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3 years ago

10

The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect

the total quantity of bricks used in the wall
True or False

2 answers:

Yakvenalex [24]3 years ago

5 0

Answer:

false

Explanation:

Lana71 [14]3 years ago

5 0

False is the answer:)!

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At steady state, air at 200 kPa, 325 K, and mass flow rate

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3 0

3 years ago

For a copper-silver alloy of composition 25 wt% Ag-75 wt% Cu and at 775C (1425F) do the following: (a) Determine the mass frac

allochka39001 [22]

Answer:

a)

mass fraction of α = 0.796

mass fraction of β = 0.204

b)

mass fraction of primary α (Wα) = 0.734

mass fraction of eutectic micro-constituents (We) = 0.266

c)

α in eutectic mixture = 0.062

Explanation:

Assumptions:

(i) the system is in thermal equilibrium with its surroundings

(ii) There are no impurities or other alloying elements present

(a) Determine the mass fractions of α and β phases

From the Cu - Ag phase diagram, Cα = 8.0 wt% Ag, Cβ = 91.2 wt% Ag, and C0 = 25 wt% Ag .Using the lever-rule:

Total mass fraction of α = (Cβ - C0) / (Cβ - Cα) = (91.2 - 25) / (91.2 - 8) = 0.796

Total mass fraction of β = (C0 - Cα) / (Cβ - Cα) = (25 - 8) / (91.2 - 8) = 0.204

(b) Determine the mass fractions of primary α and eutectic microconstituents

Ceutetic = 71.9 wt.%Ag

mass fraction of primary α (Wα) = (Ceutetic - C0) / (Ceutetic - Cα) = (71.9 - 25) / (71.9 - 8) = 0.734

mass fraction of eutectic micro-constituents (We) = (C0 - Cα) / (Ceutetic - Cα) = (25 - 8) / (71.9 - 8) = 0.266

(c) Determine the mass fraction of eutectic α.

From the eutetic reaction, L ↔ α + β

Total α = Primary α + α in eutectic mixture

Therefore: α in eutectic mixture = Total α - Primary α = 0.796 - 0.734 = 0.062

5 0

3 years ago

Read 2 more answers

11. Wet-cell batteries are commonly referred to as lead-acid batteries.<br> A) O True<br> B) O False

Alja [10]

Is true i think bye ansía

5 0

4 years ago

1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem

BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁ = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

$\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

$\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}$

$P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa$

$\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

$\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right$

$5^{0.33} = \left \dfrac{T_1}{T_2} \right$

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3× $c_v$ ×(T₂ - T₁)

$c_v$ = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86 kJ/(kg·K)

We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3× $c_v$ ×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = $U_g = h_g -pV_g$

$h_g$ = 2777.12 kJ/kg

$V_g$ = 0.194349 m³/kg

p = 1000 kPa

$U_{g1}$ = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, $U_{g1}$ + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0

3 years ago

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o

Leni [432]

Answer:

(a) The power input to the compressor: $\dot{W}=73.07 kJ/s = 73.07 kW$

(b) The volume flow rate of the refrigerant at the compressor inlet: $\dot{v}=0.209 m^{3}/s$

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

$h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg$

Now we proceed to know the work with the following expression:

$\dot{W}=\dot{m}(h_{2}-h_{1})$

Now we replace values and our result is:

$\dot{W}=73.07 kJ/s = 73.07 kW$

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

$\nu=0.1739m^{3}/kg$

With our specific volume and the mass rate, we can calculate the volume rate:

$\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s$

6 0

4 years ago