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3 years ago

Torque is a twisting force. If the required torque applied on a 3 ft wrench is 45 ft·lb, what is the force that must be applied?

Engineering

1 answer:

natka813 [3]3 years ago

8 0

Answer:

15 lbs

Explanation:

assuming you push from the end of the wrench (3ft)

torque = force(distance)

force = torque/distance

(45 ft·lb)/(3 ft)= 15 lbs

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Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E

White raven [17]

Answer:

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

Explanation:

To solve this problem we use the expression for the temperature film

$T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5$

Then, we have to compute the Reynolds number

$Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}$

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

$Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77$

but we also now that

$Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\$

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0

3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

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geniusboy [140]

Explanation:

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THE COMPUND INTEREST ON RS 30,000AT 7% PER ANNUM IS RS 4347 THE PERIOD IN YEARS

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Answer:

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