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Leya [2.2K]
3 years ago
9

Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A

ir enters the heat exchanger at 95 kPa and 20°C at a rate of 0.6 m3/s. The combustion gases (cp = 1.10 kJ/kg·°C) enter at 160°C at a rate of 0.95 kg/s and leave at 95°C. Determine the rate of heat transfer to the air and its outlet temperature.
Engineering
1 answer:
uysha [10]3 years ago
7 0

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

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The correct answer is A; True.

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How did ancient machines pave the way for improvement
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Answer:

----------

Explanation:

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4 0
3 years ago
How to find the voltage(B Aab) in series parallel circuit? ​
Sindrei [870]

Answer:

  Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

  R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

  Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

  Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

  (1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

  V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

  Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

  (Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

  (V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

  (V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0
3 years ago
(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i
kirill115 [55]

Answer:

The equation used to solve a diode is

i_d = I_se^\frac{V_d}{V_T}-1

  • i_d is the current going through the diode
  • I_s is your saturation current
  • V_D is the voltage across your diode
  • V_T is the voltage of the diode at a certain room temperature. by default, you always use V_T=25.9mV for room temperature.

If you look at the equation, i_d = I_se^\frac{V_d}{V_T}-1, you'd notice that the e^\frac{V_d}{V_T} grow exponentially fast, so we can ignore the -1 in the equation because it's so small compared to the exponential.

i_d = I_se^\frac{V_d}{V_T}-1

i_d\approx I_se^\frac{V_d}{V_T}

Therefore, use i_d= I_se^\frac{V_d}{V_T} to solve your equation.

Rearrange your equation to solve for V_D.

V_D=V_Tln(\frac{i_D}{I_s})

a.)

i.)

You're given I_s=10^{-11}A

at i_d=10\mu A,     V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{10\cdot10^{-6}}{10\cdot10^{-11}})=.298V

at i_d=100\mu A,   V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{100\cdot10^{-6}}{10\cdot10^{-11}})=.358V

at i_d=1mA,      V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{1\cdot10^{-3}}{10\cdot10^{-11}})=.417V

<em>note: always use</em>  V_T=25.9mV

ii.)

Just repeat part (i) but change to I_s=-5\cdot10^{-12}A

b.)

same process as part A. You do the rest of the problem by yourself.

4 0
3 years ago
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