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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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What was the role of the rotors in the enigma machine?.
Lisa [10]

Answer:

Implements a reordering of the letters of the alphabet.

Explanation: GIVE ME 5 STARS AND a HEART!!!                                      Those contacts are wired across the rotor so that each contact on the left connects to the contact on the right in some scrambled arrangement. Each rotor, therefore, implements a reordering of the letters of the alphabet, which mathematicians call a permutation.

3 0
3 years ago
An 1,840 W toaster, a 1,420 W electric frying pan, and a 70 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (T
Viktor [21]

Answer:

A)

Current drawn by toaster = 15.33 A

Current drawn by electric frying pan = 11.83 A

Current drawn by lamp = 0.58 A

B)

The fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

Explanation:

There are three devices plugged into the same outlet.

Toaster = 1840 W

Electric frying pan = 1420 W

Lamp = 70 W

Since the three devices are connected in parallel therefore, the voltage across them will be same but the current drawn by each will be different.

A) What current is drawn by each device?

The current flowing through the device is given by

I = P/V

Where P is the power and V is the voltage.

Current drawn by toaster:

I = 1840/120

I = 15.33 A

Current drawn by electric frying pan:

I = 1420/120

I = 11.83 A

Current drawn by lamp:

I = 70/120

I = 0.58 A

B) Will this combination blow the 15-A fuse?

The total current drawn by all three devices is

Total current = 15.33 + 11.83 + 0.58

Total current = 27.74 A

Therefore, the fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

5 0
4 years ago
Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)
IceJOKER [234]

Answer:

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

\mathbf{M_n = 49163.56431  \ g/mol }

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa)            Number- Average Molecular Weight  (g/mol)

82                                                  12,700

156                                                 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

T_S = T_{S \infty} - \dfrac{A}{M_n}

where;

T_S = Tensile Strength

T_{S \infty} = Tensile Strength (Infinity)

M_n = Number- Average Molecular Weight  (g/mol)

SO;

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

From equation (1) ; collecting the like terms; we have :

T_{S \infty} =82+ \dfrac{A}{12700}

From equation (2) ; we have:

T_{S \infty} =156+ \dfrac{A}{28500}

So; T_{S \infty} = T_{S \infty}

Then;

T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}

Solving by L.C.M

\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}

\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}

By cross multiplying ; we have:

({4446000 + A})*  {12700} ={28500} *({1041400 + A})

(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)

Collecting like terms ; we have

(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)

2.67843*10^{10}  = 15800 \ A

Dividing both sides by 15800:

\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}

A = 1695208.861

From equation (1);

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

Replacing A = 1695208.861 in the above equation; we have:

82= T_{S \infty} - \dfrac{1695208.861}{12700}

T_{S \infty}= 82 + \dfrac{1695208.861}{12700}

T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}

T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}

T_{S \infty}= \dfrac{2736608.861 }{12700}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

From equation(2);

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

Replacing A = 1695208.861 in the above equation; we have:

156= T_{S \infty} - \dfrac{1695208.861}{28500}

T_{S \infty}= 156 + \dfrac{1695208.861}{28500}

T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}

T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}

T_{S \infty}= \dfrac{6141208.861}{28500}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

T_S = T_{S \infty} - \dfrac{A}{M_n}

Then:

181 = 215.481- \dfrac{1695208.861 }{M_n}

Collecting like terms ; we have:

\dfrac{1695208.861 }{M_n} = 215.481-  181

\dfrac{1695208.861 }{M_n} =34.481

1695208.861= 34.481 M_n

Dividing both sides by 34.481; we have:

M_n = \dfrac{1695208.861}{34.481}

\mathbf{M_n = 49163.56431  \ g/mol }

5 0
3 years ago
Which of the following is NOT true about hydraulic systems?
Dmitry [639]

Answer: The answer is D

D.In hydraulic systems, the operating temperatures must be kept between 170�F and 180°F 

Explanation:

The operating temperature for hydraulic systems is 140°F and below. Anything above this temperature is too high and will reduce the useful life of hydraulic fluid.

Most often problems associated with hydraulic systems are caused by fluid contaminated with particulate matter.

7 0
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Dafna1 [17]

Answer:

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Explanation:

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