q = 815-1750.392 = -935.392 \ kW$

Therefore, the rate of heat transfer, in kW, for a control volume enclosing the turbine is - 935.392 kW

4 0

3 years ago

Using the milligrams of ascorbic acid you entered above, the ratio of total sample volume to aliquot volume, and the total milli

Dmitriy789 [7]

Answer: 7.500 2.99

Explanation:

3 0

3 years ago

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