Answer:
a) the heat exchanger area required for the evaporator is 11178.236 m²
b) the required flow rate is 1993630.38 kg/s
Explanation:
Given the data in the question;
Water temperature near the surface = 300 K
temperature at reasonable depths ( cold ) = 280 K
power plant output W' = 2 MW
efficiency η = 3% = 0.03
we know that; efficiency η = W'[tex]_{power-out[/tex] / Q[tex]_{supplied[/tex]
we substitute
0.03 = 2 / Q[tex]_{supplied[/tex]
Q[tex]_{supplied[/tex] = 2 / 0.03
Q[tex]_{supplied[/tex] = 66.667 MW = 66.667 × 10⁶ Watt
T[tex]h_{in[/tex] = 300 K T[tex]h_{out[/tex] = 292 K
T[tex]c_{in[/tex] = 290 K T[tex]c_{out[/tex] = 290 K
Now, Heat transfer in evaporator;
Q = UA( LMTD )
so
LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )
first we get ΔT₁ and ΔT₂
ΔT₁ = T[tex]h_{in[/tex] - T[tex]c_{out[/tex] = 300 - 290 = 10 K
ΔT₂ = T[tex]h_{out[/tex] - T[tex]c_{in[/tex] = 292 - 290 = 2 K
so we substitute into our equation;
LMTD = (10 - 2) / ln( 10 / 2 )
LMTD = 8 / ln( 5 )
LMTD = 8 / 1.6094379
LMTD = 4.97
a) Heat transfer Area will be;
Q[tex]_H[/tex] = UA( LMTD )
we substitute
66.667 × 10⁶ = 1200 × A × 4.97
66.667 × 10⁶ = 5964 × A
A = (66.667 × 10⁶) / 5964
A = 11178.236 m²
Therefore, the heat exchanger area required for the evaporator is 11178.236 m²
b) Flow rate
we know that;
Q[tex]_H[/tex] = m'C[tex]_P[/tex]( [tex]T_{in[/tex] - [tex]T_{out[/tex] )
specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)
we substitute
66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )
66.667 × 10⁶ = m' × 33.44
m' = ( 66.667 × 10⁶ ) / 33.44
m' = 1993630.38 kg/s
Therefore, the required flow rate is 1993630.38 kg/s
A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected
to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?
Answer:
"7.654 mm" is the correct solution.
Explanation:
According to the question,
[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]Now,
As we know,
The Elongation,
⇒ [tex]E=\frac{\sigma}{e}[/tex]
[tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]
or,
⇒ [tex]\delta=\frac{PL}{AE}[/tex]
By substituting the values, we get
[tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]
then,
⇒ [tex]D^2=58.587[/tex]
[tex]D=\sqrt{58.587}[/tex]
[tex]=7.654 \ mm[/tex]
A continuous and aligned fiber-reinforced composite is manufactured using 80 vol% aramid fiber (a kevlar-like compound) embedded nylon 6-6. A part for a high-performance aircraft utilizes this composite. If the part experiences 953 lb-f (pounds force) along the fiber alignment axis, what is the force conveyed by the fibers ?
Answer:
the force conveyed by the fibers is 947.93 lb-f
Explanation:
Given the data in the question;
V_f = 80% = 0.8
V_m = 1 - V_f = 1 - 0.8 = 0.2
Now,
length of fibre L_f = length of Nylon L_n
V_f = A_f × L_f = 0.8
V_m = A_n × L_n = 0.2
so
V_f/V_m = A_f/A_n = 0.8/0.2
A_f/A_n = 4
now, the strains in fibre is equal to strains in nylon
(P/AE)f = (P/AE)n
P_f/A_fE_f = P_n/A_nE_n
P_f = (A_f/A_n)(E_f/E_n)(P_n)
P_f = ( 4 )( 131 / 2.8 )(Pn)
P_f = 187.14Pn
and P_n = Pf / 187.14
Hence
given that P_total = 953 lb-f
P_f + P_n = 953
P_f + ( P_f / 187.14 ) = 953
P_f( 1 + ( 1 / 187.14 ) ) = 953
P_f( 1.00534359 = 953
P_f = 953 / 1.00534359
P_f = 947.93 lb-f
Therefore, the force conveyed by the fibers is 947.93 lb-f
Explain when it is appropriate to use Tier II?
????? what do u mean tier 2
A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft and the 4 in. side is positioned vertically. If the maximum load was 240 kips and the modulus of rupture was 940.3 ksi, what is the value of d
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
describe five tools used in suspension system service and repair
Answer:
Explanation:
You'll still need wrenches, sockets and screwdrivers, but there are other things that it may be necessary to buy to complete the work.
Spring Compressor. One part of suspension repair is replacing coil springs.
Hydraulic Puller. -Hydraulic pullers are used to remove shaft-fitted parts (bearings or couplings). Pullers use a controlled hydraulic force in an effective way and can quickly separate (especially compared to the manual alternative) the parts.
CV Boot Tool. he CV Boot is a ribbed, rubber flexible boot that keeps water and dirt out of the joint and the special grease inside the joint.
Torque Wrench. .A torque wrench is a tool used to apply a specific torque to a fastener such as a nut, bolt, or lag screw. ... A torque wrench is used where the tightness of screws and bolts is crucial.
Ball Joint Separator. This tool is used to separate the ball joint from the spindle support arm. It works on many domestic and import front wheel drive vehicles and is adjustable up to 2" for different size ball joints.
Strut Nuts.
Tie Rod Puller.
etc....
Consider a turbofan engine installed on an aircraft flying at an altitude of 5500m. The CPR is 12 and the inlet diameter of this engine is 2.0m The bypass ratio of this engine 8. The bypass ratio (BPR) of a turbofan engine is the ratio between the mass flow rate of the bypass stream to the mass flow rate entering the core. The inlet temperature is 253K and the outlet temperature is 233K. Determine the thrust of this engine in order to fly at the velocity of 250 m/s. Assume cold air approach. The engine is ideal.
Answer:
The thrust of the engine calculated using the cold air is 34227.35 N
Explanation:
For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as
[tex]\dot{m}=\rho AV_a[/tex]
Here
ρ is the density which is given as [tex]\dfrac{P}{RT}[/tex]P is the pressure of air at 5500 m from the ISA whose value is 50506.80 PaR is the gas constant whose value is 286.9 J/kg.KT is the temperature of the inlet which is given as 253 KA is the cross-sectional area of the inlet which is given by using the diameter of 2.0 mV_a is the velocity of the aircraft which is given as 250 m/sSo the equation becomes
[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]
Now in order to find the flow from the fan, the Bypass ratio is used.
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]
Here BPR is given as 8 so the equation becomes
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]
Now the exit velocity is calculated using the total energy balance which is given as below:
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]
Here
h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as [tex]c_pT_4[/tex] and [tex]c_pT_5[/tex] respectively.The value of T_4 is the inlet temperature which is 253 KThe value of T_5 is the outlet temperature which is 233KThe value of c_p is constant which is 1005 J/kgKV_a is the inlet velocity which is 250 m/sV_e is the outlet velocity that is to be calculated.So the equation becomes
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]
Rearranging the equation gives
[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]
Now using the cold air approach, the thrust is given as follows
[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]
So the thrust of the engine calculated using the cold air is 34227.35 N
Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.
[10 marks] What is the cost to install a 5kW PV system for a residence?
[10 marks] If the solar irradiance in Ibra is on average 800W/m2 and the installed panels have efficiency of 18%. How many panels are required if the panel’s area is 2m2?
[15 marks] Assume Ibra has an average of 10 day-hours, dusty environment which causes the efficiency of the solar system to drop by 10% on average, and 30 cloudy days/year which cause the efficiency of the solar panels drops by 50%. If electrical power cost per kWh is 0.05 OMR determine the break-even time for the 5kW PV system.
[10 marks] If the system to be off-grid, what would be the backup time if three 12-V batteries were selected each with a capacity of 200Ah. Assume that you can discharge the batteries up to 80% of their capacities.
Answer:
so hard it is
Explanation:
I don't know about this
please mark as brainleast
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Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.
Answer:
sorry i don't know
Explanation:
You do a simple experiment at home with the plastic body of a syringe, where you close the exit with one thumb, and push in the plunger with your other thumb. You are able to compress the air inside the syringe from 5ml to 1ml. Assume that you hold the plunger for long enough such that the temperature equalizes to ambient conditions.
Required:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?
Answer:
a. 89.5 N b. 1.59 N
Explanation:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
Using Boyle's law, we find the final pressure at the compressed state given that the initial pressure is atmospheric pressure
So, P₁V₁ = P₂V₂ where P₁ = initial atmospheric pressure in syringe = 1 atm = 1.013 × 10⁵ N/m²,V₁ = initial volume of syringe = 5 ml, P₂ = final pressure in syringe at compression and V₂ = final volume of syringe = 1 ml
So, making P₂ subject of the formula, we have
P₂ = P₁V₁/V₂
Substituting the values of the variables into the equation, we have
P₂ = P₁V₁/V₂
P₂ = 1.013 × 10⁵ N/m² × 5 ml/1 ml
P₂ = 1.013 × 10⁵ N/m² × 5
P₂ = 5.065 × 10⁵ N/m²
Since pressure, P = F/A where F = force and A = cross-sectional area of syringe = πd²/4 where d = diameter of syringe = 1.5 cm = 1.5 × 10⁻² m.
So, F = PA
F = P₂πd²/4
substituting the values of the variables into the equation, we have
F = P₂πd²/4
F = 5.065 × 10⁵ N/m²π(1.5 × 10⁻² m)²/4
F = 5.065 × 10⁵ N/m²π(2.25 × 10⁻⁴ m²)/4
F = 35.8 × 10/4 N
F = 8.95 × 10
F = 89.5 N
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?
Since the pressure in the syringe after compression is constant, we have
P₂ = F₁/A₁ where F₁ = force exerted on thumb and A₁ = cross-sectional area of opening of syringe = πd₁²/4 where d = diameter of opening of syringe = 2 mm = 2 × 10⁻³ m.
So, F₁ = P₂A₁
F = P₂πd₁²/4
substituting the values of the variables into the equation, we have
F = P₂πd²/4
F = 5.065 × 10⁵ N/m²π(2 × 10⁻³ m)²/4
F = 5.065 × 10⁵ N/m²π(4 × 10⁻⁶ m²)/4
F = 15.91 × 10⁻¹
F = 1.591 N
F ≅ 1.59 N
In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . What is the approximate real power if the power factor is 0.7
Answer:
1,334
Explanation:
In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . 3.87 kilowatts is the approximate real power if the power factor is 0.7.
The idea of real power in a balanced three-phase system is critical in electrical power systems. It denotes the actual power transferred and used by the system, which is usually measured in kilowatts (kW) or megawatts (MW). Understanding real power is critical for evaluating electrical system efficiency and performance. Real power is the component of power in a balanced three-phase system that does useful work, such as operating motors, generating heat, or powering appliances.
Real Power (P) = √3 × Line Current (I) × Line Voltage (V) × Power Factor (PF)
P = √3 × 5 A × 220 V × 0.7
P = 3.87 kilowatts (kW)
To know more about real power, here:
https://brainly.com/question/30163688
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3. When mixing repair adhesive, how do you know when the material is ready?
A) O The mix is uniform in color
B) O The mix has set for 2 minutes
C)The mix has no lumps
D)The mix turns blue
Answer:
O The mix is uniform in color
A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that rises 0.25 m above the channel bottom. Will a hydraulic jump occur?
Answer:
The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )
Explanation:
width of channel = 3.0 m
Flow rate = 5 m^3/s
Normal depth = 0.50 m
Flow encounters a dam rise of 0.25 m
To know if the hydraulic jump will occur
we will Determine the new normal depth
Y2 = 3.77m
Yc ( critical depth )= 0.66m
Attached below is the detailed solution