Answer:
a) the mass flow rate of the steam is [tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]
b) the exit velocity of the steam is [tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]
c) the exit area of the nozzle is [tex]A_2[/tex] = 0.0015435 m²
Explanation:
Given that:
A steam with 5 MPa and 400° C enters a nozzle steadily
So;
Inlet:
[tex]P_1 =[/tex] 5 MPa
[tex]T_1[/tex] = 400° C
Velocity V = 80 m/s
Exit:
[tex]P_2 =[/tex] 2 MPa
[tex]T_2[/tex] = 300° C
From the properties of steam tables at [tex]P_1 =[/tex] 5 MPa and [tex]T_1[/tex] = 400° C we obtain the following properties for enthalpy h and the speed v
[tex]h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg[/tex]
From the properties of steam tables at [tex]P_2 =[/tex] 2 MPa and [tex]T_1[/tex] = 300° C we obtain the following properties for enthalpy h and the speed v
[tex]h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg[/tex]
Inlet Area of the nozzle = 50 cm²
Heat lost Q = 120 kJ/s
We are to determine the following:
a) the mass flow rate of the steam.
From the system in a steady flow state;
[tex]m_1=m_2=m_3[/tex]
Thus
[tex]m_1 =\dfrac{V_1 \times A_1}{v_1}[/tex]
[tex]m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}[/tex]
[tex]m_1 =\dfrac{0.4 }{0.057838 }[/tex]
[tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]
b) the exit velocity of the steam.
Using Energy Balance equation:
[tex]\Delta E _{system} = E_{in}-E_{out}[/tex]
In a steady flow process;
[tex]\Delta E _{system} = 0[/tex]
[tex]E_{in} = E_{out}[/tex]
[tex]m(h_1 + \dfrac{V_1^2}{2})[/tex] [tex]= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})[/tex]
[tex]- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})[/tex]
[tex]- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]V_2^2 = 316631.29 \ m/s[/tex]
[tex]V_2 = \sqrt{316631.29 \ m/s[/tex]
[tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]
c) the exit area of the nozzle.
The exit of the nozzle can be determined by using the expression:
[tex]m = \dfrac{V_2A_2}{v_2}[/tex]
making [tex]A_2[/tex] the subject of the formula ; we have:
[tex]A_2 = \dfrac{ m \times v_2}{V_2}[/tex]
[tex]A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}[/tex]
[tex]A_2[/tex] = 0.0015435 m²
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]
Where:
[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]
Where:
[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.
[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]
[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]
If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:
[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]
[tex]\%V_{gr} = 10.197\,\%V[/tex]
The volume percentage of graphite is 10.197 per cent.
Following are the solution to the given points:
[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.
[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]
[tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]
Calculating the weight fraction of graphite:
[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]
[tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]
Calculating the volume percent of graphite:
[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]
[tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]
Therefore, the final answer is "0.964, 0.036, and 11.368%"
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By saying that the electrostatic field is conservative, we do not mean that:_______ The potential difference between any two points is zero. It is the gradient of scalar potential. Its circulation is identically zero along any path. Its curl is identically zero everywhere. The work done in moving a charge along closed path inside the field is zero.
Answer:
(a) The potential difference between any two points is zero.
Explanation:
A conservative field is;
i. a vector field that is the gradient of some function. Electrostatic field is the gradient of scalar potential, hence it is conservative.
ii. a vector field where the integral along every closed path is zero. This means that the work done in a closed cycle is zero. For an electrostatic field, the charge along closed path inside the field is zero. Hence, electrostatic field is conservative.
iii. a vector field if curl of its potential(vector product of the del operator and the potential) is zero. The curl of electrostatic field is identically zero everywhere.
iv. a vector field whose circulation is zero along any path.
v. a vector field whose potential difference between two points is independent of the path taken. The potential difference between any two points is not necessarily zero.
Other examples of conservative fields are;
i. gravitational field.
ii. magnetic field.
When we say that electrostatic field is conservative, we do not mean that the potential difference between any two points is zero.
What is a conservative field?A conservative field refers to a form of force between the Earth and another mass whose work is determined only by the final displacement of the object acted upon.
What we mean by saying an electrostatic field is conservative includes:
It is the gradient of scalar potentialIts circulation is identically zero along any pathIts curl is identically zero everywhereThe work done in moving a charge along closed path inside the field is zero.Hence, when we say that electrostatic field is conservative, we do not mean that the potential difference between any two points is zero.
Therefore, the Option A is correct.
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When the value of the output cannot be determined even if the value of the controllable input is known, the model is:_________
a. analog.
b. digital.
c. stochastic.
d. deterministic.
Answer:
c. stochastic.
Explanation:
A stochastic model is a tool in statistics, used to estimate the probability distributions of intended outcomes by the allowance of random variation in any number of the inputs over time. For a stochastic model, Inputs to a quantitative model are uncertain, and the value of the output from a stochastic model cannot be easily determined, even if the value of the input that can be determined is known. The distributions of the resulting outcomes of a stochastic model is usually due to the large number of simulations involved, and it is widely used as a statistical tool in the life sciences.
An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectively, and the respective atomic weights are 22.7 and 91.4 g/mol, what is the density (in g/cm3) of this material?
A. 0.438g/cm3
B. 0. 571g/cm3
C. 1.75g/cm3
D. 3.50g/cm3
Answer:
c) 1.75 g/cm³
Explanation:
Given that
Radii of the A ion, r(c) = 0.137 nm
Radii of the X ion, r(a) = 0.241 nm
Atomic weight of the A ion, A(c) = 22.7 g/mol
Atomic weight of the X ion, A(a) = 91.4 g/mol
Avogadro's number, N = 6.02*10^23 per mol
Solution is attached below
Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its terminals. True False
Answer:
false
Explanation:
the changing of a prisoner sentence or another penalty to another less severe
Anytime scaffolds are assembled or __________, a competent person must oversee the operation.
a. Drawn
b. Disassembled
c. Thought
d. Made
When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:
The competent person is also responsible for proposing whether fall protection is required for each scaffold erected. In constructing a scaffold, there are specific criteria for the ground the scaffold is constructed. On the products and components used to build the scaffold, its height in relation to the foundation. It's platform's design, and whether or not high efficiency is needed to supervise the installation.Therefore, the final answer is "Option B".
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. The job of applications engineer for which Maria was applying requires (a) excellent technical skills with respect to mechanical engineering, (b) a commitment to working in the area of pollution control, (c) the ability to deal well and confidently with customers who have engineering problems, (d) a willingness to travel worldwide, and (e) a very intelligent and well-balanced personality. List 10 questions you would ask when interviewing applicants for the job.
Answer:
Tell us about your self Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this positionExplanation:
For a job of applications engineer which require excellent technical skills, commitment to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.
The ten questions you should ask Maria to determine if she is qualified for the job are :
Tell us about your self ( functions you have )Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this positionFor a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe , the velocity of fluid, the density of fluid and the viscosity of the fluid. Show that = ∅ ൬ ൰
Answer:
Explanation:
La vaca
El pato
Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate:
a. the thermal efficiency of the cycle
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.
Answer:
a. [tex]\eta _{th}[/tex] = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)
[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
[tex]\eta _{th}[/tex] = 77.65%
b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]
Power developed is given by the relation;
[tex]\dot m \cdot w_{net, out}[/tex]
[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time have a ductility of at least 12%EL. Justify your decision
Answer:
First we determine the tensile strength using the equation;
Tₓ (MPa) = 3.45 × HB
{ Tₓ is tensile strength, HB is Brinell hardness = 225 }
therefore
Tₓ = 3.45 × 225
Tₓ = 775 Mpa
From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10
When the percentage of cold work for steel is up to 10,the ductility is 16% EL.
And 16% EL is greater than 12% EL
Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL
Calculate the camacitance-to-neutral in F/m and the admittance-to-neutral in S/km for the three-phase line in problem Neglect the effect of the earth plane.
Answer:
The answer is given below
Explanation:
A 60 Hz three-phase, three-wire overhead line has solid cylindrical conductors arranged in the form of an equilateral triangle with 4 ft conductor spacing. Conductor diameter is 0.5 in.
Given that:
The spacing between the conductors (D) = 4 ft
1 ft = 0.3048 m
D = 4 ft = 4 × 0.3048 m = 1.2192 m
The conductor diameter = 0.5 in
Radius of conductor (r) = 0.5/2 = 0.25 in = 0.00635 m
Frequency (f) = 60 Hz
The capacitance-to-neutral is given by:
[tex]C_n=\frac{2\pi \epsilon_0}{ln(\frac{D}{r} )} =\frac{2\pi *8.854*10^{-12}}{ln(1.2192/0.00635)}=1.058*10^{-11}\ F/m[/tex]
The admittance-to-neutral is given by:
[tex]Y_n=j2\pi fC_n=j*2\pi *60*1.058*10^{-11}*\frac{1000\ m}{1\ km}=j3.989*10^{-6}\ S/km[/tex]
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest
Answer:
115.2 W
Explanation:
The computation is shown below:
As we know that
Power = F . v
[tex]F_H = F cos \theta[/tex]
[tex]F_H = 30 \frac{4}{5}[/tex]
[tex]F_H = 24N[/tex]
Now we solve for V
[tex]V = V_0 + at[/tex] a = 24N ÷ 20Kg
But V_0 = 0 a = 1.2 m/s^2
F_H = ma V = 0 + (1.2) (4)
a = F_H ÷ m V = 4.8 m/s
Therefore
Power = F_Hv
= (24) (4.8)
= 115.2 W
By applying the above formuals we can get the power
Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW.
Answer:
(a) 48.2 %
(b) 0.4137
(c) 2385.9 kW
Explanation:
The given values are:
Initial pressure,
p₁ = 100 kPa
Initial temperature,
T₁ = 300 K
Mass,
M = 6 kg/s
Pressure ration,
r = 10
Inlent temperature,
T₃ = 1400 K
Specific heat ratio,
k = 1.4
At T₁ and p₁,
⇒ [tex]c_{p}=1.005 \ KJ/Kg.K[/tex]
Process 1-2 in isentropic compression, we get
⇒ [tex]\frac{T_{2}}{T_{1}}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}[/tex]
[tex]T_{2}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}. T_{1}[/tex]
On putting the estimated values, we get
[tex]=(10)^{\frac{1.4-1}{1.4}}(300)[/tex]
[tex]=579.2 \ K[/tex]
Process 3-4,
⇒ [tex]\frac{T_{4}}{T_{3}}=(\frac{p_{4}}{p_{3}})^{\frac{k-1}{k}}[/tex]
[tex]T_{4}=(\frac{1}{10})^{\frac{1.4-1}{1.4}}(1400)[/tex]
[tex]=725.13 \ K[/tex]
(a)...
The thermal efficiency will be:
⇒ [tex]\eta =\frac{\dot{W_{t}}-\dot{W_{e}}}{\dot{Q_{in}}}[/tex]
[tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]
⇒ [tex]\dot{Q_{in}}=\dot{m}(h_{1}-h_{2})[/tex]
[tex]=\dot{mc_{p}}(T_{3}-T_{2})[/tex]
[tex]=6\times 1005\times (1400-579.2)[/tex]
[tex]=4949.4 \ kJ/s[/tex]
⇒ [tex]\dot{Q_{out}}=\dot{m}(h_{4}-h_{1})[/tex]
[tex]=6\times 1.005\times (725.13-300)[/tex]
[tex]=2563.5 \ KJ/S[/tex]
As we know,
⇒ [tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]
On putting the values, we get
[tex]=1-\frac{2563.5}{4949.4}[/tex]
[tex]=0.482 \ i.e., \ 48.2 \ Percent[/tex]
(b)...
Back work ratio will be:
⇒ [tex]bwr=\frac{\dot{W_{e}}}{\dot{W_{t}}}[/tex]
Now,
⇒ [tex]\dot{W_{e}}=\dot{mc_{p}}(T_{2}-T_{1})[/tex]
On putting values, we get
[tex]=6\times 1.005\times (579.2-300)[/tex]
[tex]=1683.6 \ kJ/s[/tex]
⇒ [tex]\dot{W_{t}}=\dot{mc_{p}}(T_{3}-T_{4})[/tex]
[tex]=6\times 1.005\times (1400-725.13)[/tex]
[tex]=4069.5 \ kJ/s[/tex]
So that,
⇒ [tex]bwr=\frac{1683.6}{4069.5}=0.4137[/tex]
(c)...
Net power is equivalent to,
⇒ [tex]\dot{W}_{eyele}=\dot{W_{t}}-\dot{W_{e}}[/tex]
On substituting the values, we get
[tex]= 4069.5-1683.6[/tex]
[tex]=2385.9 \ kW[/tex]
Following are the solution to the given points:
Given :
Initial pressure [tex]p_1 = 100\ kPa \\\\[/tex]
Initial temperature [tex]T_1 = 300\ K \\\\[/tex]
Mass flow rate of air [tex]m= 6\ \frac{kg}{s}\\\\[/tex]
Compressor pressure ratio [tex]r =10\\\\[/tex]
Turbine inlet temperature [tex]T_3 = 1400\ K\\\\[/tex]
Specific heat ratio [tex]k=1.4\\\\[/tex]
Temperature [tex]\ T_1 = 300\ K[/tex]
pressure [tex]p_1 = 100\ kPa\\\\[/tex]
[tex]\to c_p=1.005\ \frac{kJ}{kg\cdot K}\\\\[/tex]
Process 1-2 is isen tropic compression
[tex]\to \frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \\\\[/tex]
[tex]\to T_2=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \ T_1 \\\\[/tex]
[tex]=(10)^{\frac{1.4-1}{1.4}} (300)\\\\ =(10)^{\frac{0.4}{1.4}} (300) \\\\[/tex]
[tex]\to T_2 = 579.2\ K \\\\[/tex]
Process 3-4 is isen tropic expansion
[tex]\to \frac{T_4}{T_3}=(\frac{P_4}{P_3})^{\frac{k-1}{k}}\\\\ \to T_4=(\frac{1}{10})^{\frac{1.4-1}{1.4}} (1400)\\\\\to T_4= 725.13\ K \\\\[/tex]
For point a:
The thermal efficiency of the cycle:
[tex]\to \eta = \frac{W_i-W_e}{Q_{in}} \\\\\to \eta = \frac{Q_{in}- Q_{out}}{Q_{in}}\\\\\to \eta =1 - \frac{Q_{out}}{Q_{in}} \\\\\to Q_{in}= m(h_3-h_1) = mc_p (T_4-T_1) =(6)(1.005)(725.13-300) = 2563 \ \frac{kJ}{S}\\\\\to \eta =1- \frac{Q_{out}}{Q_{in}}\\\\[/tex]
[tex]=1-\frac{2563.5}{4949.4}\\\\ = 0.482\\\\[/tex]
[tex]\eta = 48.2\%\\\\[/tex]
For point b:
The back work ratio
[tex]\to bwr =\frac{W_e}{W_t}[/tex]
Now
[tex]\to W_e =mc_p (T_2 -T_1)[/tex]
[tex]=(6) (1.005)(579.2 -300)\\\\ =1683.6 \ \frac{kJ}{S}\\\\[/tex]
[tex]\to W_t=mc_p(T_3-T_4)[/tex]
[tex]=(6)(1.005)(1400 - 725.13)\\\\ = 4069.5 \frac{KJ}{s}[/tex]
[tex]\to bwr =\frac{W_s}{W_t}= \frac{1683.6}{4069.5}=0.4137[/tex]
For point c:
The net power developed is equal to
[tex]\to W_{cycle} = W_t-W_e \\\\[/tex]
[tex]= ( 4069.5-1683.6)\\\\ = 2385.9 \ kW\\[/tex]
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The closed feedwater heater of a regenerative Rankine cycle is to heat 7000 kPa feedwater from 2608C to a saturated liquid. The turbine supplies bleed steam at 6000 kPa and 3258C to this unit. This steam is condensed to a saturated liquid before entering the pump. Calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
Answer:
the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
Explanation:
Given that:
Pressure of the feed water = 7000 kPa
Temperature of the closed feedwater heater = 260 ° C
Pressure of of the turbine = 6000 kPa
Temperature of the turbine = 325 ° C
The objective is to calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
From the table A-4 of saturated water temperature table at temperature 260° C at state 1 ;
Enthalpies:
[tex]h_1 = h_f = 1134.8 \ kJ/kg[/tex]
From table A-6 superheated water at state 3 ; the value of the enthalpy relating to the pressure of the turbine at 6000 kPa and temperature of 325° C is obtained by the interpolating the temperature between 300 ° C and 350 ° C
At 300° C; enthalpy = 2885.6 kJ/kg
At 325° C. enthalpy = 3043.9 kJ/kg
Thus;
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-{h_{300^0}}}{{h_{350^0}}- {h_{300^0}}}[/tex]
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]\dfrac{25}{50}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]h_{325^0} = 2885.6 + \dfrac{25}{50}({3043.9-2885.6 )[/tex]
[tex]h_{325^0} = 2885.6 + 0.5({3043.9-2885.6 )[/tex]
[tex]h_{325^0} =2964.75 \ kJ/kg[/tex]
At pressure of 7000 kPa at state 6; we obtain the enthalpies corresponding to the pressure at table A-5 of the saturated water pressure tables.
[tex]h_6 = h_f = 1267.5 \ kJ/kg[/tex]
From state 4 ;we obtain the specific volume corresponding to the pressure of 6000 kPa at table A-5 of the saturated water pressure tables.
[tex]v_4 = v_f = 0.001319\ m^3 /kg[/tex]
However; the specific work pump can be determined by using the formula;
[tex]W_p = v_4 (P_5-P_4)[/tex]
where;
[tex]P_4[/tex] = pressure at state 4
[tex]P_5[/tex] = pressure at state 5
[tex]W_p = 0.001319 (7000-6000)[/tex]
[tex]W_p = 0.001319 (1000)[/tex]
[tex]W_p =1.319 \ kJ/kg[/tex]
Using the energy balance equation of the closed feedwater heater to calculate the amount of bleed steam required to heat 1 kg of feed water ; we have:
[tex]E_{in} = E_{out} \\ \\ m_1h_1 +m_3h_3 + m_3W_p = (m_1+m_3)h_6[/tex]
where;
[tex]m_1 = 1 \ kg[/tex]
Replacing our other value as derived above into the energy balance equation ; we have:
[tex]1 \times 1134.8 +m_3 \times 2964.75 + m_3 \times 1.319 = (1+m_3)\times 1267.5[/tex]
[tex]1134.8 + 2966.069 \ m_3 = 1267.5 + 1267.5m_3[/tex]
Collect like terms
[tex]2966.069 \ m_3- 1267.5m_3 = 1267.5-1134.8[/tex]
[tex]1698.569 \ m_3 =132.7[/tex]
[tex]\ m_3 = \dfrac{132.7}{1698.569}[/tex]
[tex]\mathbf{ m_3 = 0.078 \ kg/s}[/tex]
Hence; the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compressed air tank. Assume that an air tank is being filled using a compressor to a pressure of 5 atm, and that it is being fed with air at a temperature of 25°C and 1 atm pressure. The compression process is adiabatic. Will the temperature of the air in the tank when it is done being filled i.e. once the pressure in the tank reaches 5 atm), be greater than, equal to, or less that the temperature of the 25°C air feeding the compressor?
A. Greater than 25°C
B. Unable to determine
C. Same as 25°C
D. Less than 25°C
Answer:
The temperature will be greater than 25°C
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C
After impact testing a sample at -100oC you realize that the fracture surface is very dull and fibrous. Is the sample behaving in a ductile of brittle manner at this temperature
Answer:
Ductile
Explanation:
So, from the question, we have the following information or parameters or data which is going to help us in solving this particular problem or question;
=> " impact testing a sample = -100oC shows that the fracture surface is very DULL AND FIBROUS"
TAKE NOTE: DULL AND FIBROUS.
IMPACT TESTING is used by engineers in the configuration of a sample or object.
In order to determine whether a specimen is ductile or brittle, it can be shown from its appearance for instance;
A DUCTILE SAMPLE will be DULL AND FIBROUS thus, our answer!
But a brittle sample will have a crystal shape.
why is the peak value of the rectified output less than the peak value of the ac input and by how much g
Answer:
The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Explanation:
This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Therefore this is the formula for Half wave rectifier
Vrms = Vm/2 and Vdc
= Vm/π:
Where,
Vrms = rms value of input
Vdc = Average value of input
Vm = peak value of output
Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.
Consider a double-pipe counter-flow heat exchanger. In order to enhance its heat transfer, the length of the heat exchanger is doubled. Will the effectiveness of the exchanger double?
Answer:
effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.
Because effectiveness depends on NTU and not necessarily the length of the heat exchanger
The temperature of water is 45 what does the measurement represent
Answer:
degree of hotness of coldness of a substance
The column is constructed from high-strength concrete and eight A992 steel reinforcing bars. If the column is subjected to an axial force of 200 kip.
a) Determine the average normal stress in the concrete and in each bar. Each bar has a diameter of 1 in.
b) Determine the required diameter of each bar so that 60% of the axial force is carried by concrete.
Answer:
d= 2.80inch
Explanation:
Given:
Axial force= 30kip
d= 1inch
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
A) The average normal stress in the concrete and in each bar are; σ_st = 15.52 kpi ; σ_con = 2.25 kpi
B) The required diameter of each bar so that 60% of the axial force is carried by concrete is; 0.94 inches
Concrete Column Design
We are told that;
Column has eight A992 steel reinforcing bars.
Column is subjected to an axial force of 200 kip.
A) Diameter of each bar is 1 inch.
Using equations of equilibrium, we have;
∑fy = 0;
8P_st + P_con = 200 ------(eq 1)
Using compatibility concept, we know from the image attached that;
δ_st = δ_con
where δ_st is change in length of steel and δ_con is change in length of concrete.
Thus;
δ_st = (P_st * L)/(A_st * E_st)
where;
P_st is tensile force of steel
L is length of steel = 3 ft = 36 inches
A_st is area of steel = π/4 * 1² = 0.7854 in²
E_st is young's modulus of steel = 29000 ksi
Similarly;
δ_con = (P_con * L)/(A_con * E_con)
where;
P_con is tensile force of concrete
L is length of concrete = 3 ft = 36 inches
E_con is young's modulus of concrete = 4200 ksi
A_con is area of concrete with diameter of 8 inches = (π/4 * 8²) - 6(π/4 * 1²) = 45.5531 in²
Thus;
From δ_st = δ_con;
(P_st * 36)/(0.7854 * 29000) = (P_con * 36)/(45.5531 * 4200)
Solving this gives;
P_st = 0.119P_con -----(eq 2)
Put 0.119P_con for P_st in eq 1 to get;
8(0.119P_con) + P_con = 200
1.952P_con = 200
P_con = 102.459 kip
Thus; P_st = 12.193 kip
Thus, average normal stress is;
Steel; σ_st = P_st/A_st
σ_st = 12.193/0.7854
σ_st = 15.52 kpi
Concrete; σ_con = P_con/A_con
σ_con = 102.459/45.5531
σ_con = 2.25 kpi
B) Since 60% of the axial force is carried by the concrete. Then it means that 40% will be carried by the steel.
Thus;
P_con = 60% * 200 = 120 kip
P_st = 40% * 200 = 80 kip
Using compatibility again;
δ_st = δ_con
Thus;
(P_st * L)/(A_st * E_st) = (P_con * L)/(A_con * E_con)
6(π/4 * d²)) = (80 * ((π/4 * 8²) - 6(π/4 * d²)) * 4200)/(120 * 29000)
⇒ 4.712d² = 0.09655(50.2655 - 4.712d²)
⇒ 4.712d²/0.09655 = 50.2655 - 4.712d²
⇒ 48.8037d² = 50.2655 - 4.712d²
Solving this gives;
d = 0.94 inches
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Before you attempt to change a tire yourself, you should _____.
A. put on a pair of gloves
B. read your vehicle owner's manual for any special directions or warnings.
C. always call for emergency assistance first
D. let the remaining air out of the flat tire
Answer: read your vehicle owner's manual for any special directions or warnings.
Answer:
B. read your vehicle owner's manual for any special directions or warnings.
Explanation:
Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are the advantages and disadvantages of using 6 tube passes instead of just 2 of the same diameter?
Answer:
Please check explanation for answer
Explanation:
Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:
Advantages
* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface
* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too
Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.
Disadvantages
* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.
* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs
which solution causes cells to shrink
It is to be noted that a hypertonic solution have the capacity to make cells to shrink.
What happens in a hypertonic solution?In a hypertonic solution, the concentration of solutes (e.g., salts, sugars) outside the cell is higher than inside the cell.
As a result, water moves out of the cell through osmosis, trying to equalize the concentration, causing the cell to lose water and shrink.
This process is commonly observed in biology when examining the effect of different solutions on cells, such as in red blood cells or plant cells.
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Aggregate blend composed of 65% coarse aggregate (SG 2.701), 35% fine aggregate (SG 2.625)
Compacted specimen weight in air = 1257.9 g, submerged weight = 740.0 g, SSD weight = 1258.7 g
Compacted specimen contains 5.0% asphalt by total weight of the mix with Gb = 1.030
Theoretical maximum specific gravity = 2.511
Bulk specific gravity of the aggregate __________
Bulk specific gravity of the compacted specimen__________
Percent stone __________
Effective specific gravity of the stone__________
Percent voids in total mix__________
Percent voids in mineral aggregate__________
Percent voids filled with asphalt__________
Answer:
2.6742.42891.695%2.5923.305%11.786%78.1%Explanation:
coarse aggregate (ca) = 65%, SG = 2.701
Fine aggregate = 35%, SG = 2.625
A) Bulk specific gravity of aggregate
= [tex]\frac{65*2.701 + 35*2.625}{100} = 2.674[/tex]
B) Wm = 1257.9 g { weight in air }
Ww = 740 g { submerged weight }
therefore Bulk specific gravity of compacted specimen
= [tex]\frac{Wm}{Wm-Ww}[/tex] = [tex]\frac{1257.9}{1257.9 - 740 }[/tex] = 2.428
Theoretical specific gravity = 2.511
Percent stone
= 100 - asphalt content - Vv
= 100 - 5 - 3.305 = 91.695%
c) percent of void
= [tex]\frac{9.511-2.428}{2.511} * 100[/tex] Vv = 3.305%
d) let effective specific gravity in stone
= [tex]\frac{91.695*unstone+ 5 *1.030 }{96.695} = 2.511[/tex]
= Instone = 2.592 effective specific gravity of stone
e) Vv = 3.305%
f ) volume filled with asphalt (Vb) = [tex]\frac{\frac{Wb}{lnb} }{\frac{Wm}{Inm} } * 100[/tex]
Vb = [tex]\frac{5 * 2.428}{1.030 * 100} * 100[/tex]
Vb = 11.786 %
Volume of mineral aggregate = Vb + Vv
VMA = 11.786 + 3.305 = 15.091
g) percent void filled with alphalt
= Vb / VMA * 100
VMA = 11.786 + 3.305 = 15.091
percent void filled with alphalt
= Vb / VMA * 100 = (11.786 / 15.091) * 100 = 78.1%
The first choice for how to reduce or eliminate a hazard is: a) Engineering controls b) Workplace controls c) Personal protective equipment d) Administrative controls
Answer:
The correct answer would be a) Engineering Controls.
Explanation:
If the controls are handled correctly, you can reduce and eliminate hazards so no one gets hurt. Engineering controls are absolutely necessary to prevent hazards.
Hope this helped! :)
Personal protective equipment (PPE) is appropriate for controlling hazards
PPE are used for exposure to hazards when safe work practices and other forms of administrative controls cannot provide sufficient additional protection, a supplementary method of control is the use of protective clothing or equipment. PPE may also be appropriate for controlling hazards while engineering and work practice controls are being installed.
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For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using the Maximum Shear Stress Theory the material will __________
a. fail / not fail
b. fail /fail
c. not fail/fail
d. not fail/not fail
Answer:
Option A - fail/ not fail
Explanation:
For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______
A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 in. wide by 12 in. deep. The beam carries a 9-kip concentrated load at midspan. Which point has the largest Q value at section a–a?
Answer:
point B where [tex]Q_B = 101.25 \ in^3[/tex] has the largest Q value at section a–a
Explanation:
The missing diagram that is suppose to be attached to this question can be found in the attached file below.
So from the given information ;we are to determine the point that has the largest Q value at section a–a
In order to do that; we will work hand in hand with the image attached below.
From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.
We also have block partitioned into different point segments . i,e A,B,C, D
For point A ;
Let Q be the moment of the Area A;
SO ; [tex]Q_A = Area \times y_1[/tex]
where ;
[tex]y_1 = (6 - \dfrac{1.5}{2})[/tex]
[tex]y_1 = (6- 0.75)[/tex]
[tex]y_1 = 5.25 \ in[/tex]
[tex]Q_A =(L \times B) \times y_1[/tex]
[tex]Q_A =(6 \times 1.5) \times 5.25[/tex]
[tex]Q_A =47.25 \ in^3[/tex]
For point B ;
Let Q be the moment of the Area B;
SO ; [tex]Q_B = Area \times y_2[/tex]
where ;
[tex]y_2 = (6 - \dfrac{1.5 \times 3}{2})[/tex]
[tex]y_2= (6 - \dfrac{4.5}{2}})[/tex]
[tex]y_2 = (6 -2.25})[/tex]
[tex]y_2 = 3.75 \ in[/tex]
[tex]Q_B =(L \times B) \times y_1[/tex]
[tex]Q_B=(6 \times 4.5) \times 3.75[/tex]
[tex]Q_B = 101.25 \ in^3[/tex]
For point C ;
Let Q be the moment of the Area C;
SO ; [tex]Q_C = Area \times y_3[/tex]
where ;
[tex]y_3 = (6 - \dfrac{1.5 \times 2}{2})[/tex]
[tex]y_3 = (6 - 1.5})[/tex]
[tex]y_3= 4.5 \ in[/tex]
[tex]Q_C =(L \times B) \times y_1[/tex]
[tex]Q_C =(6 \times 3) \times 4.5[/tex]
[tex]Q_C=81 \ in^3[/tex]
For point D ;
Let Q be the moment of the Area D;
SO ; [tex]Q_D = Area \times y_4[/tex]
since there is no area about point D
Area = 0
[tex]Q_D =0 \times y_4[/tex]
[tex]Q_D = 0[/tex]
Thus; from the foregoing ; point B where [tex]Q_B = 101.25 \ in^3[/tex] has the largest Q value at section a–a
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 mL of the base. Assuming complete neutralization of the acid,
1) What was the normality of the acid solution?
2) What was the molarity of the acid solution?
Answer:
a. 0.4544 N
b. [tex]5.112 \times 10^{-5 M}[/tex]
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
[tex]H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O[/tex]
[tex]NaOH\ Mass = Normality \times equivalent\ weight \times\ volume[/tex]
[tex]= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL[/tex]
= 0.27264 g
[tex]NaOH\ mass = \frac{mass}{molecular\ weight}[/tex]
[tex]= \frac{0.27264\ g}{40g/mol}[/tex]
= 0.006816 mol
Now
Moles of [tex]H_2SO_4[/tex] needed is
[tex]= \frac{0.006816}{2}[/tex]
= 0.003408 mol
[tex]Mass\ of\ H_2SO_4 = moles \times molecular\ weight[/tex]
[tex]= 0.003408\ mol \times 98g/mol[/tex]
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
[tex]= \frac{acid\ mass}{equivalent\ weight \times volume}[/tex]
[tex]= \frac{0.333984 g}{49 \times 0.015}[/tex]
= 0.4544 N
b. And, the acid solution molarity is
[tex]= \frac{moles}{Volume}[/tex]
[tex]= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}[/tex]
= 0.00005112
=[tex]5.112 \times 10^{-5 M}[/tex]
We simply applied the above formulas
The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is
21.30 mL, which gives the following acid solution approximate values;
1) Normality of the acid solution is 0.4544 N
2) The molarity of the acid is 0.2272
How can the normality, molarity of the solution be found?Molarity of the NaOH = 0.3200 M
Volume of NaOH required = 21.30 mL
1) The normality of the acid solution is found as follows;
The chemical reaction is presented as follows;
H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O
Number of moles of NaOH in the reaction is found as follows;
[tex]n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}[/tex]
Therefore;
The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M
[tex]Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}[/tex]
Which gives;
[tex]Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}[/tex]
The normality of the acid solution, H₂SO₄(aq), N ≈ 0.45442) The molarity is found as follows;
[tex]Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}[/tex]
The molarity of the acid solution is 0.2272 MLearn more about the normality and the molarity of a solution here:
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The effectiveness of a heat exchanger is defined as the ratio of the maximum possible heat transfer rate to the actual heat transfer rate.
a. True
b. False
Answer:
False
Explanation:
Because
The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.
Solid solution strengthening is achieved byGroup of answer choicesstrain hardening restricting the dislocation motion increasing the dislocation motion increasing the grain boundary g
Answer:
B. restricting the dislocation motion
Explanation:
Solid solution strengthening is a type of alloying that is carried out by the addition of the atoms of the element used for the alloying to the crystallized lattice structure of the base metal, which the metal that would be strengthened. The purpose of this act is to increase the strength of metals. It actually works by impeding or restricting the motion in the crystal lattice structure of metals thus making them more difficult to deform.
The solute atoms used for strengthening could be interstitial or substitutional. The interstitial solute atoms work by moving in between the space in the atoms of the base metal while the substitutional solute atoms make a replacement with the solvent atoms in the base metal.