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3 years ago

Derive the following conversion factors:

Engineering

1 answer:

zimovet [89]3 years ago

8 0

Answer:

A. 0.0283 mm3/min

B. 15850.2 gal/min

C. 0.2642 gal/min

D. 1.7 m3/hour

Explanation:

[(1 in)3/min *(25.4mm)3/(1 in)]

= 0.02832 mm3/min

[(1m)3/sec*(264.173gal)/(1m)3]*(60secs)/1min

= 15850.2 gal/min

[(Liter/min)*(0.264172gal/liter)]

=0.2642 gal/min

[(1ft)3/min*(0.3048m)3/(1ft)3*(60mins/1hour)]

=1.7 m3/hour

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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

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The scale of the blueprint tells us the<br> of drawing to real space?

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A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

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<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

$\\$

Explanation:

<h2>Given:</h2>

$a_1 \:=\:0.4\:m/s²$

$t_1 \:=\:60\:s$

$v_{i1} \:=\:0\:m/s$

$a_2 \:=\:0\:m/s²$

$t_2 \:=\:25\:min\:=\:1500\:s$

$a_3 \:=\:-0.8\:m/s²$

$v_{f3} \:=\:0\:m/s$

$\\$

<h2>Required:</h2>

Distance from Station A to Station B

$\\$

<h2>Equation:</h2>

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v\:=\:\frac{d}{t}$

$\\$

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for $v_{f1}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}$

$24\:m/s\:=\:v_f\:-\:0\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave1}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_1$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{60\:s}$

$720\:m\:=\:d$

$d_1\:=\:720\:m$

$\\$

<h3>Distance when a = 0 m/s²</h3>

$v_{f1}\:=\:v_{i2}$

$v_{i2}\:=\:24\:m/s$

$\\$

Solve for $v_{f2}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}$

$0\:=\:v_f\:-\:24\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave2}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:24\:m/s$

$\\$

Solve for $d_2$

$v\:=\:\frac{d}{t}$

$24\:m/s\:=\:\frac{d}{1500\:s}$

$36,000\:m\:=\:d$

$d_2\:=\:36,000\:m$

$\\$

<h3>Distance when a = -0.8 m/s²</h3>

$v_{f2}\:=\:v_{i3}$

$v_{i3}\:=\:24\:m/s$

$\\$

Solve for $v_{f3}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}$

$(t)(-0.8\:m/s²)\:=\:-24\:m/s$

$t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}$

$t\:=\:30\:s$

$\\$

Solve for $v_{ave3}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_3$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{30\:s}$

$360\:m\:=\:d$

$d_3\:=\:360\:m$

$\\$

<h3>Total Distance from Station A to Station B</h3>

$d\:= \:d_1\:+\:d_2\:+\:d_3$

$d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m$

$d\:= \:37,080\:m$

$d\:= \:37.08\:km$

$\\$

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

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