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Anna71 [15]
3 years ago
14

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

m-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace
Engineering
1 answer:
zvonat [6]3 years ago
8 0

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

\frac{T_{0}-T_{x}  }{T_{1}-T_{x}  } = C_{1} e^{(-0.4888^{2}*Fo )}

= 0.4167 = 1.0396e^{-0.4888*Fo}

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = (\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

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Elenna [48]

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

  Q1 - Q2 + Q + W = 0

b)  rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

<u>Write the First law balance around the system and rate of work input to the system</u>

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T  = 1.65 * 1.005 * 293 = 485.86 Kw

             Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

              Q = 18 Kw

Insert values into equation 2 above

W = 6.88 Kw

5 0
3 years ago
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jeka94

Answer:

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Explanation:

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5 0
3 years ago
Read 2 more answers
A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a
bazaltina [42]

Answer:

Q_{cv}=-339.347kJ

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

P_iV_i=m_iRT_i (i could be 1 for initial and 2 for the end)

State1

1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295

m_1=232kg

State2

8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350

m_2=11.946

So, the total mass of the aire entered is

m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

u_1=210.49kJ/kg

For Specific enthalpy

h_1=295.17kJ/kg

For the second state the Specific internal Energy (6bar, 350K)

u_2=250.02kJ/kg

At the end we make a Energy balance, so

U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

No work done there is here, so clearing the equation for Q

Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)

Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)

Q_{cv}=-339.347kJ

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0
3 years ago
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
4 years ago
The underground cafe has an operating cash flow of $187,000 and a cash flow to creditors of $71,400 for the past year. During th
Serggg [28]

Answer:

cash flow to stockholders = $39,700

Explanation:

Operating cash flow = $187,000

cash flow to creditors = $71,400

Net working capital = $28,000

Net capital spending =  $47,900

Cash flow to stockholders = ?

CFF = operating cash flow - net working capital - net capital spending

CFF = $187,000 - $28,000 - $47,900 = $111,100

CFF = cash flow to creditors + cash flow to stockholders

cash flow to stockholders = CFF - cash flow to creditors

cash flow to stockholders = $111,100 - $71,400 = $39,700

Hence $39,700 is the amount of the cash flow to stockholders for the last year.

3 0
3 years ago
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