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2 years ago

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The air conditioner in a house or a car has a cooler that brings atmospheric air from 30C to 10C, with both states at 101KPa. If

the flow rate is 0.75kg/s, find the rate of heat transfer using constant specific heat of 1.004kj/kg.K

1 answer:

torisob [31]2 years ago

6 0

The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.

<h3>What is the rate of heat transfer?</h3>

Rate of heat transfer is the power rating of the machine.

Work done and changes in potential and kinetic energy are neglected since it is a steady state process.

The specific heat in terms of specific heat capacity and temperature change is given as:

$q_{out} = Cp(Ti - Te)$

$q_{out} = 1.004(30 - 10) = 20.08 kJ/kg \\$

The rate of heat transfer, is then determined as follows:

Qout = flow rate × specific heat

Qout = 0.75 × 20.08 = 15.06 kW

Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.

Learn more about rate of heat transfer at: brainly.com/question/17152804

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Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

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3 years ago

Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)

ArbitrLikvidat [17]

Answer:

$\mathbf{P_x =25 \ watts}$

$\mathbf{x_{rmx} = 5 \ unit}$

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:

$P= \dfrac{A^2}{2}$

For the number of sinosoidial signals;

Power can be expressed as:

$P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...$

As such,

For x(t), Power $P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}$

$P_x = \dfrac{25}{2}+ \dfrac{25}{2}$

$P_x = \dfrac{50}{2}$

$\mathbf{P_x =25 \ watts}$

For the number of sinosoidial signals;

$RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...$

For x(t), the RMS value is as follows:

$x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }$

$x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }$

$x_{rmx }=\sqrt{(\dfrac{50}{2} )}$

$x_{rmx} =\sqrt{25}$

$\mathbf{x_{rmx} = 5 \ unit}$

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