If you make the shape parameter to equal 1, the weibull model will reduce to an exponential distribution and will have a constant hazard rate.

When shape parameter equals to one then Weibull model will have constant hazard rate.

Clearly, the early ("infant mortality") "phase" of the bathtub can be approximated by a Weibull hazard function with shape parameter c1.

See this URL please

http://documentation.statsoft.com/STATISTICAHelp.aspx?path=Process/ProcessAnalysis/Overview/WeibullandReliabilityFailureTimeAnalysisTheWeibullDistribution

Since the shape parameter (B) depends on the standard deviation of the response variable Y (Sigma(y)), [where Y=ln(-ln(R(t))=b0+Bx] and on the standard deviation of the logarithm of the analyzed lifetime data (Sigma(x); [x=ln(t)], then B=1 (exponential case) ocurs only when Sigma(x) equals Sigmal(y). Please observe that ones the sample size n was determined sigma(Y) is constant. (the relation of B with sigma(y) and Sigma(x) is given in equation (14) of my paper "Weibull accelerated life testing analysis with several variables using multiple linear regression" .

On the other hand,iIn practice it occurs when the analyzed component presents several competitve  failure modes (electronic divices), thus, the component fails randomly for any of the failure modes. 

Finally, found that B=1 when the scale parameter eta is far away of the designed time t (eta is several times higher than t). It is to say, to constant R(t) index, say R(t)=0.96, with a desired design time t, say t=1500hrs, then with B=1, eta=24.4966*1500. Clearly higher the difference between eta and t, higher Sigma(x), and thus, from the mentioned equation (14) BSigma(y) implying that specific factors are causing the failures.