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Physics Numerical Examples

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1). A coil is used as an evaporator in an air-conditioner. The loop is used to chill 20C air by evaporating refrigerant o C. the face area of the circle is Ar =0.4, and the coil is L= 10cm deep in the airflow direction the air properties are and cp=1000. Neglect property variation. For this surface, the cross-sectional flow area after the air squeezes into the coil reduced to 0.2, and the hydraulic diameter inside the heat exchange is Dn=3mm. The air-side heat transfer area is 24.

 

  • For air volume flow rate Q=1.28, what is the air mass flow rate, kg/s?

We know that mass flow rate formula which is:

m=ρ×volume flow rate

m=1×1.28

m=1.28kgs

  • Options are: 0.64 (2) 1.0 (3)​​ 1.28​​ (4) 2.56

1.28 is the correct answer.

b) If the heat transfer rate is 16Kw at​​ m=1.6kg/s, what are the exits air temperature, C?

temperature=161.6​​ =10

T2-T1=20-10=10

  • Options are: ​​ 0 (2)  ​​​​ 10​​ (3) 20 (4) 40

The correct answer is 10. When heat transfer is divided by mass, then the answer minus with the given temperature.

  • At m=1.2kg/s, what is the velocity​​ pressure inside the heat exchanger, Pa, at this flow rate?

Velocity pressure Pa=12ρv2=?

m= ρAv ​​ v= mρA=A​​ is sued for area after the air squeezes which is 0.2m2

v= 1.21*6= 6ms-1

Velocity pressure Pa= 12ρv2=​​  121*(6)2= 18

  • Options are :​​ 18 ​​ ​​​​ (2) ​​ 24 ​​ (3) 10 (4) 36

18 is the correct answer

  • If the​​ velocity pressure is 20Pa, and the finning friction factor is 0.02 what is the pressure drop Pa?

Pressure drop, Pd=​​ 2flρV2D

If the Pa =20Pa = V=​​ 2*Paρ

V=​​ 2*201

= 6.32m/s

Pd=​​ 2*0.02*0.1*1*(6.32)20.003

Pd= 53.33Pa

  • Options are: 0.4 (2)13.3 ​​ (3)20.0 ​​ (4)53.3

53.3 is the pressure drop in pa.

e) ​​ At mair​​ =1.2, the overall heat transfer coefficient U=25. For the heat transfer area, given what is effective?​​ 

Effectiveness= ϵ​​ =​​ qqmax

qmax=mCpT-0

qmax=1.2*100020-0

qmax=24,000=24KW

ϵ​​ =​​ 2425

ϵ​​ = 0.39

Options are:​​ (1) 1.2 (2) 0.9 (3) 0.59​​ (4)0.39

0.39 is the correct answer

f) if the effectiveness is 50% at​​ m=1.2 what is the heat transfer in Kw?

Q= mCpT

Q= 1.2 * 1000 * 20

Q = 24,000 W = 24KW

Options are: (1) 6 (2) 12 (3)24​​ (10)

24 is the correct Answer

2). A room is ventilated with m=0.8 of air at temperature 25 C. initially, there is no heating load in the​​ room, and the air in the room is To=30. The mass m of air in the room can be calculated using the ideal gas law, and m is=500kg. Then at time t=0, the room's equipment is switched on in the room qequiped= 21kj while a cooling system starts that removed qcool=14kw. Cp=1000. K=1.4.

a). Which one of the following is a correct differeiental equation and initial condition for this problem?​​ 

=mCp(T1-T)+qequip-qcocl ​​​​ T (0) = 30

The correct answer is B

Because in the air room T(0) =30​​ 

b). How hot​​ ​​ will the well missed air in the room eventually get if the equipment and cooling system remain on?

=mCp(T1-T)+qequip-qcocl ​​​​ T (0) = 30

dTdt=​​ (mCp(T1-T)+qequip-qcocl)kmCp

dTdt=(T1-T)+qequip-qcocl)k

dTdt=​​ ((30-25)+21-14))1.4

dTdt=16.8

integration of both side

T= 33.6

The correct answer is c (33.6)

c). How hot​​ ​​ will the well missed air in the room eventually get if the equipment remain same cooling system fail.

=mCp(T1-T)+qequip-qcocl ​​​​ T​​ (0) = 30

dTdt=​​ (mCp(T1-T)+qequip-qcocl)kmCp

dTdt=(T1-T)+qequip-0)k

dTdt=​​ ((30-25)+21-0))1.4

dTdt=36.4

integration of both side

T= 36.4

T= 38.8

c). The correct answer is c

If the time to reach steady states is 4 times constant, how long will it take to do so, minutes?​​ 

The correct answer is a 41.7​​ 

3). Recall that the pump power required to pump water through a closed constant diameter water piping loop can be calculated using Hazen-Williams for head loss as​​ 

(a) What is the normalized sensitivity coefficient for C (NSC)

NSC= y-yoyos-x0x0

power= γhLQη

hL=ALeffQ1.852C1.852D4.8704

Leff=L+ Leq

power= γALeffQ1.852C1.852D4.8704Qη

 

Under a given set of assumptions the in what way an independent value of a​​ variable is a collision on the dependent value of the variable, this technique is used for normalized sensitivity. The correct answer is -1.852

(b) What is the normalized sensitivity coefficient for Leff​​ (NSL)

power= γALeffQ1.852C1.852D4.8704Qη

=γQη.​​ AQ1.852C1.852D4.8704.​​ Leff

 

Under a given set of assumptions in what way an independent value of a variable is a collision on the variable's dependent value, this technique is used for normalized sensitivity.

The correct answer is d.1.

c). What is the normalized sensitivity coefficient for Q (NSQ)

power= γALeffQ1.852C1.852D4.8704Qη

 

 

Under a given set of​​ how an independent value of a variable is a collision on the variable's dependent value, this technique is used for normalized sensitivity. The correct answer is b. 2.852

 

d). The description of the Leff​​ is missing

Total uncertainty:

The total delay is 46.2

  • Water from the cooling pond at Ta​​ =25 is used to cool the condenser in a chiller plant operating at Tc​​ =45. The required heat condenser heat transfer rate q is at least 30kw, and the pump head Hp​​ cannot exceed 18m.  ​​ ​​ ​​ ​​ ​​​​ 

 

  • What is the objective​​ function?​​ 

Total cost = Fixed cost +variable Cost

Ctot=2000*A+8*1.7E6*mw3

The correct answer for this statement is C, which is

Ctot=2000*A+8*1.7E6*mw3

  • what are the constraints in the given option? I chose only one.

100mw2​​ 18m, MW​​ Cpw(Tc-TA​​ )30KW

For maximum heat exchanging

A= 100m2, ​​ mW= 0.4​​ Tc=TB

The​​ correct answer to this statement is B.

100mw2​​ 18m, MW​​ Cpw(Tc-TA​​ )30KW

  •  ​​​​ what is the heat transfer rate in​​ 

kWT

Q= MCP∆

T=Change in temperature=45-25=20

Q=0.4*4.19*20

Q=33.52

the correct answer to the given option is A, which is 33.5

  • what is the present value of the total cost.

Present value at total cost

Ctot=2000*100+ (8*1.7E6*(0.4)3

Ctot=$1.07 E6

The correct answer is A.

Answer sheet provided . Write names on ALL andwer sheet Now. Only the answer sheets will be graded and your name must be on each one . You muist trun this sheet in with your work​​ 

1). Recall that the fan power required pushing air​​ through a car radiator like heat exchanger

What is the normalized sensitivity coefficient for Q?​​ 

J=power =​​ 4FfLDn.γ2g.PAf2.Q3η​​ = K. Ff. Q3η

dJdQ= K. Ff. Q2η​​ , K=​​ 4LDn.​​ γ2g.PAf2

i.e​​ J= K. Ff. 3Q2η. for small proportions in Q

= , =​​ K. Ff. 3Q2η.Q.​​ . ηK. Ff3Q3= 3

NSCFf=​​ K. Ff. 3Q3η= 3

Correct answer is 3

What is the normalized sensitivity​​ coefficient forfF?​​ 

Sff=KQ3η=NSCFf=KQ3η.ηKQ3.Ff

=KQ3η.ηKQ3.Ff=1

Correct answer is 1

What is the normalized sensitivity coefficient forη?

Jη=Ff.-KQ3η2=NSCFf=-KQ3η.ηKQ3.Ff

Jη=-1

Correct answer is -1

If Ns=2.4,0.8 ……% uncertainty in power.

The description of the Ns is missing​​ 

What uncertainty contributes most to uncertainty in power?​​ 

With higher exponents the pentameter, t is on, which contributes most to an uncertainty, hence where the uncertainty​​ Q contributes​​ most to a delay in J​​ 

2).​​ A 2m2 insulated reservoirs is used to store chilled water for cooling machinery in a plant. After being chilled all night with no machinery running, the reservoir initially contains water at 15 C. At time t = 0, the morning shift starts and machinery needing cooling starts extracting 200 L/min from the tank while 200 L/min of 75 C water flows from the machinery back into the reservoir, keeping the reservoir full. At the same time, a​​ cooling coil that removes 500 kW from the reservoir is running. The cold-water temperature leaving the reservoir starts to rise due to the energy in the water returning from the machinery. Use​​ \rhowater = 1.0 kg/L and Cp = 4.2 kJ/kg-K if needed.

a.) Write a differential and initial condition equation that could be solved for the temperature T of the water leaving the reservoir as a function of time. Assume the water in the​​ reservoir is well mixed. Do not solve

=mCp(T1-T)+qequip-qcocl ​​​​ T (0) = 15

dTdt=​​ (mCp(T1-T)+qequip-qcocl)kmCp

b.) What is the eventual steady water exit temperature if the flow continues?

The flow rate difference in substrate concentration among reactor

F([S]0​​ – [S] =​​ Vmax[S]Km+[S]

VmaxF]=​​ KmS0-[S][S]​​ +([S]0​​ – [S]

S0-[S][S]= X1-X

c.) Roughly how​​ long until steady state is reached? This can be done without solving the ODE in (a.)

To reach the steady state the rime is defined as “by the removing of the drug half-life, after the 1 half-life we reached the 50% of the stable conditions, or after the 2​​ half-life’s we reached at the 75% stable states ​​ 

3). Water from cooling ​​ pond at Ta​​ =30 is used to cool the condenser in a chiller plant operating at Tc​​ =45. The rquired heat condenser heat transfer rate q is at least 60kw and the pump head Hp​​ cannot exeed 25m.  ​​ ​​ ​​ ​​​​ 

a). What is objective function?

The target work is a way to augment (or limit) something. This something is a numeric esteem: in reality it could be the cost of a task, a generation amount, benefit esteem, or else even materials spared from a streamlined procedure. With the goal work, you are endeavoring to land at an objective for yield, benefit, asset utilization, and so on.

We have to take a gander at the connections amongst imperatives along with any constraints inside the business itself. These can incorporate creation limit limits, asset accessibility, or else even innovation.​​ 

Minimze or maximize=i=1nciXi

b). What are constraints?

The subsystems, factor along with the element which works for the bottleneck. It confines individual projects and the systems​​ where from it getting their potentials along with sources for its purpose.

The three constraints used in the projects are time table, scope, and the cost, which is sometimes also called the triple rules in the projects management’s triangles.​​ 

c). For A=800 and mW=0.8 is the present value of the cot, $?

The present value​​ =8000.8

=10.00$E

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