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3 years ago

What is the never repeat rule

1 answer:

4 0

Don't repeat yourself (DRY, or sometimes do not repeat yourself) is a principle of software development aimed at reducing repetition of software patterns,[1] replacing it with abstractions or using data normalization to avoid redundancy.

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Determine if the fluid is satisfied

guapka [62]

Is there a picture or something?

8 0

3 years ago

Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h a

Mazyrski [523]

Answer:

a) $\mathbf{Q_c = -3730.8684 \ Btu/hr}$

b) $\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}$

Explanation:

From the properties of Super-heated Refrigerant 134a Vapor at $T_1 = 40^0 F$ , $P_1 = 30 \ lbf/in^2$ ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy $h_1 = 109.12 \ Btu/lb$

specific entropy $s_1 = 0.2315 \ Btu/lb.^0R$

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at $P_2 = 160 \ lbf/in^2$ ; we obtain the following properties:

$h_2 = 115.91 \ Btu/lb\\\\ s_2 = 0.2157 \ Btu/lb.^0R$

Given that the power input to the compressor is 2 hp;

Then converting to Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr

2 hp = 2 × 2544.4342 Btu/hr

2 hp = 5088.8684 Btu/hr

The steady state energy for a compressor can be expressed by the formula:

$0 = Q_c -W_c+m((h_1-h_e) + \dfrac{v_i^2-v_e^2}{2}+g(\bar \omega_i - \bar \omega_e)$

By neglecting kinetic and potential energy effects; we have:

$0 = Q_c -W_c+m(h_1-h_2) \\ \\ Q_c = -W_c+m(h_2-h_1)$

$Q_c = -5088.8684 \ Btu/hr +200 \ lb/hr( 115.91 -109.12) Btu/lb \\ \\$

$\mathbf{Q_c = -3730.8684 \ Btu/hr}$

b) To determine the entropy generation; we employ the formula:

$\dfrac{dS}{dt} =\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c$

In a steady state condition $\dfrac{dS}{dt} =0$

Hence;

$0=\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c$

$\sigma _c = m( s_1 -s_2) - \dfrac{Qc}{T}$

$\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}]$

$\sigma _c = [(-3.16 ) \ Btu/hr .^0R + (7.4667 ) Btu/hr ^0R}]$

$\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}$

5 0

3 years ago

Write a SELECT statement that returns the same result set as this SELECT statement. Substitute a subquery in a WHERE clause for

sergiy2304 [10]

Answer:

SELECT distinct VendorName FROM Vendors

WHERE VendorID IN (

SELECT VendorID FROM Invoices

)

Explanation:

6 0

3 years ago

A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v

In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0

3 years ago

A mason requires 12 mason-hours per 100 ft2 for the bottom 4 ft of wall, and a hod carrier requires 14.5 helper hours to mix the

mote1985 [20]

harden you could either me or leave

GO WATCH AFTER OUT NOW RADDED R

8 0

3 years ago